k is , 3 n On Sequences of the form, P(x, y) , modolu , 2, Evaluated at, {x = 1, y = 1} For all Polynomials that are Sums of Monomials taken from, 2 2 2 2 2 2 {1, x, y, x , y , x y, x y , x y, x y } By Shalosh B. Ekhad In this webbook, we will consider the sequences described in the title, that\ after normalization and weeding out obvious symmetry, concerns the following set of, 28, polynomials 2 2 2 {1 + x + y, x + x + 1, y + x + 1, y + x + y, x y + y + 1, x y + x + y, 2 2 2 2 2 2 2 2 x y + y + 1, x y + y + x, x + y + 1, x + y + y, x + x y + y , 2 2 2 2 2 2 2 x y + y + 1, x y + x + 1, x y + x + y, x y + y + 1, x y + y + x, 2 2 2 2 2 2 2 2 x y + x y + 1, x y + x + 1, x y + x + y, x y + x + y , 2 2 2 2 2 2 2 2 2 x y + x y + 1, x y + y + 1, x y + x + y, x y + y + 1, 2 2 2 2 2 2 2 2 2 2 2 2 x y + y + x, x y + x y + 1, x y + x + y , x y + x y + 1} by finding enumerative automata with at most, 200, states . ----------------------------------------------------------------------------\ ---- Theorem Number, 1, : Let C(n) be n (1 + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.013, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 2, : Let C(n) be 2 n (x + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 5, 3, 9, 5, 11, 3, 9, 9, 15, 5, 15, 11, 21, 3, 9, 9, 15, 9, 27, 15, 33, 5, 15, 15, 25, 11, 33, 21, 43, 3, 9, 9, 15, 9, 27, 15, 33, 9] Just for kicks C(googol) equals , 67491179529985179890010057158074951171875 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, 5726623061, 11453246123, 22906492245, 45812984491, 91625968981, 183251937963, 366503875925, 733007751851, 1466015503701] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - ----------------- (t + 1) (2 t - 1) and in Maple notation -(2*t+1)/(t+1)/(2*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 3, : Let C(n) be 2 n (y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 4, : Let C(n) be 2 n (y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 5, : Let C(n) be n (x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 6, : Let C(n) be n (x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 7, : Let C(n) be 2 n (x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 8, : Let C(n) be 2 n (x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 9, : Let C(n) be 2 2 n (x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 10, : Let C(n) be 2 2 n (x + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 11, : Let C(n) be 2 2 n (x + x y + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 5, 3, 9, 5, 11, 3, 9, 9, 15, 5, 15, 11, 21, 3, 9, 9, 15, 9, 27, 15, 33, 5, 15, 15, 25, 11, 33, 21, 43, 3, 9, 9, 15, 9, 27, 15, 33, 9] Just for kicks C(googol) equals , 67491179529985179890010057158074951171875 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, 5726623061, 11453246123, 22906492245, 45812984491, 91625968981, 183251937963, 366503875925, 733007751851, 1466015503701] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - ----------------- (t + 1) (2 t - 1) and in Maple notation -(2*t+1)/(t+1)/(2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 12, : Let C(n) be 2 n (x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 13, : Let C(n) be 2 n (x y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 14, : Let C(n) be 2 n (x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 15, : Let C(n) be 2 2 n (x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 16, : Let C(n) be 2 2 n (x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 17, : Let C(n) be 2 n (x y + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 18, : Let C(n) be 2 2 n (x y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 19, : Let C(n) be 2 2 n (x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 20, : Let C(n) be 2 2 2 n (x y + x + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 21, : Let C(n) be 2 2 n (x y + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 22, : Let C(n) be 2 2 n (x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 23, : Let C(n) be 2 2 n (x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.001, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 24, : Let C(n) be 2 2 2 n (x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 25, : Let C(n) be 2 2 2 n (x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 26, : Let C(n) be 2 2 n (x y + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 5, 3, 9, 5, 11, 3, 9, 9, 15, 5, 15, 11, 21, 3, 9, 9, 15, 9, 27, 15, 33, 5, 15, 15, 25, 11, 33, 21, 43, 3, 9, 9, 15, 9, 27, 15, 33, 9] Just for kicks C(googol) equals , 67491179529985179890010057158074951171875 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, 5726623061, 11453246123, 22906492245, 45812984491, 91625968981, 183251937963, 366503875925, 733007751851, 1466015503701] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - ----------------- (t + 1) (2 t - 1) and in Maple notation -(2*t+1)/(t+1)/(2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 27, : Let C(n) be 2 2 2 2 n (x y + x + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 28, : Let C(n) be 2 2 2 n (x y + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9] Just for kicks C(googol) equals , 125236737537878753441860054533045969266612127846243 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 3 t - 1 and in Maple notation -1/(3*t-1) This ends this theorem, that took, 0.003, seconds. ------------------------------------------------------------------ This concludes this webbook, that took, 0.136, seconds. to generate. k is , 4 n On Sequences of the form, P(x, y) , modolu , 2, Evaluated at, {x = 1, y = 1} For all Polynomials that are Sums of Monomials taken from, 2 2 2 2 2 2 {1, x, y, x , y , x y, x y , x y, x y } By Shalosh B. Ekhad In this webbook, we will consider the sequences described in the title, that\ after normalization and weeding out obvious symmetry, concerns the following set of, 54, polynomials 2 2 2 {y + x + y + 1, x y + x + y + 1, x y + y + y + 1, x y + y + x + 1, 2 2 2 2 2 2 2 x y + y + x + y, x + y + y + 1, x + y + x + y, x + x y + y + 1, 2 2 2 2 2 2 2 x + x y + y + y, x y + x + y + 1, x y + y + y + 1, x y + y + x + 1, 2 2 2 2 x y + y + x + y, x y + x y + y + 1, x y + x y + x + 1, 2 2 2 2 2 x y + x y + x + y, x y + x y + y + 1, x y + x y + y + x, 2 2 2 2 2 2 2 2 2 x y + x + y + 1, x y + x + x + 1, x y + x + x + y, x y + x + y + 1, 2 2 2 2 2 2 2 2 x y + x + y + y, x y + x + y + x, x y + x + x y + 1, 2 2 2 2 2 2 2 x y + x + x y + y, x y + x + x y + y , x y + x y + y + 1, 2 2 2 2 2 2 2 2 x y + x y + x + y, x y + x y + y + 1, x y + x y + y + x, 2 2 2 2 2 2 2 2 x y + x y + x y + 1, x y + x y + x + y , x y + x + y + 1, 2 2 2 2 2 2 2 2 2 x y + y + y + 1, x y + y + x + 1, x y + y + x + y, 2 2 2 2 2 2 2 x y + x y + y + 1, x y + x y + x + y, x y + x y + y + 1, 2 2 2 2 2 2 2 2 2 2 2 x y + x y + y + x, x y + x + y + 1, x y + x + y + y, 2 2 2 2 2 2 2 2 2 2 x y + x + x y + y , x y + x y + y + 1, x y + x y + x + 1, 2 2 2 2 2 2 2 2 2 2 2 x y + x y + x + y, x y + x y + y + 1, x y + x y + y + x, 2 2 2 2 2 2 2 2 2 2 2 x y + x y + x y + 1, x y + x y + x + 1, x y + x y + x + y, 2 2 2 2 2 2 2 2 2 x y + x y + x + y , x y + x y + x y + 1} by finding enumerative automata with at most, 200, states . ----------------------------------------------------------------------------\ ---- Theorem Number, 1, : Let C(n) be 2 n (y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 2, : Let C(n) be n (x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 3, : Let C(n) be 2 n (x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 4, : Let C(n) be 2 n (x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 5, : Let C(n) be 2 n (x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 6, : Let C(n) be 2 2 n (x + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 7, : Let C(n) be 2 2 n (x + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 8, : Let C(n) be 2 2 n (x + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 9, : Let C(n) be 2 2 n (x + x y + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.004, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 10, : Let C(n) be 2 n (x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 11, : Let C(n) be 2 2 n (x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 12, : Let C(n) be 2 2 n (x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.004, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 13, : Let C(n) be 2 2 n (x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 14, : Let C(n) be 2 n (x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 15, : Let C(n) be 2 n (x y + x y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 16, : Let C(n) be 2 n (x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.001, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 17, : Let C(n) be 2 2 n (x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 18, : Let C(n) be 2 2 n (x y + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 19, : Let C(n) be 2 2 n (x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 58, 4, 16, 16, 64, 16, 64, 58, 204, 4, 16, 16, 64, 16, 64, 64, 232, 16, 64, 64, 256, 58, 232, 204, 714, 4, 16, 16, 64, 16, 64, 64, 232 , 16] Just for kicks C(googol) equals , 171219100609297807343357223578748840991942042613831206271188992 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 58, 204, 714, 2492, 8682, 30228, 105226, 366276, 1274922, 4437692, 15446554, 53765916, 187147146, 651418116, 2267444842, 7892485300, 27472040138, 95624259340, 332847482970, 1158570537292, 4032735032490, 14037083980308, 48860072772074, 170071413502180, 591982043090090, 2060562279111580, 7172374493538586, 24965494321148156, 86899520830961866, 302478557946941732, 1052862859805769450, 3664789362535367700, 12756344234833402634, 44402093037343122988, 154554144181302100058, 537969763351565659308, 1872557140499791061866, 6517969007386764142132] Using the found enumerative automaton with, 8, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 3 2 (2 t - 2 t + 4 t + t - t + t - 1) (t - 1) - ------------------------------------------------------------ 8 7 6 5 4 3 2 4 t - 10 t + 20 t - 12 t + 3 t + 4 t - 10 t + 6 t - 1 and in Maple notation -(2*t^6-2*t^5+4*t^4+t^3-t^2+t-1)*(t-1)/(4*t^8-10*t^7+20*t^6-12*t^5+3*t^4+4*t^3-\ 10*t^2+6*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 20, : Let C(n) be 2 2 n (x y + x + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 21, : Let C(n) be 2 2 n (x y + x + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 60, 4, 16, 16, 64, 16, 64, 60, 216, 4, 16, 16, 64, 16, 64, 64, 240, 16, 64, 64, 256, 60, 240, 216, 768, 4, 16, 16, 64, 16, 64, 64, 240 , 16] Just for kicks C(googol) equals , 297962263182225803415527030257798638760010520316071641088000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 60, 216, 768, 2728, 9704, 34552, 123064, 438328, 1561176, 5560248, 19803096, 70529656, 251194904, 894643768, 3186321112, 11348251384, 40417397400, 143948695992, 512680882776, 1825940038264, 6503181876248, 23161425701176, 82490640856024, 293794773978616, 1046365603664280, 3726686358966456, 13272790284227928, 47271743570583928, 168360811280142104, 599625920981528632, 2135599385504870104, 7606049998477798648, 27089348766443398296, 96480146296286054328, 343619136421744959576, 1223817701857806463096, 4358691378416908259864, 15523709538966336657208] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 2 t + 2 t - t + 1 - --------------------------------- 2 2 (2 t + 3 t - 1) (2 t - 2 t + 1) and in Maple notation -(2*t^3+2*t^2-t+1)/(2*t^2+3*t-1)/(2*t^2-2*t+1) This ends this theorem, that took, 0.011, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 22, : Let C(n) be 2 2 2 n (x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 23, : Let C(n) be 2 2 2 n (x y + x + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 60, 4, 16, 16, 64, 16, 64, 60, 216, 4, 16, 16, 64, 16, 64, 64, 240, 16, 64, 64, 256, 60, 240, 216, 768, 4, 16, 16, 64, 16, 64, 64, 240 , 16] Just for kicks C(googol) equals , 297962263182225803415527030257798638760010520316071641088000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 60, 216, 768, 2728, 9704, 34552, 123064, 438328, 1561176, 5560248, 19803096, 70529656, 251194904, 894643768, 3186321112, 11348251384, 40417397400, 143948695992, 512680882776, 1825940038264, 6503181876248, 23161425701176, 82490640856024, 293794773978616, 1046365603664280, 3726686358966456, 13272790284227928, 47271743570583928, 168360811280142104, 599625920981528632, 2135599385504870104, 7606049998477798648, 27089348766443398296, 96480146296286054328, 343619136421744959576, 1223817701857806463096, 4358691378416908259864, 15523709538966336657208] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 2 t + 2 t - t + 1 - --------------------------------- 2 2 (2 t + 3 t - 1) (2 t - 2 t + 1) and in Maple notation -(2*t^3+2*t^2-t+1)/(2*t^2+3*t-1)/(2*t^2-2*t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 24, : Let C(n) be 2 2 2 n (x y + x + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 25, : Let C(n) be 2 2 n (x y + x + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 56, 4, 16, 16, 64, 16, 64, 56, 196, 4, 16, 16, 64, 16, 64, 64, 224, 16, 64, 64, 256, 56, 224, 196, 680, 4, 16, 16, 64, 16, 64, 64, 224 , 16] Just for kicks C(googol) equals , 109429817869262809017100907387585740502094373007612411956428800 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 56, 196, 680, 2348, 8096, 27892, 96056, 330748, 1138768, 3920644, 13498088, 46471180, 159990272, 550811156, 1896319640, 6528602140, 22476505520, 77381536036, 266407155784, 917179667500, 3157642420064, 10871049557044, 37426567849976, 128851218332732, 443605636686608, 1527233994485572, 5257921633190312, 18101836391571532, 62320533397378240, 214555517953618516, 738666179101191512, 2543060786090104924, 8755183795762478192, 30142119966942950756, 103772509783424348936, 357265308430869068908, 1229983748823116112416, 4234556187426974879860] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 (2 t + t + 1) (t - 1) - -------------------------- 4 3 2 4 t + 2 t + t - 4 t + 1 and in Maple notation -(2*t^2+t+1)*(t-1)/(4*t^4+2*t^3+t^2-4*t+1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 26, : Let C(n) be 2 2 n (x y + x + x y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 27, : Let C(n) be 2 2 2 n (x y + x + x y + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 28, : Let C(n) be 2 2 n (x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 60, 4, 16, 16, 64, 16, 64, 60, 216, 4, 16, 16, 64, 16, 64, 64, 240, 16, 64, 64, 256, 60, 240, 216, 768, 4, 16, 16, 64, 16, 64, 64, 240 , 16] Just for kicks C(googol) equals , 297962263182225803415527030257798638760010520316071641088000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 60, 216, 768, 2728, 9704, 34552, 123064, 438328, 1561176, 5560248, 19803096, 70529656, 251194904, 894643768, 3186321112, 11348251384, 40417397400, 143948695992, 512680882776, 1825940038264, 6503181876248, 23161425701176, 82490640856024, 293794773978616, 1046365603664280, 3726686358966456, 13272790284227928, 47271743570583928, 168360811280142104, 599625920981528632, 2135599385504870104, 7606049998477798648, 27089348766443398296, 96480146296286054328, 343619136421744959576, 1223817701857806463096, 4358691378416908259864, 15523709538966336657208] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 2 t + 2 t - t + 1 - --------------------------------- 2 2 (2 t + 3 t - 1) (2 t - 2 t + 1) and in Maple notation -(2*t^3+2*t^2-t+1)/(2*t^2+3*t-1)/(2*t^2-2*t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 29, : Let C(n) be 2 2 n (x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 30, : Let C(n) be 2 2 2 n (x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 58, 4, 16, 16, 64, 16, 64, 58, 204, 4, 16, 16, 64, 16, 64, 64, 232, 16, 64, 64, 256, 58, 232, 204, 714, 4, 16, 16, 64, 16, 64, 64, 232 , 16] Just for kicks C(googol) equals , 171219100609297807343357223578748840991942042613831206271188992 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 58, 204, 714, 2492, 8682, 30228, 105226, 366276, 1274922, 4437692, 15446554, 53765916, 187147146, 651418116, 2267444842, 7892485300, 27472040138, 95624259340, 332847482970, 1158570537292, 4032735032490, 14037083980308, 48860072772074, 170071413502180, 591982043090090, 2060562279111580, 7172374493538586, 24965494321148156, 86899520830961866, 302478557946941732, 1052862859805769450, 3664789362535367700, 12756344234833402634, 44402093037343122988, 154554144181302100058, 537969763351565659308, 1872557140499791061866, 6517969007386764142132] Using the found enumerative automaton with, 8, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 3 2 (2 t - 2 t + 4 t + t - t + t - 1) (t - 1) - ------------------------------------------------------------ 8 7 6 5 4 3 2 4 t - 10 t + 20 t - 12 t + 3 t + 4 t - 10 t + 6 t - 1 and in Maple notation -(2*t^6-2*t^5+4*t^4+t^3-t^2+t-1)*(t-1)/(4*t^8-10*t^7+20*t^6-12*t^5+3*t^4+4*t^3-\ 10*t^2+6*t-1) This ends this theorem, that took, 0.014, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 31, : Let C(n) be 2 2 2 n (x y + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 60, 4, 16, 16, 64, 16, 64, 60, 216, 4, 16, 16, 64, 16, 64, 64, 240, 16, 64, 64, 256, 60, 240, 216, 768, 4, 16, 16, 64, 16, 64, 64, 240 , 16] Just for kicks C(googol) equals , 297962263182225803415527030257798638760010520316071641088000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 60, 216, 768, 2728, 9704, 34552, 123064, 438328, 1561176, 5560248, 19803096, 70529656, 251194904, 894643768, 3186321112, 11348251384, 40417397400, 143948695992, 512680882776, 1825940038264, 6503181876248, 23161425701176, 82490640856024, 293794773978616, 1046365603664280, 3726686358966456, 13272790284227928, 47271743570583928, 168360811280142104, 599625920981528632, 2135599385504870104, 7606049998477798648, 27089348766443398296, 96480146296286054328, 343619136421744959576, 1223817701857806463096, 4358691378416908259864, 15523709538966336657208] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 2 t + 2 t - t + 1 - --------------------------------- 2 2 (2 t + 3 t - 1) (2 t - 2 t + 1) and in Maple notation -(2*t^3+2*t^2-t+1)/(2*t^2+3*t-1)/(2*t^2-2*t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 32, : Let C(n) be 2 2 n (x y + x y + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 33, : Let C(n) be 2 2 2 2 n (x y + x y + x + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 34, : Let C(n) be 2 2 n (x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 60, 4, 16, 16, 64, 16, 64, 60, 216, 4, 16, 16, 64, 16, 64, 64, 240, 16, 64, 64, 256, 60, 240, 216, 768, 4, 16, 16, 64, 16, 64, 64, 240 , 16] Just for kicks C(googol) equals , 297962263182225803415527030257798638760010520316071641088000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 60, 216, 768, 2728, 9704, 34552, 123064, 438328, 1561176, 5560248, 19803096, 70529656, 251194904, 894643768, 3186321112, 11348251384, 40417397400, 143948695992, 512680882776, 1825940038264, 6503181876248, 23161425701176, 82490640856024, 293794773978616, 1046365603664280, 3726686358966456, 13272790284227928, 47271743570583928, 168360811280142104, 599625920981528632, 2135599385504870104, 7606049998477798648, 27089348766443398296, 96480146296286054328, 343619136421744959576, 1223817701857806463096, 4358691378416908259864, 15523709538966336657208] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 2 t + 2 t - t + 1 - --------------------------------- 2 2 (2 t + 3 t - 1) (2 t - 2 t + 1) and in Maple notation -(2*t^3+2*t^2-t+1)/(2*t^2+3*t-1)/(2*t^2-2*t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 35, : Let C(n) be 2 2 2 n (x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 36, : Let C(n) be 2 2 2 n (x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 37, : Let C(n) be 2 2 2 n (x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 58, 4, 16, 16, 64, 16, 64, 58, 204, 4, 16, 16, 64, 16, 64, 64, 232, 16, 64, 64, 256, 58, 232, 204, 714, 4, 16, 16, 64, 16, 64, 64, 232 , 16] Just for kicks C(googol) equals , 171219100609297807343357223578748840991942042613831206271188992 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 58, 204, 714, 2492, 8682, 30228, 105226, 366276, 1274922, 4437692, 15446554, 53765916, 187147146, 651418116, 2267444842, 7892485300, 27472040138, 95624259340, 332847482970, 1158570537292, 4032735032490, 14037083980308, 48860072772074, 170071413502180, 591982043090090, 2060562279111580, 7172374493538586, 24965494321148156, 86899520830961866, 302478557946941732, 1052862859805769450, 3664789362535367700, 12756344234833402634, 44402093037343122988, 154554144181302100058, 537969763351565659308, 1872557140499791061866, 6517969007386764142132] Using the found enumerative automaton with, 8, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 3 2 (2 t - 2 t + 4 t + t - t + t - 1) (t - 1) - ------------------------------------------------------------ 8 7 6 5 4 3 2 4 t - 10 t + 20 t - 12 t + 3 t + 4 t - 10 t + 6 t - 1 and in Maple notation -(2*t^6-2*t^5+4*t^4+t^3-t^2+t-1)*(t-1)/(4*t^8-10*t^7+20*t^6-12*t^5+3*t^4+4*t^3-\ 10*t^2+6*t-1) This ends this theorem, that took, 0.012, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 38, : Let C(n) be 2 2 n (x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 39, : Let C(n) be 2 2 n (x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 1, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 40, : Let C(n) be 2 2 2 n (x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 41, : Let C(n) be 2 2 2 n (x y + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 56, 4, 16, 16, 64, 16, 64, 56, 196, 4, 16, 16, 64, 16, 64, 64, 224, 16, 64, 64, 256, 56, 224, 196, 680, 4, 16, 16, 64, 16, 64, 64, 224 , 16] Just for kicks C(googol) equals , 109429817869262809017100907387585740502094373007612411956428800 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 56, 196, 680, 2348, 8096, 27892, 96056, 330748, 1138768, 3920644, 13498088, 46471180, 159990272, 550811156, 1896319640, 6528602140, 22476505520, 77381536036, 266407155784, 917179667500, 3157642420064, 10871049557044, 37426567849976, 128851218332732, 443605636686608, 1527233994485572, 5257921633190312, 18101836391571532, 62320533397378240, 214555517953618516, 738666179101191512, 2543060786090104924, 8755183795762478192, 30142119966942950756, 103772509783424348936, 357265308430869068908, 1229983748823116112416, 4234556187426974879860] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 (2 t + t + 1) (t - 1) - -------------------------- 4 3 2 4 t + 2 t + t - 4 t + 1 and in Maple notation -(2*t^2+t+1)*(t-1)/(4*t^4+2*t^3+t^2-4*t+1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 42, : Let C(n) be 2 2 2 2 n (x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 43, : Let C(n) be 2 2 2 2 n (x y + x + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 44, : Let C(n) be 2 2 2 2 n (x y + x + x y + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 45, : Let C(n) be 2 2 2 n (x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 46, : Let C(n) be 2 2 2 n (x y + x y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 4, 16, 16, 64, 16, 64, 64, 256, 16] Just for kicks C(googol) equals , 1645504557321206042154969182557350504982735865633579863348609024 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736 , 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976, 4611686018427387904 , 18446744073709551616, 73786976294838206464, 295147905179352825856, 1180591620717411303424, 4722366482869645213696, 18889465931478580854784, 75557863725914323419136, 302231454903657293676544, 1208925819614629174706176] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 1 - ------- 4 t - 1 and in Maple notation -1/(4*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 47, : Let C(n) be 2 2 2 n (x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 60, 4, 16, 16, 64, 16, 64, 60, 216, 4, 16, 16, 64, 16, 64, 64, 240, 16, 64, 64, 256, 60, 240, 216, 768, 4, 16, 16, 64, 16, 64, 64, 240 , 16] Just for kicks C(googol) equals , 297962263182225803415527030257798638760010520316071641088000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 60, 216, 768, 2728, 9704, 34552, 123064, 438328, 1561176, 5560248, 19803096, 70529656, 251194904, 894643768, 3186321112, 11348251384, 40417397400, 143948695992, 512680882776, 1825940038264, 6503181876248, 23161425701176, 82490640856024, 293794773978616, 1046365603664280, 3726686358966456, 13272790284227928, 47271743570583928, 168360811280142104, 599625920981528632, 2135599385504870104, 7606049998477798648, 27089348766443398296, 96480146296286054328, 343619136421744959576, 1223817701857806463096, 4358691378416908259864, 15523709538966336657208] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 2 t + 2 t - t + 1 - --------------------------------- 2 2 (2 t + 3 t - 1) (2 t - 2 t + 1) and in Maple notation -(2*t^3+2*t^2-t+1)/(2*t^2+3*t-1)/(2*t^2-2*t+1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 48, : Let C(n) be 2 2 2 2 n (x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 49, : Let C(n) be 2 2 2 2 n (x y + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 50, : Let C(n) be 2 2 2 n (x y + x y + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 51, : Let C(n) be 2 2 2 2 n (x y + x y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 14, 4, 16, 14, 46, 4, 16, 16, 56, 14, 56, 46, 146, 4, 16, 16, 56, 16, 64, 56, 184, 14, 56, 56, 196, 46, 184, 146, 454, 4, 16, 16, 56, 16, 64, 56, 184 , 16] Just for kicks C(googol) equals , 869984117320134395230202571950639929975664405551389683482624 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454 , 5083664547794, 15251060752246, 45753316474466, 137260217858854, 411781190447474, 1235344645084246, 3706036082736386, 11118112543176454, 33354346219463954, 100063055838261046, 300189201874521506, 900567674343041254, 2701703160468077234, 8105109756282138646, 24315329818602229826] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (3 t - 1) (2 t - 1) and in Maple notation -(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 52, : Let C(n) be 2 2 2 2 n (x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 58, 4, 16, 16, 64, 16, 64, 58, 204, 4, 16, 16, 64, 16, 64, 64, 232, 16, 64, 64, 256, 58, 232, 204, 714, 4, 16, 16, 64, 16, 64, 64, 232 , 16] Just for kicks C(googol) equals , 171219100609297807343357223578748840991942042613831206271188992 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 58, 204, 714, 2492, 8682, 30228, 105226, 366276, 1274922, 4437692, 15446554, 53765916, 187147146, 651418116, 2267444842, 7892485300, 27472040138, 95624259340, 332847482970, 1158570537292, 4032735032490, 14037083980308, 48860072772074, 170071413502180, 591982043090090, 2060562279111580, 7172374493538586, 24965494321148156, 86899520830961866, 302478557946941732, 1052862859805769450, 3664789362535367700, 12756344234833402634, 44402093037343122988, 154554144181302100058, 537969763351565659308, 1872557140499791061866, 6517969007386764142132] Using the found enumerative automaton with, 8, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 3 2 (2 t - 2 t + 4 t + t - t + t - 1) (t - 1) - ------------------------------------------------------------ 8 7 6 5 4 3 2 4 t - 10 t + 20 t - 12 t + 3 t + 4 t - 10 t + 6 t - 1 and in Maple notation -(2*t^6-2*t^5+4*t^4+t^3-t^2+t-1)*(t-1)/(4*t^8-10*t^7+20*t^6-12*t^5+3*t^4+4*t^3-\ 10*t^2+6*t-1) This ends this theorem, that took, 0.013, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 53, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 12, 4, 16, 12, 40, 4, 16, 16, 48, 12, 48, 40, 128, 4, 16, 16, 48, 16, 64, 48, 160, 12, 48, 48, 144, 40, 160, 128, 416, 4, 16, 16, 48, 16, 64, 48, 160 , 16] Just for kicks C(googol) equals , 31545810853354993842838632067457252985987728312107008000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888, 6590694424576, 21327935176704, 69018648051712, 223349036810240, 722772665827328 , 2338941478895616, 7568973621100544, 24493713157783552, 79263320799969280, 256501494231072768, 830056271662022656, 2686118520248336384, 8692462127144763392, 28129398335282872320, 91028645179144798208, 294574883699421085696] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 4 t + 2 t - 1 and in Maple notation -(2*t+1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 54, : Let C(n) be 2 2 2 2 n (x y + x y + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 4, 4, 16, 4, 16, 16, 60, 4, 16, 16, 64, 16, 64, 60, 216, 4, 16, 16, 64, 16, 64, 64, 240, 16, 64, 64, 256, 60, 240, 216, 768, 4, 16, 16, 64, 16, 64, 64, 240 , 16] Just for kicks C(googol) equals , 297962263182225803415527030257798638760010520316071641088000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 4, 16, 60, 216, 768, 2728, 9704, 34552, 123064, 438328, 1561176, 5560248, 19803096, 70529656, 251194904, 894643768, 3186321112, 11348251384, 40417397400, 143948695992, 512680882776, 1825940038264, 6503181876248, 23161425701176, 82490640856024, 293794773978616, 1046365603664280, 3726686358966456, 13272790284227928, 47271743570583928, 168360811280142104, 599625920981528632, 2135599385504870104, 7606049998477798648, 27089348766443398296, 96480146296286054328, 343619136421744959576, 1223817701857806463096, 4358691378416908259864, 15523709538966336657208] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 2 t + 2 t - t + 1 - --------------------------------- 2 2 (2 t + 3 t - 1) (2 t - 2 t + 1) and in Maple notation -(2*t^3+2*t^2-t+1)/(2*t^2+3*t-1)/(2*t^2-2*t+1) This ends this theorem, that took, 0.010, seconds. ------------------------------------------------------------------ This concludes this webbook, that took, 0.525, seconds. to generate. k is , 5 n On Sequences of the form, P(x, y) , modolu , 2, Evaluated at, {x = 1, y = 1} For all Polynomials that are Sums of Monomials taken from, 2 2 2 2 2 2 {1, x, y, x , y , x y, x y , x y, x y } By Shalosh B. Ekhad In this webbook, we will consider the sequences described in the title, that\ after normalization and weeding out obvious symmetry, concerns the following set of, 63, polynomials 2 2 2 2 2 {x y + y + x + y + 1, x + y + x + y + 1, x + x y + y + y + 1, 2 2 2 2 2 x + x y + y + x + y, x y + y + x + y + 1, x y + x y + x + y + 1, 2 2 2 2 2 2 x y + x y + y + y + 1, x y + x y + y + x + 1, x y + x y + y + x + y, 2 2 2 2 2 2 2 2 x y + x + x + y + 1, x y + x + y + y + 1, x y + x + y + x + 1, 2 2 2 2 2 2 2 x y + x + y + x + y, x y + x + x y + y + 1, x y + x + x y + x + 1, 2 2 2 2 2 2 2 2 x y + x + x y + x + y, x y + x + x y + y + 1, x y + x + x y + y + y, 2 2 2 2 2 2 2 2 x y + x + x y + y + x, x y + x y + x + y + 1, x y + x y + y + y + 1, 2 2 2 2 2 2 x y + x y + y + x + 1, x y + x y + y + x + y, 2 2 2 2 x y + x y + x y + y + 1, x y + x y + x y + x + y, 2 2 2 2 2 2 x y + x y + x y + y + 1, x y + x y + x y + y + x, 2 2 2 2 2 2 2 2 x y + x y + x + y + 1, x y + x y + x + y + y, 2 2 2 2 2 2 2 x y + x y + x + x y + y , x y + y + x + y + 1, 2 2 2 2 2 2 2 2 x y + x y + x + y + 1, x y + x y + y + y + 1, x y + x y + y + x + 1, 2 2 2 2 2 2 2 2 2 2 2 x y + x y + y + x + y, x y + x + y + y + 1, x y + x + y + x + y, 2 2 2 2 2 2 2 2 x y + x + x y + y + 1, x y + x + x y + y + y, 2 2 2 2 2 2 2 x y + x y + x + y + 1, x y + x y + y + y + 1, 2 2 2 2 2 2 2 2 x y + x y + y + x + 1, x y + x y + y + x + y, 2 2 2 2 2 2 x y + x y + x y + y + 1, x y + x y + x y + x + 1, 2 2 2 2 2 2 2 x y + x y + x y + x + y, x y + x y + x y + y + 1, 2 2 2 2 2 2 2 2 x y + x y + x y + y + x, x y + x y + x + y + 1, 2 2 2 2 2 2 2 2 x y + x y + x + x + 1, x y + x y + x + x + y, 2 2 2 2 2 2 2 2 2 2 x y + x y + x + y + 1, x y + x y + x + y + y, 2 2 2 2 2 2 2 2 2 x y + x y + x + y + x, x y + x y + x + x y + 1, 2 2 2 2 2 2 2 2 2 x y + x y + x + x y + y, x y + x y + x + x y + y , 2 2 2 2 2 2 2 2 x y + x y + x y + y + 1, x y + x y + x y + x + y, 2 2 2 2 2 2 2 2 2 2 x y + x y + x y + y + 1, x y + x y + x y + y + x, 2 2 2 2 2 2 2 2 2 2 x y + x y + x y + x y + 1, x y + x y + x y + x + y } by finding enumerative automata with at most, 200, states . ----------------------------------------------------------------------------\ ---- Theorem Number, 1, : Let C(n) be 2 n (x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 2, : Let C(n) be 2 2 n (x + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 15, 5, 25, 15, 51, 5, 25, 25, 75, 15, 75, 51, 153, 5, 25, 25, 75, 25, 125, 75, 255, 15, 75, 75, 225, 51, 255, 153, 477, 5, 25, 25, 75, 25, 125, 75, 255, 25] Just for kicks C(googol) equals , 18991393753931390188376409637260877616427023895084857940673828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 15, 51, 153, 477, 1431, 4347, 13041, 39285, 117855, 354051, 1062153, 3187917, 9563751, 28695627, 86086881, 258273765, 774821295, 2324503251, 6973509753, 20920647357, 62761942071, 188286180507, 564858541521, 1694576687445 , 5083730062335, 15251193375651, 45753580126953, 137260749946797, 411782249840391, 1235346778218987, 3706040334656961, 11118121090064325, 33354363270192975, 100063090068859251, 300189270206577753, 900567811394574237, 2701703434183722711, 8105110304875691067, 24315330914627073201] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (3 t + 1) (t - 1) - -------------------- 2 (3 t - 1) (3 t - 1) and in Maple notation -(3*t+1)*(t-1)/(3*t-1)/(3*t^2-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 3, : Let C(n) be 2 2 n (x + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 15, 5, 25, 15, 51, 5, 25, 25, 75, 15, 75, 51, 153, 5, 25, 25, 75, 25, 125, 75, 255, 15, 75, 75, 225, 51, 255, 153, 477, 5, 25, 25, 75, 25, 125, 75, 255, 25] Just for kicks C(googol) equals , 18991393753931390188376409637260877616427023895084857940673828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 15, 51, 153, 477, 1431, 4347, 13041, 39285, 117855, 354051, 1062153, 3187917, 9563751, 28695627, 86086881, 258273765, 774821295, 2324503251, 6973509753, 20920647357, 62761942071, 188286180507, 564858541521, 1694576687445 , 5083730062335, 15251193375651, 45753580126953, 137260749946797, 411782249840391, 1235346778218987, 3706040334656961, 11118121090064325, 33354363270192975, 100063090068859251, 300189270206577753, 900567811394574237, 2701703434183722711, 8105110304875691067, 24315330914627073201] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (3 t + 1) (t - 1) - -------------------- 2 (3 t - 1) (3 t - 1) and in Maple notation -(3*t+1)*(t-1)/(3*t-1)/(3*t^2-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 4, : Let C(n) be 2 2 n (x + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 5, : Let C(n) be 2 2 n (x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 61, 5, 25, 25, 85, 17, 85, 61, 217, 5, 25, 25, 85, 25, 125, 85, 305, 17, 85, 85, 289, 61, 305, 217, 773, 5, 25, 25, 85, 25, 125, 85, 305, 25] Just for kicks C(googol) equals , 2876006182730628489294771910743936557642955449409782886505126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 61, 217, 773, 2753, 9805, 34921, 124373, 442961, 1577629, 5618809, 20011685, 71272673, 253841389, 904069513, 3219891317, 11467812977, 40843221565, 145465290649, 518082315077, 1845177526529, 6571697209741, 23405446682281, 83359734466325, 296890096763537, 1057389759223261, 3765949471196857, 13412627932037093, 47769782738504993, 170134604079589165, 605943377715777481, 2158099341306510773, 7686184779351087281, 27374753020666283389, 97496628620701024729, 347239391903435640965, 1236711432951708972353, 4404613082661998198989, 15687262113889412541673] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 2 t + 3 t - 1 and in Maple notation -(2*t+1)/(2*t^2+3*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 6, : Let C(n) be 2 n (x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 7, : Let C(n) be 2 2 n (x y + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 8, : Let C(n) be 2 2 n (x y + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 61, 5, 25, 25, 85, 17, 85, 61, 217, 5, 25, 25, 85, 25, 125, 85, 305, 17, 85, 85, 289, 61, 305, 217, 773, 5, 25, 25, 85, 25, 125, 85, 305, 25] Just for kicks C(googol) equals , 2876006182730628489294771910743936557642955449409782886505126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 61, 217, 773, 2753, 9805, 34921, 124373, 442961, 1577629, 5618809, 20011685, 71272673, 253841389, 904069513, 3219891317, 11467812977, 40843221565, 145465290649, 518082315077, 1845177526529, 6571697209741, 23405446682281, 83359734466325, 296890096763537, 1057389759223261, 3765949471196857, 13412627932037093, 47769782738504993, 170134604079589165, 605943377715777481, 2158099341306510773, 7686184779351087281, 27374753020666283389, 97496628620701024729, 347239391903435640965, 1236711432951708972353, 4404613082661998198989, 15687262113889412541673] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 2 t + 3 t - 1 and in Maple notation -(2*t+1)/(2*t^2+3*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 9, : Let C(n) be 2 2 n (x y + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 10, : Let C(n) be 2 2 n (x y + x + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 73, 5, 25, 25, 95, 19, 95, 73, 263, 5, 25, 25, 95, 25, 125, 95, 365, 19, 95, 95, 361, 73, 365, 263, 973, 5, 25, 25, 95, 25, 125, 95, 365, 25] Just for kicks C(googol) equals , 107067836886053863857891855428572931230758040328510105609893798828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 73, 263, 973, 3531, 12965, 47251, 173009, 631631, 2310273, 8440503, 30859717, 112776995, 412263309, 1506786763, 5507802233, 20131419959, 73585107241, 268963530095, 983115443149, 3593443967243, 13134702768181, 48009561539555, 175483552150433, 641422737413919, 2344513580442321, 8569606120531879, 31323418577252629, 114492577146357555, 418490478101645149, 1529656037199376891, 5591161254718610057, 20436674729055515463, 74699631405808296057, 273040253115772225503, 998010019384523821213, 3647901659180242072987, 13333720414125860919557, 48737086752902132101907] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 2 8 t - 4 t + 14 t - 12 t + 9 t - 5 t - t + 1 - ------------------------------------------------------------- 8 7 6 5 4 3 2 8 t - 20 t + 46 t - 56 t + 51 t - 20 t - 6 t + 6 t - 1 and in Maple notation -(8*t^7-4*t^6+14*t^5-12*t^4+9*t^3-5*t^2-t+1)/(8*t^8-20*t^7+46*t^6-56*t^5+51*t^4 -20*t^3-6*t^2+6*t-1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 11, : Let C(n) be 2 2 2 n (x y + x + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 73, 5, 25, 25, 95, 19, 95, 73, 271, 5, 25, 25, 95, 25, 125, 95, 365, 19, 95, 95, 361, 73, 365, 271, 993, 5, 25, 25, 95, 25, 125, 95, 365, 25] Just for kicks C(googol) equals , 122157569584608411577942394826877806529805602622218430042266845703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 73, 271, 993, 3643, 13309, 48655, 177805, 649699, 2374129, 8674999, 31698625, 115826659, 423229469, 1546477863, 5650815053, 20648028987, 75447710625, 275685244735, 1007351382097, 3680852748715, 13449802390397, 49145455249183, 179577045144493, 656172884958899, 2397649735785777, 8760990263119911, 32012578502499905, 116973669824702227, 427420722490916541, 1561791420989452407, 5706771605422366317, 20852491388259007563, 76194813313405705761, 278415152787650743407, 1017326428544807383953, 3717301489711406891547, 13582985733670941871837, 49632105964972931002927] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 (2 t + 4 t + 7 t + 2 t + 1) (t - 1) ---------------------------------------------------------------- 9 8 7 6 5 4 3 2 4 t - 2 t - 14 t - 11 t + 26 t - 7 t - 8 t + t + 4 t - 1 and in Maple notation (2*t^6+4*t^5+7*t^4+2*t+1)*(t-1)/(4*t^9-2*t^8-14*t^7-11*t^6+26*t^5-7*t^4-8*t^3+t ^2+4*t-1) This ends this theorem, that took, 0.017, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 12, : Let C(n) be 2 2 2 n (x y + x + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 13, : Let C(n) be 2 2 2 n (x y + x + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 77, 5, 25, 25, 105, 21, 105, 77, 277, 5, 25, 25, 105, 25, 125, 105, 385, 21, 105, 105, 441, 77, 385, 277, 1005, 5, 25, 25, 105, 25, 125, 105, 385, 25] Just for kicks C(googol) equals , 666077427771157268916900567838401442979245985043235123157501220703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 77, 277, 1005, 3669, 13421, 49109, 179693, 657493, 2405741, 8802517, 32208109, 117848405, 431203437, 1577759189, 5772968941, 21123103317, 77288739693, 282796954325, 1034744746733, 3786096966485, 13853204169325, 50688418034645, 185467253023213, 678618573585493, 2483043021926765, 9085372680213717, 33243079563849965, 121635333825674581, 445059682466127981, 1628458727631992277, 5958476878665891821, 21801870757402017365, 79772327425542705005, 291884320097964988117, 1067995118966176780013, 3907758983945264791381, 14298361486321123229293, 52317234003798734518229] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 2 t + t - 1 - ------------------------------- 3 2 (4 t - 6 t + 5 t - 1) (t - 1) and in Maple notation -(4*t^3-2*t^2+t-1)/(4*t^3-6*t^2+5*t-1)/(t-1) This ends this theorem, that took, 0.035, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 14, : Let C(n) be 2 2 n (x y + x + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 23, 5, 25, 23, 93, 5, 25, 25, 115, 23, 115, 93, 359, 5, 25, 25, 115, 25, 125, 115, 465, 23, 115, 115, 529, 93, 465, 359, 1335, 5, 25, 25, 115, 25, 125, 115, 465, 25] Just for kicks C(googol) equals , 26558134128640705502054849209301010895619254370103590190410614013671875 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 23, 93, 359, 1335, 4873, 17535, 62601, 222181, 785855, 2772717, 9768351, 34378167, 120910529, 425062511, 1493898001, 5249371781, 18443445415, 64795091709, 227625068503, 799619495287, 2808906276921, 9866994688223, 34659998140825, 121750158651877, 427670046315727, 1502266603229837, 5276968090316303, 18536231688176247, 65111563355182961, 228714972444138447, 803398355314114081, 2822066188904200197, 9912960994888418871, 34820865610603543453, 122313869380578977351, 429647047899475776183, 1509204004977591524457, 5301320480826218438463, 18621736010632618462889] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 (t - t + t + 1) (t - 1) ------------------------------- 5 4 3 17 t + 15 t - 24 t + 5 t - 1 and in Maple notation (t^3-t^2+t+1)*(t-1)/(17*t^5+15*t^4-24*t^3+5*t-1) This ends this theorem, that took, 0.011, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 15, : Let C(n) be 2 2 n (x y + x + x y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 65, 5, 25, 25, 85, 17, 85, 65, 229, 5, 25, 25, 85, 25, 125, 85, 325, 17, 85, 85, 289, 65, 325, 229, 813, 5, 25, 25, 85, 25, 125, 85, 325, 25] Just for kicks C(googol) equals , 5548091733590341811874582661507331368966333684511482715606689453125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 65, 229, 813, 2945, 10513, 37701, 135261, 484609, 1737665, 6229413, 22330829, 80057281, 286996657, 1028861637, 3688409853, 13222664897, 47402353633 , 169934149285, 609201913325, 2183946525185, 7829295473489, 28067476697413, 100619943566493, 360715462666625, 1293139722607873, 4635815526277797, 16619074628961549, 59578220978784577, 213583758002183921, 765682844011136709, 2744919478730247485, 9840344475718721857, 35276947157358116641, 126465389887529234853, 453369583483133780781, 1625298268650589782401, 5826580693354037214097, 20887884538471765618501] Using the found enumerative automaton with, 7, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 (2 t + 1) (4 t + 3 t + 2 t + 1) - ---------------------------------- 5 4 3 2 8 t + 6 t + 13 t + 5 t + t - 1 and in Maple notation -(2*t+1)*(4*t^3+3*t^2+2*t+1)/(8*t^5+6*t^4+13*t^3+5*t^2+t-1) This ends this theorem, that took, 0.012, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 16, : Let C(n) be 2 2 n (x y + x + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 85, 5, 25, 25, 105, 21, 105, 85, 333, 5, 25, 25, 105, 25, 125, 105, 425, 21, 105, 105, 441, 85, 425, 333, 1293, 5, 25, 25, 105, 25, 125, 105, 425, 25] Just for kicks C(googol) equals , 3926896116798394746365847178957652259388169113663025200366973876953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 85, 333, 1293, 4997, 19269, 74237, 285885, 1100725, 4237685, 16314029, 62803821, 241772773, 930737317, 3582994525, 13793193757, 53098638101, 204409869525, 786901384973, 3029275306957, 11661574029893, 44892686930437, 172820009444029, 665292226131325, 2561125574008437, 9859372988945461, 37954888549712749, 146112087082712109, 562476740332711845, 2165324510303144805, 8335687324816740637, 32089270151577563101, 123531655967252719061, 475550548638896318357, 1830691271317993814989, 7047474848829240197261, 27130135224341347578117, 104440846271811484702917, 402058090745726375080829] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 -------------------------- 4 3 2 8 t + 4 t - t - 4 t + 1 and in Maple notation (t+1)/(8*t^4+4*t^3-t^2-4*t+1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 17, : Let C(n) be 2 2 2 n (x y + x + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 79, 5, 25, 25, 105, 21, 105, 79, 291, 5, 25, 25, 105, 25, 125, 105, 395, 21, 105, 105, 441, 79, 395, 291, 1043, 5, 25, 25, 105, 25, 125, 105, 395, 25] Just for kicks C(googol) equals , 937528355092813527820166281406489661277666527894325554370880126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 79, 291, 1043, 3725, 13205, 46793, 165471, 585191, 2068315, 7310897, 25837453, 91315845, 322715687, 1140515523, 4030652683, 14244666301, 50341596733 , 177910950553, 628749519543, 2222045769223, 7852864484723, 27752577642625, 98079551982773, 346620034979541, 1224979511026399, 4329163636559235, 15299568080670371, 54069747804545357, 191086283816500421, 675312341587546505, 2386601217999499311, 8434416239961774983, 29807819052431215723, 105342924959126484593, 372289291499985868285, 1315696489718894970501, 4649763752106301098743, 16432591498413660089571] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 (t + 1) (t - t - 1) (t - 1) - ------------------------------------- 2 3 2 (t + t - 1) (11 t - 5 t - 3 t + 1) and in Maple notation -(t+1)*(t^2-t-1)*(t-1)/(t^2+t-1)/(11*t^3-5*t^2-3*t+1) This ends this theorem, that took, 0.011, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 18, : Let C(n) be 2 2 2 n (x y + x + x y + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 85, 5, 25, 25, 105, 21, 105, 85, 333, 5, 25, 25, 105, 25, 125, 105, 425, 21, 105, 105, 441, 85, 425, 333, 1293, 5, 25, 25, 105, 25, 125, 105, 425, 25] Just for kicks C(googol) equals , 3926896116798394746365847178957652259388169113663025200366973876953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 85, 333, 1293, 4997, 19269, 74237, 285885, 1100725, 4237685, 16314029, 62803821, 241772773, 930737317, 3582994525, 13793193757, 53098638101, 204409869525, 786901384973, 3029275306957, 11661574029893, 44892686930437, 172820009444029, 665292226131325, 2561125574008437, 9859372988945461, 37954888549712749, 146112087082712109, 562476740332711845, 2165324510303144805, 8335687324816740637, 32089270151577563101, 123531655967252719061, 475550548638896318357, 1830691271317993814989, 7047474848829240197261, 27130135224341347578117, 104440846271811484702917, 402058090745726375080829] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 -------------------------- 4 3 2 8 t + 4 t - t - 4 t + 1 and in Maple notation (t+1)/(8*t^4+4*t^3-t^2-4*t+1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 19, : Let C(n) be 2 2 2 n (x y + x + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 61, 5, 25, 25, 85, 17, 85, 61, 217, 5, 25, 25, 85, 25, 125, 85, 305, 17, 85, 85, 289, 61, 305, 217, 773, 5, 25, 25, 85, 25, 125, 85, 305, 25] Just for kicks C(googol) equals , 2876006182730628489294771910743936557642955449409782886505126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 61, 217, 773, 2753, 9805, 34921, 124373, 442961, 1577629, 5618809, 20011685, 71272673, 253841389, 904069513, 3219891317, 11467812977, 40843221565, 145465290649, 518082315077, 1845177526529, 6571697209741, 23405446682281, 83359734466325, 296890096763537, 1057389759223261, 3765949471196857, 13412627932037093, 47769782738504993, 170134604079589165, 605943377715777481, 2158099341306510773, 7686184779351087281, 27374753020666283389, 97496628620701024729, 347239391903435640965, 1236711432951708972353, 4404613082661998198989, 15687262113889412541673] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 2 t + 3 t - 1 and in Maple notation -(2*t+1)/(2*t^2+3*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 20, : Let C(n) be 2 2 n (x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 77, 5, 25, 25, 105, 21, 105, 77, 277, 5, 25, 25, 105, 25, 125, 105, 385, 21, 105, 105, 441, 77, 385, 277, 1005, 5, 25, 25, 105, 25, 125, 105, 385, 25] Just for kicks C(googol) equals , 666077427771157268916900567838401442979245985043235123157501220703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 77, 277, 1005, 3669, 13421, 49109, 179693, 657493, 2405741, 8802517, 32208109, 117848405, 431203437, 1577759189, 5772968941, 21123103317, 77288739693, 282796954325, 1034744746733, 3786096966485, 13853204169325, 50688418034645, 185467253023213, 678618573585493, 2483043021926765, 9085372680213717, 33243079563849965, 121635333825674581, 445059682466127981, 1628458727631992277, 5958476878665891821, 21801870757402017365, 79772327425542705005, 291884320097964988117, 1067995118966176780013, 3907758983945264791381, 14298361486321123229293, 52317234003798734518229] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 2 t + t - 1 - ------------------------------- 3 2 (4 t - 6 t + 5 t - 1) (t - 1) and in Maple notation -(4*t^3-2*t^2+t-1)/(4*t^3-6*t^2+5*t-1)/(t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 21, : Let C(n) be 2 2 2 n (x y + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 73, 5, 25, 25, 95, 19, 95, 73, 263, 5, 25, 25, 95, 25, 125, 95, 365, 19, 95, 95, 361, 73, 365, 263, 973, 5, 25, 25, 95, 25, 125, 95, 365, 25] Just for kicks C(googol) equals , 107067836886053863857891855428572931230758040328510105609893798828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 73, 263, 973, 3531, 12965, 47251, 173009, 631631, 2310273, 8440503, 30859717, 112776995, 412263309, 1506786763, 5507802233, 20131419959, 73585107241, 268963530095, 983115443149, 3593443967243, 13134702768181, 48009561539555, 175483552150433, 641422737413919, 2344513580442321, 8569606120531879, 31323418577252629, 114492577146357555, 418490478101645149, 1529656037199376891, 5591161254718610057, 20436674729055515463, 74699631405808296057, 273040253115772225503, 998010019384523821213, 3647901659180242072987, 13333720414125860919557, 48737086752902132101907] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 2 8 t - 4 t + 14 t - 12 t + 9 t - 5 t - t + 1 - ------------------------------------------------------------- 8 7 6 5 4 3 2 8 t - 20 t + 46 t - 56 t + 51 t - 20 t - 6 t + 6 t - 1 and in Maple notation -(8*t^7-4*t^6+14*t^5-12*t^4+9*t^3-5*t^2-t+1)/(8*t^8-20*t^7+46*t^6-56*t^5+51*t^4 -20*t^3-6*t^2+6*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 22, : Let C(n) be 2 2 2 n (x y + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 25, 5, 25, 25, 101, 5, 25, 25, 125, 25, 125, 101, 361, 5, 25, 25, 125 , 25, 125, 125, 505, 25, 125, 125, 625, 101, 505, 361, 1205, 5, 25, 25, 125, 25 , 125, 125, 505, 25] Just for kicks C(googol) equals , 84173718217807518475510932258638885361534676121664233505725860595703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 25, 101, 361, 1205, 3865, 12101, 37321, 114005, 346105, 1046501, 3155881 , 9500405, 28566745, 85831301, 257756041, 773792405, 2322425785, 6969374501, 20912317801, 62745342005, 188252803225, 564791964101, 1694443001161, 5083463221205, 15250658099065, 45752511168101, 137258607246121, 411777969222005 , 1235338202633305, 3706023197834501, 11118086773372681, 33354294679856405, 100062952759045945, 300188995716091301, 900567262026180841, 2701702335834356405 , 8105108107014696985, 24315326520067346501, 72945983958248550601] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 6 t - t + 1 - --------------------------- (t - 1) (3 t - 1) (2 t - 1) and in Maple notation -(6*t^2-t+1)/(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 23, : Let C(n) be 2 2 2 n (x y + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 77, 5, 25, 25, 105, 21, 105, 77, 277, 5, 25, 25, 105, 25, 125, 105, 385, 21, 105, 105, 441, 77, 385, 277, 1005, 5, 25, 25, 105, 25, 125, 105, 385, 25] Just for kicks C(googol) equals , 666077427771157268916900567838401442979245985043235123157501220703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 77, 277, 1005, 3669, 13421, 49109, 179693, 657493, 2405741, 8802517, 32208109, 117848405, 431203437, 1577759189, 5772968941, 21123103317, 77288739693, 282796954325, 1034744746733, 3786096966485, 13853204169325, 50688418034645, 185467253023213, 678618573585493, 2483043021926765, 9085372680213717, 33243079563849965, 121635333825674581, 445059682466127981, 1628458727631992277, 5958476878665891821, 21801870757402017365, 79772327425542705005, 291884320097964988117, 1067995118966176780013, 3907758983945264791381, 14298361486321123229293, 52317234003798734518229] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 2 t + t - 1 - ------------------------------- 3 2 (4 t - 6 t + 5 t - 1) (t - 1) and in Maple notation -(4*t^3-2*t^2+t-1)/(4*t^3-6*t^2+5*t-1)/(t-1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 24, : Let C(n) be 2 2 n (x y + x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 85, 5, 25, 25, 105, 21, 105, 85, 333, 5, 25, 25, 105, 25, 125, 105, 425, 21, 105, 105, 441, 85, 425, 333, 1293, 5, 25, 25, 105, 25, 125, 105, 425, 25] Just for kicks C(googol) equals , 3926896116798394746365847178957652259388169113663025200366973876953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 85, 333, 1293, 4997, 19269, 74237, 285885, 1100725, 4237685, 16314029, 62803821, 241772773, 930737317, 3582994525, 13793193757, 53098638101, 204409869525, 786901384973, 3029275306957, 11661574029893, 44892686930437, 172820009444029, 665292226131325, 2561125574008437, 9859372988945461, 37954888549712749, 146112087082712109, 562476740332711845, 2165324510303144805, 8335687324816740637, 32089270151577563101, 123531655967252719061, 475550548638896318357, 1830691271317993814989, 7047474848829240197261, 27130135224341347578117, 104440846271811484702917, 402058090745726375080829] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 -------------------------- 4 3 2 8 t + 4 t - t - 4 t + 1 and in Maple notation (t+1)/(8*t^4+4*t^3-t^2-4*t+1) This ends this theorem, that took, 0.012, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 25, : Let C(n) be 2 2 n (x y + x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 61, 5, 25, 25, 85, 17, 85, 61, 217, 5, 25, 25, 85, 25, 125, 85, 305, 17, 85, 85, 289, 61, 305, 217, 773, 5, 25, 25, 85, 25, 125, 85, 305, 25] Just for kicks C(googol) equals , 2876006182730628489294771910743936557642955449409782886505126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 61, 217, 773, 2753, 9805, 34921, 124373, 442961, 1577629, 5618809, 20011685, 71272673, 253841389, 904069513, 3219891317, 11467812977, 40843221565, 145465290649, 518082315077, 1845177526529, 6571697209741, 23405446682281, 83359734466325, 296890096763537, 1057389759223261, 3765949471196857, 13412627932037093, 47769782738504993, 170134604079589165, 605943377715777481, 2158099341306510773, 7686184779351087281, 27374753020666283389, 97496628620701024729, 347239391903435640965, 1236711432951708972353, 4404613082661998198989, 15687262113889412541673] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 2 t + 3 t - 1 and in Maple notation -(2*t+1)/(2*t^2+3*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 26, : Let C(n) be 2 2 2 n (x y + x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 23, 5, 25, 23, 93, 5, 25, 25, 115, 23, 115, 93, 359, 5, 25, 25, 115, 25, 125, 115, 465, 23, 115, 115, 529, 93, 465, 359, 1335, 5, 25, 25, 115, 25, 125, 115, 465, 25] Just for kicks C(googol) equals , 26558134128640705502054849209301010895619254370103590190410614013671875 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 23, 93, 359, 1335, 4873, 17535, 62601, 222181, 785855, 2772717, 9768351, 34378167, 120910529, 425062511, 1493898001, 5249371781, 18443445415, 64795091709, 227625068503, 799619495287, 2808906276921, 9866994688223, 34659998140825, 121750158651877, 427670046315727, 1502266603229837, 5276968090316303, 18536231688176247, 65111563355182961, 228714972444138447, 803398355314114081, 2822066188904200197, 9912960994888418871, 34820865610603543453, 122313869380578977351, 429647047899475776183, 1509204004977591524457, 5301320480826218438463, 18621736010632618462889] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 (t - t + t + 1) (t - 1) ------------------------------- 5 4 3 17 t + 15 t - 24 t + 5 t - 1 and in Maple notation (t^3-t^2+t+1)*(t-1)/(17*t^5+15*t^4-24*t^3+5*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 27, : Let C(n) be 2 2 2 n (x y + x y + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 85, 5, 25, 25, 105, 21, 105, 85, 333, 5, 25, 25, 105, 25, 125, 105, 425, 21, 105, 105, 441, 85, 425, 333, 1293, 5, 25, 25, 105, 25, 125, 105, 425, 25] Just for kicks C(googol) equals , 3926896116798394746365847178957652259388169113663025200366973876953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 85, 333, 1293, 4997, 19269, 74237, 285885, 1100725, 4237685, 16314029, 62803821, 241772773, 930737317, 3582994525, 13793193757, 53098638101, 204409869525, 786901384973, 3029275306957, 11661574029893, 44892686930437, 172820009444029, 665292226131325, 2561125574008437, 9859372988945461, 37954888549712749, 146112087082712109, 562476740332711845, 2165324510303144805, 8335687324816740637, 32089270151577563101, 123531655967252719061, 475550548638896318357, 1830691271317993814989, 7047474848829240197261, 27130135224341347578117, 104440846271811484702917, 402058090745726375080829] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 -------------------------- 4 3 2 8 t + 4 t - t - 4 t + 1 and in Maple notation (t+1)/(8*t^4+4*t^3-t^2-4*t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 28, : Let C(n) be 2 2 2 2 n (x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 257, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 257, 925, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 73606661197199861757816780443770926797560605336911976337432861328125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 257, 925, 3323, 11919, 42729, 153173, 549075, 1968247, 7055601, 25292621, 90668587, 325028895, 1165167673, 4176914949, 14973497539, 53677355783 , 192423938625, 689806295677, 2472835597211, 8864685923631, 31778359388937, 113919900383925, 408383064188083, 1463982386475159, 5248122704415889, 18813608822615213, 67443521618241035, 241773316967213887, 866715370142188761, 3107024142696604261, 11138142181492144547, 39928306173024688935, 143136046198521240033, 513117877643462669789, 1839438515665766891899, 6594067757794661727823, 23638588201825352540585] Using the found enumerative automaton with, 8, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 3 2 (6 t + 2 t + t - 1) (t - 1) - ---------------------------------------------- 6 5 4 3 2 10 t + 10 t - 15 t + 8 t - 10 t + 6 t - 1 and in Maple notation -(6*t^4+2*t^3+t^2-1)*(t-1)/(10*t^6+10*t^5-15*t^4+8*t^3-10*t^2+6*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 29, : Let C(n) be 2 2 2 2 n (x y + x y + x + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 77, 5, 25, 25, 105, 21, 105, 77, 277, 5, 25, 25, 105, 25, 125, 105, 385, 21, 105, 105, 441, 77, 385, 277, 1005, 5, 25, 25, 105, 25, 125, 105, 385, 25] Just for kicks C(googol) equals , 666077427771157268916900567838401442979245985043235123157501220703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 77, 277, 1005, 3669, 13421, 49109, 179693, 657493, 2405741, 8802517, 32208109, 117848405, 431203437, 1577759189, 5772968941, 21123103317, 77288739693, 282796954325, 1034744746733, 3786096966485, 13853204169325, 50688418034645, 185467253023213, 678618573585493, 2483043021926765, 9085372680213717, 33243079563849965, 121635333825674581, 445059682466127981, 1628458727631992277, 5958476878665891821, 21801870757402017365, 79772327425542705005, 291884320097964988117, 1067995118966176780013, 3907758983945264791381, 14298361486321123229293, 52317234003798734518229] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 2 t + t - 1 - ------------------------------- 3 2 (4 t - 6 t + 5 t - 1) (t - 1) and in Maple notation -(4*t^3-2*t^2+t-1)/(4*t^3-6*t^2+5*t-1)/(t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 30, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 31, : Let C(n) be 2 2 2 n (x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 73, 5, 25, 25, 95, 19, 95, 73, 271, 5, 25, 25, 95, 25, 125, 95, 365, 19, 95, 95, 361, 73, 365, 271, 993, 5, 25, 25, 95, 25, 125, 95, 365, 25] Just for kicks C(googol) equals , 122157569584608411577942394826877806529805602622218430042266845703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 73, 271, 993, 3643, 13309, 48655, 177805, 649699, 2374129, 8674999, 31698625, 115826659, 423229469, 1546477863, 5650815053, 20648028987, 75447710625, 275685244735, 1007351382097, 3680852748715, 13449802390397, 49145455249183, 179577045144493, 656172884958899, 2397649735785777, 8760990263119911, 32012578502499905, 116973669824702227, 427420722490916541, 1561791420989452407, 5706771605422366317, 20852491388259007563, 76194813313405705761, 278415152787650743407, 1017326428544807383953, 3717301489711406891547, 13582985733670941871837, 49632105964972931002927] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 (2 t + 4 t + 7 t + 2 t + 1) (t - 1) ---------------------------------------------------------------- 9 8 7 6 5 4 3 2 4 t - 2 t - 14 t - 11 t + 26 t - 7 t - 8 t + t + 4 t - 1 and in Maple notation (2*t^6+4*t^5+7*t^4+2*t+1)*(t-1)/(4*t^9-2*t^8-14*t^7-11*t^6+26*t^5-7*t^4-8*t^3+t ^2+4*t-1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 32, : Let C(n) be 2 2 n (x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 85, 5, 25, 25, 105, 21, 105, 85, 333, 5, 25, 25, 105, 25, 125, 105, 425, 21, 105, 105, 441, 85, 425, 333, 1293, 5, 25, 25, 105, 25, 125, 105, 425, 25] Just for kicks C(googol) equals , 3926896116798394746365847178957652259388169113663025200366973876953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 85, 333, 1293, 4997, 19269, 74237, 285885, 1100725, 4237685, 16314029, 62803821, 241772773, 930737317, 3582994525, 13793193757, 53098638101, 204409869525, 786901384973, 3029275306957, 11661574029893, 44892686930437, 172820009444029, 665292226131325, 2561125574008437, 9859372988945461, 37954888549712749, 146112087082712109, 562476740332711845, 2165324510303144805, 8335687324816740637, 32089270151577563101, 123531655967252719061, 475550548638896318357, 1830691271317993814989, 7047474848829240197261, 27130135224341347578117, 104440846271811484702917, 402058090745726375080829] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 -------------------------- 4 3 2 8 t + 4 t - t - 4 t + 1 and in Maple notation (t+1)/(8*t^4+4*t^3-t^2-4*t+1) This ends this theorem, that took, 0.041, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 33, : Let C(n) be 2 2 2 n (x y + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 15, 5, 25, 15, 51, 5, 25, 25, 75, 15, 75, 51, 153, 5, 25, 25, 75, 25, 125, 75, 255, 15, 75, 75, 225, 51, 255, 153, 477, 5, 25, 25, 75, 25, 125, 75, 255, 25] Just for kicks C(googol) equals , 18991393753931390188376409637260877616427023895084857940673828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 15, 51, 153, 477, 1431, 4347, 13041, 39285, 117855, 354051, 1062153, 3187917, 9563751, 28695627, 86086881, 258273765, 774821295, 2324503251, 6973509753, 20920647357, 62761942071, 188286180507, 564858541521, 1694576687445 , 5083730062335, 15251193375651, 45753580126953, 137260749946797, 411782249840391, 1235346778218987, 3706040334656961, 11118121090064325, 33354363270192975, 100063090068859251, 300189270206577753, 900567811394574237, 2701703434183722711, 8105110304875691067, 24315330914627073201] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (3 t + 1) (t - 1) - -------------------- 2 (3 t - 1) (3 t - 1) and in Maple notation -(3*t+1)*(t-1)/(3*t-1)/(3*t^2-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 34, : Let C(n) be 2 2 2 n (x y + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 79, 5, 25, 25, 105, 21, 105, 79, 291, 5, 25, 25, 105, 25, 125, 105, 395, 21, 105, 105, 441, 79, 395, 291, 1043, 5, 25, 25, 105, 25, 125, 105, 395, 25] Just for kicks C(googol) equals , 937528355092813527820166281406489661277666527894325554370880126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 79, 291, 1043, 3725, 13205, 46793, 165471, 585191, 2068315, 7310897, 25837453, 91315845, 322715687, 1140515523, 4030652683, 14244666301, 50341596733 , 177910950553, 628749519543, 2222045769223, 7852864484723, 27752577642625, 98079551982773, 346620034979541, 1224979511026399, 4329163636559235, 15299568080670371, 54069747804545357, 191086283816500421, 675312341587546505, 2386601217999499311, 8434416239961774983, 29807819052431215723, 105342924959126484593, 372289291499985868285, 1315696489718894970501, 4649763752106301098743, 16432591498413660089571] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 (t + 1) (t - t - 1) (t - 1) - ------------------------------------- 2 3 2 (t + t - 1) (11 t - 5 t - 3 t + 1) and in Maple notation -(t+1)*(t^2-t-1)*(t-1)/(t^2+t-1)/(11*t^3-5*t^2-3*t+1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 35, : Let C(n) be 2 2 2 n (x y + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 23, 5, 25, 23, 93, 5, 25, 25, 115, 23, 115, 93, 359, 5, 25, 25, 115, 25, 125, 115, 465, 23, 115, 115, 529, 93, 465, 359, 1335, 5, 25, 25, 115, 25, 125, 115, 465, 25] Just for kicks C(googol) equals , 26558134128640705502054849209301010895619254370103590190410614013671875 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 23, 93, 359, 1335, 4873, 17535, 62601, 222181, 785855, 2772717, 9768351, 34378167, 120910529, 425062511, 1493898001, 5249371781, 18443445415, 64795091709, 227625068503, 799619495287, 2808906276921, 9866994688223, 34659998140825, 121750158651877, 427670046315727, 1502266603229837, 5276968090316303, 18536231688176247, 65111563355182961, 228714972444138447, 803398355314114081, 2822066188904200197, 9912960994888418871, 34820865610603543453, 122313869380578977351, 429647047899475776183, 1509204004977591524457, 5301320480826218438463, 18621736010632618462889] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 (t - t + t + 1) (t - 1) ------------------------------- 5 4 3 17 t + 15 t - 24 t + 5 t - 1 and in Maple notation (t^3-t^2+t+1)*(t-1)/(17*t^5+15*t^4-24*t^3+5*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 36, : Let C(n) be 2 2 2 2 n (x y + x + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 61, 5, 25, 25, 85, 17, 85, 61, 217, 5, 25, 25, 85, 25, 125, 85, 305, 17, 85, 85, 289, 61, 305, 217, 773, 5, 25, 25, 85, 25, 125, 85, 305, 25] Just for kicks C(googol) equals , 2876006182730628489294771910743936557642955449409782886505126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 61, 217, 773, 2753, 9805, 34921, 124373, 442961, 1577629, 5618809, 20011685, 71272673, 253841389, 904069513, 3219891317, 11467812977, 40843221565, 145465290649, 518082315077, 1845177526529, 6571697209741, 23405446682281, 83359734466325, 296890096763537, 1057389759223261, 3765949471196857, 13412627932037093, 47769782738504993, 170134604079589165, 605943377715777481, 2158099341306510773, 7686184779351087281, 27374753020666283389, 97496628620701024729, 347239391903435640965, 1236711432951708972353, 4404613082661998198989, 15687262113889412541673] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 2 t + 3 t - 1 and in Maple notation -(2*t+1)/(2*t^2+3*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 37, : Let C(n) be 2 2 2 2 n (x y + x + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 257, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 257, 925, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 73606661197199861757816780443770926797560605336911976337432861328125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 257, 925, 3323, 11919, 42729, 153173, 549075, 1968247, 7055601, 25292621, 90668587, 325028895, 1165167673, 4176914949, 14973497539, 53677355783 , 192423938625, 689806295677, 2472835597211, 8864685923631, 31778359388937, 113919900383925, 408383064188083, 1463982386475159, 5248122704415889, 18813608822615213, 67443521618241035, 241773316967213887, 866715370142188761, 3107024142696604261, 11138142181492144547, 39928306173024688935, 143136046198521240033, 513117877643462669789, 1839438515665766891899, 6594067757794661727823, 23638588201825352540585] Using the found enumerative automaton with, 8, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 3 2 (6 t + 2 t + t - 1) (t - 1) - ---------------------------------------------- 6 5 4 3 2 10 t + 10 t - 15 t + 8 t - 10 t + 6 t - 1 and in Maple notation -(6*t^4+2*t^3+t^2-1)*(t-1)/(10*t^6+10*t^5-15*t^4+8*t^3-10*t^2+6*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 38, : Let C(n) be 2 2 2 2 n (x y + x + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 61, 5, 25, 25, 85, 17, 85, 61, 217, 5, 25, 25, 85, 25, 125, 85, 305, 17, 85, 85, 289, 61, 305, 217, 773, 5, 25, 25, 85, 25, 125, 85, 305, 25] Just for kicks C(googol) equals , 2876006182730628489294771910743936557642955449409782886505126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 61, 217, 773, 2753, 9805, 34921, 124373, 442961, 1577629, 5618809, 20011685, 71272673, 253841389, 904069513, 3219891317, 11467812977, 40843221565, 145465290649, 518082315077, 1845177526529, 6571697209741, 23405446682281, 83359734466325, 296890096763537, 1057389759223261, 3765949471196857, 13412627932037093, 47769782738504993, 170134604079589165, 605943377715777481, 2158099341306510773, 7686184779351087281, 27374753020666283389, 97496628620701024729, 347239391903435640965, 1236711432951708972353, 4404613082661998198989, 15687262113889412541673] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 2 t + 3 t - 1 and in Maple notation -(2*t+1)/(2*t^2+3*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 39, : Let C(n) be 2 2 2 2 n (x y + x + x y + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 79, 5, 25, 25, 105, 21, 105, 79, 291, 5, 25, 25, 105, 25, 125, 105, 395, 21, 105, 105, 441, 79, 395, 291, 1043, 5, 25, 25, 105, 25, 125, 105, 395, 25] Just for kicks C(googol) equals , 937528355092813527820166281406489661277666527894325554370880126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 79, 291, 1043, 3725, 13205, 46793, 165471, 585191, 2068315, 7310897, 25837453, 91315845, 322715687, 1140515523, 4030652683, 14244666301, 50341596733 , 177910950553, 628749519543, 2222045769223, 7852864484723, 27752577642625, 98079551982773, 346620034979541, 1224979511026399, 4329163636559235, 15299568080670371, 54069747804545357, 191086283816500421, 675312341587546505, 2386601217999499311, 8434416239961774983, 29807819052431215723, 105342924959126484593, 372289291499985868285, 1315696489718894970501, 4649763752106301098743, 16432591498413660089571] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 (t + 1) (t - t - 1) (t - 1) - ------------------------------------- 2 3 2 (t + t - 1) (11 t - 5 t - 3 t + 1) and in Maple notation -(t+1)*(t^2-t-1)*(t-1)/(t^2+t-1)/(11*t^3-5*t^2-3*t+1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 40, : Let C(n) be 2 2 2 n (x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 77, 5, 25, 25, 105, 21, 105, 77, 277, 5, 25, 25, 105, 25, 125, 105, 385, 21, 105, 105, 441, 77, 385, 277, 1005, 5, 25, 25, 105, 25, 125, 105, 385, 25] Just for kicks C(googol) equals , 666077427771157268916900567838401442979245985043235123157501220703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 77, 277, 1005, 3669, 13421, 49109, 179693, 657493, 2405741, 8802517, 32208109, 117848405, 431203437, 1577759189, 5772968941, 21123103317, 77288739693, 282796954325, 1034744746733, 3786096966485, 13853204169325, 50688418034645, 185467253023213, 678618573585493, 2483043021926765, 9085372680213717, 33243079563849965, 121635333825674581, 445059682466127981, 1628458727631992277, 5958476878665891821, 21801870757402017365, 79772327425542705005, 291884320097964988117, 1067995118966176780013, 3907758983945264791381, 14298361486321123229293, 52317234003798734518229] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 2 t + t - 1 - ------------------------------- 3 2 (4 t - 6 t + 5 t - 1) (t - 1) and in Maple notation -(4*t^3-2*t^2+t-1)/(4*t^3-6*t^2+5*t-1)/(t-1) This ends this theorem, that took, 0.011, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 41, : Let C(n) be 2 2 2 2 n (x y + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 15, 5, 25, 15, 51, 5, 25, 25, 75, 15, 75, 51, 153, 5, 25, 25, 75, 25, 125, 75, 255, 15, 75, 75, 225, 51, 255, 153, 477, 5, 25, 25, 75, 25, 125, 75, 255, 25] Just for kicks C(googol) equals , 18991393753931390188376409637260877616427023895084857940673828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 15, 51, 153, 477, 1431, 4347, 13041, 39285, 117855, 354051, 1062153, 3187917, 9563751, 28695627, 86086881, 258273765, 774821295, 2324503251, 6973509753, 20920647357, 62761942071, 188286180507, 564858541521, 1694576687445 , 5083730062335, 15251193375651, 45753580126953, 137260749946797, 411782249840391, 1235346778218987, 3706040334656961, 11118121090064325, 33354363270192975, 100063090068859251, 300189270206577753, 900567811394574237, 2701703434183722711, 8105110304875691067, 24315330914627073201] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (3 t + 1) (t - 1) - -------------------- 2 (3 t - 1) (3 t - 1) and in Maple notation -(3*t+1)*(t-1)/(3*t-1)/(3*t^2-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 42, : Let C(n) be 2 2 2 2 n (x y + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 43, : Let C(n) be 2 2 2 2 n (x y + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 73, 5, 25, 25, 95, 19, 95, 73, 263, 5, 25, 25, 95, 25, 125, 95, 365, 19, 95, 95, 361, 73, 365, 263, 973, 5, 25, 25, 95, 25, 125, 95, 365, 25] Just for kicks C(googol) equals , 107067836886053863857891855428572931230758040328510105609893798828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 73, 263, 973, 3531, 12965, 47251, 173009, 631631, 2310273, 8440503, 30859717, 112776995, 412263309, 1506786763, 5507802233, 20131419959, 73585107241, 268963530095, 983115443149, 3593443967243, 13134702768181, 48009561539555, 175483552150433, 641422737413919, 2344513580442321, 8569606120531879, 31323418577252629, 114492577146357555, 418490478101645149, 1529656037199376891, 5591161254718610057, 20436674729055515463, 74699631405808296057, 273040253115772225503, 998010019384523821213, 3647901659180242072987, 13333720414125860919557, 48737086752902132101907] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 2 8 t - 4 t + 14 t - 12 t + 9 t - 5 t - t + 1 - ------------------------------------------------------------- 8 7 6 5 4 3 2 8 t - 20 t + 46 t - 56 t + 51 t - 20 t - 6 t + 6 t - 1 and in Maple notation -(8*t^7-4*t^6+14*t^5-12*t^4+9*t^3-5*t^2-t+1)/(8*t^8-20*t^7+46*t^6-56*t^5+51*t^4 -20*t^3-6*t^2+6*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 44, : Let C(n) be 2 2 2 n (x y + x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 45, : Let C(n) be 2 2 2 n (x y + x y + x y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 61, 5, 25, 25, 85, 17, 85, 61, 217, 5, 25, 25, 85, 25, 125, 85, 305, 17, 85, 85, 289, 61, 305, 217, 773, 5, 25, 25, 85, 25, 125, 85, 305, 25] Just for kicks C(googol) equals , 2876006182730628489294771910743936557642955449409782886505126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 61, 217, 773, 2753, 9805, 34921, 124373, 442961, 1577629, 5618809, 20011685, 71272673, 253841389, 904069513, 3219891317, 11467812977, 40843221565, 145465290649, 518082315077, 1845177526529, 6571697209741, 23405446682281, 83359734466325, 296890096763537, 1057389759223261, 3765949471196857, 13412627932037093, 47769782738504993, 170134604079589165, 605943377715777481, 2158099341306510773, 7686184779351087281, 27374753020666283389, 97496628620701024729, 347239391903435640965, 1236711432951708972353, 4404613082661998198989, 15687262113889412541673] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 2 t + 3 t - 1 and in Maple notation -(2*t+1)/(2*t^2+3*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 46, : Let C(n) be 2 2 2 n (x y + x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 85, 5, 25, 25, 105, 21, 105, 85, 333, 5, 25, 25, 105, 25, 125, 105, 425, 21, 105, 105, 441, 85, 425, 333, 1293, 5, 25, 25, 105, 25, 125, 105, 425, 25] Just for kicks C(googol) equals , 3926896116798394746365847178957652259388169113663025200366973876953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 85, 333, 1293, 4997, 19269, 74237, 285885, 1100725, 4237685, 16314029, 62803821, 241772773, 930737317, 3582994525, 13793193757, 53098638101, 204409869525, 786901384973, 3029275306957, 11661574029893, 44892686930437, 172820009444029, 665292226131325, 2561125574008437, 9859372988945461, 37954888549712749, 146112087082712109, 562476740332711845, 2165324510303144805, 8335687324816740637, 32089270151577563101, 123531655967252719061, 475550548638896318357, 1830691271317993814989, 7047474848829240197261, 27130135224341347578117, 104440846271811484702917, 402058090745726375080829] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 -------------------------- 4 3 2 8 t + 4 t - t - 4 t + 1 and in Maple notation (t+1)/(8*t^4+4*t^3-t^2-4*t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 47, : Let C(n) be 2 2 2 2 n (x y + x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 15, 5, 25, 15, 51, 5, 25, 25, 75, 15, 75, 51, 153, 5, 25, 25, 75, 25, 125, 75, 255, 15, 75, 75, 225, 51, 255, 153, 477, 5, 25, 25, 75, 25, 125, 75, 255, 25] Just for kicks C(googol) equals , 18991393753931390188376409637260877616427023895084857940673828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 15, 51, 153, 477, 1431, 4347, 13041, 39285, 117855, 354051, 1062153, 3187917, 9563751, 28695627, 86086881, 258273765, 774821295, 2324503251, 6973509753, 20920647357, 62761942071, 188286180507, 564858541521, 1694576687445 , 5083730062335, 15251193375651, 45753580126953, 137260749946797, 411782249840391, 1235346778218987, 3706040334656961, 11118121090064325, 33354363270192975, 100063090068859251, 300189270206577753, 900567811394574237, 2701703434183722711, 8105110304875691067, 24315330914627073201] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (3 t + 1) (t - 1) - -------------------- 2 (3 t - 1) (3 t - 1) and in Maple notation -(3*t+1)*(t-1)/(3*t-1)/(3*t^2-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 48, : Let C(n) be 2 2 2 2 n (x y + x y + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 65, 5, 25, 25, 85, 17, 85, 65, 229, 5, 25, 25, 85, 25, 125, 85, 325, 17, 85, 85, 289, 65, 325, 229, 813, 5, 25, 25, 85, 25, 125, 85, 325, 25] Just for kicks C(googol) equals , 5548091733590341811874582661507331368966333684511482715606689453125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 65, 229, 813, 2945, 10513, 37701, 135261, 484609, 1737665, 6229413, 22330829, 80057281, 286996657, 1028861637, 3688409853, 13222664897, 47402353633 , 169934149285, 609201913325, 2183946525185, 7829295473489, 28067476697413, 100619943566493, 360715462666625, 1293139722607873, 4635815526277797, 16619074628961549, 59578220978784577, 213583758002183921, 765682844011136709, 2744919478730247485, 9840344475718721857, 35276947157358116641, 126465389887529234853, 453369583483133780781, 1625298268650589782401, 5826580693354037214097, 20887884538471765618501] Using the found enumerative automaton with, 7, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 (2 t + 1) (4 t + 3 t + 2 t + 1) - ---------------------------------- 5 4 3 2 8 t + 6 t + 13 t + 5 t + t - 1 and in Maple notation -(2*t+1)*(4*t^3+3*t^2+2*t+1)/(8*t^5+6*t^4+13*t^3+5*t^2+t-1) This ends this theorem, that took, 0.012, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 49, : Let C(n) be 2 2 2 2 n (x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 257, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 257, 925, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 73606661197199861757816780443770926797560605336911976337432861328125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 257, 925, 3323, 11919, 42729, 153173, 549075, 1968247, 7055601, 25292621, 90668587, 325028895, 1165167673, 4176914949, 14973497539, 53677355783 , 192423938625, 689806295677, 2472835597211, 8864685923631, 31778359388937, 113919900383925, 408383064188083, 1463982386475159, 5248122704415889, 18813608822615213, 67443521618241035, 241773316967213887, 866715370142188761, 3107024142696604261, 11138142181492144547, 39928306173024688935, 143136046198521240033, 513117877643462669789, 1839438515665766891899, 6594067757794661727823, 23638588201825352540585] Using the found enumerative automaton with, 8, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 3 2 (6 t + 2 t + t - 1) (t - 1) - ---------------------------------------------- 6 5 4 3 2 10 t + 10 t - 15 t + 8 t - 10 t + 6 t - 1 and in Maple notation -(6*t^4+2*t^3+t^2-1)*(t-1)/(10*t^6+10*t^5-15*t^4+8*t^3-10*t^2+6*t-1) This ends this theorem, that took, 0.012, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 50, : Let C(n) be 2 2 2 2 n (x y + x y + x + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 51, : Let C(n) be 2 2 2 2 n (x y + x y + x + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 25, 5, 25, 25, 101, 5, 25, 25, 125, 25, 125, 101, 361, 5, 25, 25, 125 , 25, 125, 125, 505, 25, 125, 125, 625, 101, 505, 361, 1205, 5, 25, 25, 125, 25 , 125, 125, 505, 25] Just for kicks C(googol) equals , 84173718217807518475510932258638885361534676121664233505725860595703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 25, 101, 361, 1205, 3865, 12101, 37321, 114005, 346105, 1046501, 3155881 , 9500405, 28566745, 85831301, 257756041, 773792405, 2322425785, 6969374501, 20912317801, 62745342005, 188252803225, 564791964101, 1694443001161, 5083463221205, 15250658099065, 45752511168101, 137258607246121, 411777969222005 , 1235338202633305, 3706023197834501, 11118086773372681, 33354294679856405, 100062952759045945, 300188995716091301, 900567262026180841, 2701702335834356405 , 8105108107014696985, 24315326520067346501, 72945983958248550601] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 6 t - t + 1 - --------------------------- (t - 1) (3 t - 1) (2 t - 1) and in Maple notation -(6*t^2-t+1)/(t-1)/(3*t-1)/(2*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 52, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 17, 5, 25, 17, 61, 5, 25, 25, 85, 17, 85, 61, 217, 5, 25, 25, 85, 25, 125, 85, 305, 17, 85, 85, 289, 61, 305, 217, 773, 5, 25, 25, 85, 25, 125, 85, 305, 25] Just for kicks C(googol) equals , 2876006182730628489294771910743936557642955449409782886505126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 17, 61, 217, 773, 2753, 9805, 34921, 124373, 442961, 1577629, 5618809, 20011685, 71272673, 253841389, 904069513, 3219891317, 11467812977, 40843221565, 145465290649, 518082315077, 1845177526529, 6571697209741, 23405446682281, 83359734466325, 296890096763537, 1057389759223261, 3765949471196857, 13412627932037093, 47769782738504993, 170134604079589165, 605943377715777481, 2158099341306510773, 7686184779351087281, 27374753020666283389, 97496628620701024729, 347239391903435640965, 1236711432951708972353, 4404613082661998198989, 15687262113889412541673] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - -------------- 2 2 t + 3 t - 1 and in Maple notation -(2*t+1)/(2*t^2+3*t-1) This ends this theorem, that took, 0.043, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 53, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 73, 5, 25, 25, 95, 19, 95, 73, 271, 5, 25, 25, 95, 25, 125, 95, 365, 19, 95, 95, 361, 73, 365, 271, 993, 5, 25, 25, 95, 25, 125, 95, 365, 25] Just for kicks C(googol) equals , 122157569584608411577942394826877806529805602622218430042266845703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 73, 271, 993, 3643, 13309, 48655, 177805, 649699, 2374129, 8674999, 31698625, 115826659, 423229469, 1546477863, 5650815053, 20648028987, 75447710625, 275685244735, 1007351382097, 3680852748715, 13449802390397, 49145455249183, 179577045144493, 656172884958899, 2397649735785777, 8760990263119911, 32012578502499905, 116973669824702227, 427420722490916541, 1561791420989452407, 5706771605422366317, 20852491388259007563, 76194813313405705761, 278415152787650743407, 1017326428544807383953, 3717301489711406891547, 13582985733670941871837, 49632105964972931002927] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 (2 t + 4 t + 7 t + 2 t + 1) (t - 1) ---------------------------------------------------------------- 9 8 7 6 5 4 3 2 4 t - 2 t - 14 t - 11 t + 26 t - 7 t - 8 t + t + 4 t - 1 and in Maple notation (2*t^6+4*t^5+7*t^4+2*t+1)*(t-1)/(4*t^9-2*t^8-14*t^7-11*t^6+26*t^5-7*t^4-8*t^3+t ^2+4*t-1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 54, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 71, 5, 25, 25, 95, 19, 95, 71, 265, 5, 25, 25, 95, 25, 125, 95, 355, 19, 95, 95, 361, 71, 355, 265, 989, 5, 25, 25, 95, 25, 125, 95, 355, 25] Just for kicks C(googol) equals , 99372322788069899795871763666960332596065200050361454486846923828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085, 1014133226193379, 3784796725797431, 14125053676996345, 52715417982187949, 196736618251755451, 734231055024833855, 2740187601847579969, 10226519352365486021, 38165889807614364115, 142437039878091970439, 531582269704753517641, 1983892038940922100125, 7403985886058934882859, 27632051505294817431311, 103124220135120334842385] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 ------------ 2 t - 4 t + 1 and in Maple notation (t+1)/(t^2-4*t+1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 55, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 79, 5, 25, 25, 105, 21, 105, 79, 291, 5, 25, 25, 105, 25, 125, 105, 395, 21, 105, 105, 441, 79, 395, 291, 1043, 5, 25, 25, 105, 25, 125, 105, 395, 25] Just for kicks C(googol) equals , 937528355092813527820166281406489661277666527894325554370880126953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 79, 291, 1043, 3725, 13205, 46793, 165471, 585191, 2068315, 7310897, 25837453, 91315845, 322715687, 1140515523, 4030652683, 14244666301, 50341596733 , 177910950553, 628749519543, 2222045769223, 7852864484723, 27752577642625, 98079551982773, 346620034979541, 1224979511026399, 4329163636559235, 15299568080670371, 54069747804545357, 191086283816500421, 675312341587546505, 2386601217999499311, 8434416239961774983, 29807819052431215723, 105342924959126484593, 372289291499985868285, 1315696489718894970501, 4649763752106301098743, 16432591498413660089571] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 (t + 1) (t - t - 1) (t - 1) - ------------------------------------- 2 3 2 (t + t - 1) (11 t - 5 t - 3 t + 1) and in Maple notation -(t+1)*(t^2-t-1)*(t-1)/(t^2+t-1)/(11*t^3-5*t^2-3*t+1) This ends this theorem, that took, 0.011, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 56, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 23, 5, 25, 23, 93, 5, 25, 25, 115, 23, 115, 93, 359, 5, 25, 25, 115, 25, 125, 115, 465, 23, 115, 115, 529, 93, 465, 359, 1335, 5, 25, 25, 115, 25, 125, 115, 465, 25] Just for kicks C(googol) equals , 26558134128640705502054849209301010895619254370103590190410614013671875 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 23, 93, 359, 1335, 4873, 17535, 62601, 222181, 785855, 2772717, 9768351, 34378167, 120910529, 425062511, 1493898001, 5249371781, 18443445415, 64795091709, 227625068503, 799619495287, 2808906276921, 9866994688223, 34659998140825, 121750158651877, 427670046315727, 1502266603229837, 5276968090316303, 18536231688176247, 65111563355182961, 228714972444138447, 803398355314114081, 2822066188904200197, 9912960994888418871, 34820865610603543453, 122313869380578977351, 429647047899475776183, 1509204004977591524457, 5301320480826218438463, 18621736010632618462889] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 (t - t + t + 1) (t - 1) ------------------------------- 5 4 3 17 t + 15 t - 24 t + 5 t - 1 and in Maple notation (t^3-t^2+t+1)*(t-1)/(17*t^5+15*t^4-24*t^3+5*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 57, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + x y + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 15, 5, 25, 15, 51, 5, 25, 25, 75, 15, 75, 51, 153, 5, 25, 25, 75, 25, 125, 75, 255, 15, 75, 75, 225, 51, 255, 153, 477, 5, 25, 25, 75, 25, 125, 75, 255, 25] Just for kicks C(googol) equals , 18991393753931390188376409637260877616427023895084857940673828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 15, 51, 153, 477, 1431, 4347, 13041, 39285, 117855, 354051, 1062153, 3187917, 9563751, 28695627, 86086881, 258273765, 774821295, 2324503251, 6973509753, 20920647357, 62761942071, 188286180507, 564858541521, 1694576687445 , 5083730062335, 15251193375651, 45753580126953, 137260749946797, 411782249840391, 1235346778218987, 3706040334656961, 11118121090064325, 33354363270192975, 100063090068859251, 300189270206577753, 900567811394574237, 2701703434183722711, 8105110304875691067, 24315330914627073201] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (3 t + 1) (t - 1) - -------------------- 2 (3 t - 1) (3 t - 1) and in Maple notation -(3*t+1)*(t-1)/(3*t-1)/(3*t^2-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 58, : Let C(n) be 2 2 2 2 n (x y + x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 77, 5, 25, 25, 105, 21, 105, 77, 277, 5, 25, 25, 105, 25, 125, 105, 385, 21, 105, 105, 441, 77, 385, 277, 1005, 5, 25, 25, 105, 25, 125, 105, 385, 25] Just for kicks C(googol) equals , 666077427771157268916900567838401442979245985043235123157501220703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 77, 277, 1005, 3669, 13421, 49109, 179693, 657493, 2405741, 8802517, 32208109, 117848405, 431203437, 1577759189, 5772968941, 21123103317, 77288739693, 282796954325, 1034744746733, 3786096966485, 13853204169325, 50688418034645, 185467253023213, 678618573585493, 2483043021926765, 9085372680213717, 33243079563849965, 121635333825674581, 445059682466127981, 1628458727631992277, 5958476878665891821, 21801870757402017365, 79772327425542705005, 291884320097964988117, 1067995118966176780013, 3907758983945264791381, 14298361486321123229293, 52317234003798734518229] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 2 t + t - 1 - ------------------------------- 3 2 (4 t - 6 t + 5 t - 1) (t - 1) and in Maple notation -(4*t^3-2*t^2+t-1)/(4*t^3-6*t^2+5*t-1)/(t-1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 59, : Let C(n) be 2 2 2 2 n (x y + x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 77, 5, 25, 25, 105, 21, 105, 77, 277, 5, 25, 25, 105, 25, 125, 105, 385, 21, 105, 105, 441, 77, 385, 277, 1005, 5, 25, 25, 105, 25, 125, 105, 385, 25] Just for kicks C(googol) equals , 666077427771157268916900567838401442979245985043235123157501220703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 77, 277, 1005, 3669, 13421, 49109, 179693, 657493, 2405741, 8802517, 32208109, 117848405, 431203437, 1577759189, 5772968941, 21123103317, 77288739693, 282796954325, 1034744746733, 3786096966485, 13853204169325, 50688418034645, 185467253023213, 678618573585493, 2483043021926765, 9085372680213717, 33243079563849965, 121635333825674581, 445059682466127981, 1628458727631992277, 5958476878665891821, 21801870757402017365, 79772327425542705005, 291884320097964988117, 1067995118966176780013, 3907758983945264791381, 14298361486321123229293, 52317234003798734518229] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 2 t + t - 1 - ------------------------------- 3 2 (4 t - 6 t + 5 t - 1) (t - 1) and in Maple notation -(4*t^3-2*t^2+t-1)/(4*t^3-6*t^2+5*t-1)/(t-1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 60, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 73, 5, 25, 25, 95, 19, 95, 73, 271, 5, 25, 25, 95, 25, 125, 95, 365, 19, 95, 95, 361, 73, 365, 271, 993, 5, 25, 25, 95, 25, 125, 95, 365, 25] Just for kicks C(googol) equals , 122157569584608411577942394826877806529805602622218430042266845703125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 73, 271, 993, 3643, 13309, 48655, 177805, 649699, 2374129, 8674999, 31698625, 115826659, 423229469, 1546477863, 5650815053, 20648028987, 75447710625, 275685244735, 1007351382097, 3680852748715, 13449802390397, 49145455249183, 179577045144493, 656172884958899, 2397649735785777, 8760990263119911, 32012578502499905, 116973669824702227, 427420722490916541, 1561791420989452407, 5706771605422366317, 20852491388259007563, 76194813313405705761, 278415152787650743407, 1017326428544807383953, 3717301489711406891547, 13582985733670941871837, 49632105964972931002927] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 (2 t + 4 t + 7 t + 2 t + 1) (t - 1) ---------------------------------------------------------------- 9 8 7 6 5 4 3 2 4 t - 2 t - 14 t - 11 t + 26 t - 7 t - 8 t + t + 4 t - 1 and in Maple notation (2*t^6+4*t^5+7*t^4+2*t+1)*(t-1)/(4*t^9-2*t^8-14*t^7-11*t^6+26*t^5-7*t^4-8*t^3+t ^2+4*t-1) This ends this theorem, that took, 0.014, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 61, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 19, 5, 25, 19, 73, 5, 25, 25, 95, 19, 95, 73, 263, 5, 25, 25, 95, 25, 125, 95, 365, 19, 95, 95, 361, 73, 365, 263, 973, 5, 25, 25, 95, 25, 125, 95, 365, 25] Just for kicks C(googol) equals , 107067836886053863857891855428572931230758040328510105609893798828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 19, 73, 263, 973, 3531, 12965, 47251, 173009, 631631, 2310273, 8440503, 30859717, 112776995, 412263309, 1506786763, 5507802233, 20131419959, 73585107241, 268963530095, 983115443149, 3593443967243, 13134702768181, 48009561539555, 175483552150433, 641422737413919, 2344513580442321, 8569606120531879, 31323418577252629, 114492577146357555, 418490478101645149, 1529656037199376891, 5591161254718610057, 20436674729055515463, 74699631405808296057, 273040253115772225503, 998010019384523821213, 3647901659180242072987, 13333720414125860919557, 48737086752902132101907] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 2 8 t - 4 t + 14 t - 12 t + 9 t - 5 t - t + 1 - ------------------------------------------------------------- 8 7 6 5 4 3 2 8 t - 20 t + 46 t - 56 t + 51 t - 20 t - 6 t + 6 t - 1 and in Maple notation -(8*t^7-4*t^6+14*t^5-12*t^4+9*t^3-5*t^2-t+1)/(8*t^8-20*t^7+46*t^6-56*t^5+51*t^4 -20*t^3-6*t^2+6*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 62, : Let C(n) be 2 2 2 2 n (x y + x y + x y + x y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 21, 5, 25, 21, 85, 5, 25, 25, 105, 21, 105, 85, 333, 5, 25, 25, 105, 25, 125, 105, 425, 21, 105, 105, 441, 85, 425, 333, 1293, 5, 25, 25, 105, 25, 125, 105, 425, 25] Just for kicks C(googol) equals , 3926896116798394746365847178957652259388169113663025200366973876953125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 21, 85, 333, 1293, 4997, 19269, 74237, 285885, 1100725, 4237685, 16314029, 62803821, 241772773, 930737317, 3582994525, 13793193757, 53098638101, 204409869525, 786901384973, 3029275306957, 11661574029893, 44892686930437, 172820009444029, 665292226131325, 2561125574008437, 9859372988945461, 37954888549712749, 146112087082712109, 562476740332711845, 2165324510303144805, 8335687324816740637, 32089270151577563101, 123531655967252719061, 475550548638896318357, 1830691271317993814989, 7047474848829240197261, 27130135224341347578117, 104440846271811484702917, 402058090745726375080829] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t + 1 -------------------------- 4 3 2 8 t + 4 t - t - 4 t + 1 and in Maple notation (t+1)/(8*t^4+4*t^3-t^2-4*t+1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 63, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 5, 5, 15, 5, 25, 15, 51, 5, 25, 25, 75, 15, 75, 51, 153, 5, 25, 25, 75, 25, 125, 75, 255, 15, 75, 75, 225, 51, 255, 153, 477, 5, 25, 25, 75, 25, 125, 75, 255, 25] Just for kicks C(googol) equals , 18991393753931390188376409637260877616427023895084857940673828125 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 5, 15, 51, 153, 477, 1431, 4347, 13041, 39285, 117855, 354051, 1062153, 3187917, 9563751, 28695627, 86086881, 258273765, 774821295, 2324503251, 6973509753, 20920647357, 62761942071, 188286180507, 564858541521, 1694576687445 , 5083730062335, 15251193375651, 45753580126953, 137260749946797, 411782249840391, 1235346778218987, 3706040334656961, 11118121090064325, 33354363270192975, 100063090068859251, 300189270206577753, 900567811394574237, 2701703434183722711, 8105110304875691067, 24315330914627073201] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (3 t + 1) (t - 1) - -------------------- 2 (3 t - 1) (3 t - 1) and in Maple notation -(3*t+1)*(t-1)/(3*t-1)/(3*t^2-1) This ends this theorem, that took, 0.003, seconds. ------------------------------------------------------------------ This concludes this webbook, that took, 0.966, seconds. to generate. k is , 6 n On Sequences of the form, P(x, y) , modolu , 2, Evaluated at, {x = 1, y = 1} For all Polynomials that are Sums of Monomials taken from, 2 2 2 2 2 2 {1, x, y, x , y , x y, x y , x y, x y } By Shalosh B. Ekhad In this webbook, we will consider the sequences described in the title, that\ after normalization and weeding out obvious symmetry, concerns the following set of, 46, polynomials 2 2 2 2 {x + x y + y + x + y + 1, x y + x y + y + x + y + 1, 2 2 2 2 2 x y + x + y + x + y + 1, x y + x + x y + x + y + 1, 2 2 2 2 2 2 x y + x + x y + y + y + 1, x y + x + x y + y + x + 1, 2 2 2 2 2 2 x y + x + x y + y + x + y, x y + x y + y + x + y + 1, 2 2 2 2 2 x y + x y + x y + x + y + 1, x y + x y + x y + y + y + 1, 2 2 2 2 2 2 x y + x y + x y + y + x + 1, x y + x y + x y + y + x + y, 2 2 2 2 2 2 2 2 x y + x y + x + y + y + 1, x y + x y + x + y + x + y, 2 2 2 2 2 2 2 2 x y + x y + x + x y + y + 1, x y + x y + x + x y + y + y, 2 2 2 2 2 2 2 x y + x y + y + x + y + 1, x y + x + y + x + y + 1, 2 2 2 2 2 2 2 2 x y + x + x y + y + y + 1, x y + x + x y + y + x + y, 2 2 2 2 2 2 2 x y + x y + y + x + y + 1, x y + x y + x y + x + y + 1, 2 2 2 2 2 2 2 2 x y + x y + x y + y + y + 1, x y + x y + x y + y + x + 1, 2 2 2 2 2 2 2 2 x y + x y + x y + y + x + y, x y + x y + x + x + y + 1, 2 2 2 2 2 2 2 2 2 2 x y + x y + x + y + y + 1, x y + x y + x + y + x + 1, 2 2 2 2 2 2 2 2 2 x y + x y + x + y + x + y, x y + x y + x + x y + y + 1, 2 2 2 2 2 2 2 2 x y + x y + x + x y + x + 1, x y + x y + x + x y + x + y, 2 2 2 2 2 2 2 2 2 2 x y + x y + x + x y + y + 1, x y + x y + x + x y + y + y, 2 2 2 2 2 2 2 2 2 x y + x y + x + x y + y + x, x y + x y + x y + x + y + 1, 2 2 2 2 2 2 2 2 2 2 x y + x y + x y + y + y + 1, x y + x y + x y + y + x + 1, 2 2 2 2 2 2 2 2 2 x y + x y + x y + y + x + y, x y + x y + x y + x y + y + 1, 2 2 2 2 2 2 2 2 2 x y + x y + x y + x y + x + y, x y + x y + x y + x y + y + 1, 2 2 2 2 2 2 2 2 2 2 2 x y + x y + x y + x y + y + x, x y + x y + x y + x + y + 1, 2 2 2 2 2 2 2 2 2 2 2 2 x y + x y + x y + x + y + y, x y + x y + x y + x + x y + y } by finding enumerative automata with at most, 200, states . ----------------------------------------------------------------------------\ ---- Theorem Number, 1, : Let C(n) be 2 2 n (x + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 90, 6, 36, 36, 132, 22, 132, 90, 358, 6, 36, 36, 132, 36, 216, 132, 540, 22, 132, 132, 484, 90, 540, 358, 1434, 6, 36, 36, 132, 36, 216, 132, 540, 36] Just for kicks C(googol) equals , 14056209326117951659557069010655383121710308703415637491300865933312000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 90, 358, 1434, 5734, 22938, 91750, 367002, 1468006, 5872026, 23488102, 93952410, 375809638, 1503238554, 6012954214, 24051816858, 96207267430 , 384829069722, 1539316278886, 6157265115546, 24629060462182, 98516241848730, 394064967394918, 1576259869579674, 6305039478318694, 25220157913274778, 100880631653099110, 403522526612396442, 1614090106449585766, 6456360425798343066, 25825441703193372262, 103301766812773489050, 413207067251093956198, 1652828269004375824794, 6611313076017503299174, 26445252304070013196698, 105781009216280052786790, 423124036865120211147162, 1692496147460480844588646] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 - ----------------- (t + 1) (4 t - 1) and in Maple notation -(3*t+1)/(t+1)/(4*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 2, : Let C(n) be 2 2 n (x y + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 20, 6, 36, 20, 88, 6, 36, 36, 120, 20, 120, 88, 336, 6, 36, 36, 120, 36, 216, 120, 528, 20, 120, 120, 400, 88, 528, 336, 1376, 6, 36, 36, 120, 36, 216, 120, 528, 36] Just for kicks C(googol) equals , 2737767495721500889251655914073374844297767225771330581299200000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 20, 88, 336, 1376, 5440, 21888, 87296, 349696, 1397760, 5593088, 22368256, 89481216, 357908480, 1431666688, 5726601216, 22906535936, 91625881600 , 366504050688, 1466015154176, 5864062713856, 23456246661120, 93824995033088, 375299963355136, 1501199886974976, 6004799480791040, 24019198057381888, 96076791961092096, 384307168381239296, 1537228672451215360, 6148914691952345088 , 24595658763514413056, 98382635062647586816, 393530540233410478080, 1574122160968001650688, 6296488643803287126016, 25185954575350587457536, 100743818301127471923200, 402975273205059643506688, 1611901092819139062398976] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 t + 1 - ------------------- (2 t + 1) (4 t - 1) and in Maple notation -(4*t+1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.005, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 3, : Let C(n) be 2 2 2 n (x y + x + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 20, 6, 36, 20, 84, 6, 36, 36, 120, 20, 120, 84, 296, 6, 36, 36, 120, 36, 216, 120, 504, 20, 120, 120, 400, 84, 504, 296, 1124, 6, 36, 36, 120, 36, 216, 120, 504, 36] Just for kicks C(googol) equals , 799605762186889740692889634076881341731181793016126185144320000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 20, 84, 296, 1124, 4056, 15100, 55288, 204740, 753336, 2782588, 10252952 , 37831684, 139470616, 514443580, 1896954168, 6996160196, 25799583352, 95147186940, 350881763288, 1294007162500, 4772059430872, 17598637988284, 64900766401656, 239343741625668, 882659972822840, 3255107458681212, 12004301666699032, 44269913022747396, 163260197361058968, 602077001213815356, 2220361636142223288, 8188331534303637444, 30197229928670023672, 111362456662997249020, 410686563525906396760, 1514545026074115320708, 5585394869142171484504, 20598024753141397814972, 75962153600605611299576] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 8 7 6 5 4 3 2 8 t + 16 t - 20 t + 34 t - 2 t - 15 t + 9 t - t - 1 - ------------------------------------------------------------------- 9 8 7 6 5 4 3 2 16 t + 24 t - 40 t + 56 t + 2 t - 44 t + 25 t + t - 5 t + 1 and in Maple notation -(8*t^8+16*t^7-20*t^6+34*t^5-2*t^4-15*t^3+9*t^2-t-1)/(16*t^9+24*t^8-40*t^7+56*t ^6+2*t^5-44*t^4+25*t^3+t^2-5*t+1) This ends this theorem, that took, 0.017, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 4, : Let C(n) be 2 2 n (x y + x + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 94, 6, 36, 36, 144, 24, 144, 94, 356, 6, 36, 36, 144, 36, 216, 144, 564, 24, 144, 144, 576, 94, 564, 356, 1338, 6, 36, 36, 144, 36, 216, 144, 564, 36] Just for kicks C(googol) equals , 42757024268292254411190836399528995947490826964920299882746213199528329216 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 94, 356, 1338, 5036, 18942, 71264, 268114, 1008728, 3795162, 14278632, 53720914, 202115688, 760425734, 2860971568, 10763916406, 40497394188, 152364521390, 573245456832, 2156738015294, 8114358001856, 30528884544598, 114859708062884, 432140012215626, 1625852906053008, 6116993561285206, 23014142354936060, 86586775518891606, 325767937779248176, 1225646164201790634, 4611284124710132760, 17349168055096974798, 65273278345832840520, 245579548972188856786, 923951062391070531508, 3476207889730692435414, 13078637802908965906948, 49206138472038541615246, 185129682449871663049116] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 8 7 6 5 4 3 2 (2 t + 1) (6 t - 7 t + 2 t - 7 t - 5 t - t - 2 t + t + 1) ------------------------------------------------------------------------ 10 9 8 7 6 5 4 3 2 8 t + 8 t + 4 t - 15 t - 2 t + 7 t + 9 t + 9 t - 6 t - 3 t + 1 and in Maple notation (2*t+1)*(6*t^8-7*t^7+2*t^6-7*t^5-5*t^4-t^3-2*t^2+t+1)/(8*t^10+8*t^9+4*t^8-15*t^ 7-2*t^6+7*t^5+9*t^4+9*t^3-6*t^2-3*t+1) This ends this theorem, that took, 0.017, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 5, : Let C(n) be 2 2 2 n (x y + x + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 26, 6, 36, 26, 110, 6, 36, 36, 156, 26, 156, 110, 450, 6, 36, 36, 156 , 36, 216, 156, 660, 26, 156, 156, 676, 110, 660, 450, 1822, 6, 36, 36, 156, 36 , 216, 156, 660, 36] Just for kicks C(googol) equals , 134039144168938693310529048640833965397591\ 6267283481061109845196800000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 26, 110, 450, 1822, 7330, 29406, 117794, 471518, 1886754, 7548382, 30196258, 120790494, 483172898, 1932713438, 7730897442, 30923677150, 123694883362, 494779882974, 1979120230946, 7916482321886, 31665932083746, 126663733927390, 506654946894370, 2026619809947102, 8106479284527650, 32425917227589086, 129703669089313314, 518814676715167198, 2075258707576496674, 8301034831737642462, 33204139329813881378, 132816557324982148574, 531266229311381840418, 2125064917268433853918, 8500259669119548400162, 34001038676569819569630, 136004154706462530216482, 544016618826216624741854, 2176066475305599506719266] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (2 t + 1) (t - 1) - --------------------------- (4 t - 1) (2 t - 1) (t + 1) and in Maple notation -(2*t+1)*(t-1)/(4*t-1)/(2*t-1)/(t+1) This ends this theorem, that took, 0.049, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 6, : Let C(n) be 2 2 2 n (x y + x + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 82, 6, 36, 36, 132, 22, 132, 82, 302, 6, 36, 36, 132, 36, 216, 132, 492, 22, 132, 132, 484, 82, 492, 302, 1106, 6, 36, 36, 132, 36, 216, 132, 492, 36] Just for kicks C(googol) equals , 2413602479613831581859142085056807815258942153585992997640886289759731712 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 82, 302, 1106, 4066, 14902, 54678, 200578, 735770, 2699182, 9901550, 36323050, 133247570, 488805718, 1793137798, 6577952882, 24130592458, 88520767614, 324729961566, 1191240790586, 4369952806274, 16030753627238, 58807285300086, 215728897446594, 791380812129402, 2903104763095054, 10649761956573134, 39067632408628746, 143315870180817330, 525740552457316726, 1928628896079822694, 7074990508936834578, 25953925508046213930, 95209491578317150878, 349266906996069982270, 1281252218664080409818, 4700151130694245039842, 17242054553786609505670, 63250826828592359807062] Using the found enumerative automaton with, 7, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 5 3 2 (2 t + 1) (2 t + t - t + t + 1) - ------------------------------------------------ 7 6 5 4 3 2 8 t - 8 t + 10 t - 6 t - t + 3 t + 3 t - 1 and in Maple notation -(2*t+1)*(2*t^5+t^3-t^2+t+1)/(8*t^7-8*t^6+10*t^5-6*t^4-t^3+3*t^2+3*t-1) This ends this theorem, that took, 0.013, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 7, : Let C(n) be 2 2 2 n (x y + x + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 100, 6, 36, 36, 144, 24, 144, 100, 396, 6, 36, 36, 144 , 36, 216, 144, 600, 24, 144, 144, 576, 100, 600, 396, 1596, 6, 36, 36, 144, 36 , 216, 144, 600, 36] Just for kicks C(googol) equals , 137888714657610667256694059187392359170292799328425980226174976000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 100, 396, 1596, 6364, 25500, 101916, 407836, 1631004, 6524700, 26097436, 104392476, 417564444, 1670268700, 6681052956, 26724255516, 106896934684, 427587913500, 1710351304476, 6841405916956, 27365622269724, 109462491875100, 437849961907996, 1751399858816796, 7005599412897564, 28022397696329500, 112089590695839516, 448358362962315036, 1793433451491346204, 7173733806681212700, 28694935225293195036, 114779740904036091676, 459118963610417743644, 1836475854453124220700, 7345903417789590390556, 29383613671204174546716, 117534454684725072217884, 470137818739083540809500, 1880551274955967659362076] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 8, : Let C(n) be 2 2 2 n (x y + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 88, 6, 36, 36, 144, 24, 144, 88, 336, 6, 36, 36, 144, 36, 216, 144, 528, 24, 144, 144, 576, 88, 528, 336, 1280, 6, 36, 36, 144, 36, 216, 144, 528, 36] Just for kicks C(googol) equals , 22961890025021468049088246083148323519502246213219980004482861654815539200 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 88, 336, 1280, 4928, 19072, 74240, 290304, 1139712, 4489216, 17731584, 70197248, 278429696, 1106083840, 4399628288, 17518559232, 69815500800 , 278424715264, 1110989340672, 4435189170176, 17712382214144, 70757707153408, 282733687341056, 1129973180006400, 4516781016219648, 18057054379835392, 72195631334031360, 288677074225332224, 1154367049938501632, 4616363901385179136 , 18461882020951752704, 73835963721153773568, 295306431820952764416, 1181104623705873448960, 4724026595412969259008, 18894838168519077527552, 75575248650168612945920, 302287713700327859421184, 1209107876904987464302592] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 16 t + 8 t - 1 - -------------------------- 2 (4 t - 1) (4 t + 2 t - 1) and in Maple notation -(16*t^3+8*t^2-1)/(4*t-1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 9, : Let C(n) be 2 2 n (x y + x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 100, 6, 36, 36, 144, 24, 144, 100, 396, 6, 36, 36, 144 , 36, 216, 144, 600, 24, 144, 144, 576, 100, 600, 396, 1596, 6, 36, 36, 144, 36 , 216, 144, 600, 36] Just for kicks C(googol) equals , 137888714657610667256694059187392359170292799328425980226174976000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 100, 396, 1596, 6364, 25500, 101916, 407836, 1631004, 6524700, 26097436, 104392476, 417564444, 1670268700, 6681052956, 26724255516, 106896934684, 427587913500, 1710351304476, 6841405916956, 27365622269724, 109462491875100, 437849961907996, 1751399858816796, 7005599412897564, 28022397696329500, 112089590695839516, 448358362962315036, 1793433451491346204, 7173733806681212700, 28694935225293195036, 114779740904036091676, 459118963610417743644, 1836475854453124220700, 7345903417789590390556, 29383613671204174546716, 117534454684725072217884, 470137818739083540809500, 1880551274955967659362076] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 10, : Let C(n) be 2 2 2 n (x y + x y + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 94, 6, 36, 36, 144, 24, 144, 94, 356, 6, 36, 36, 144, 36, 216, 144, 564, 24, 144, 144, 576, 94, 564, 356, 1338, 6, 36, 36, 144, 36, 216, 144, 564, 36] Just for kicks C(googol) equals , 42757024268292254411190836399528995947490826964920299882746213199528329216 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 94, 356, 1338, 5036, 18942, 71264, 268114, 1008728, 3795162, 14278632, 53720914, 202115688, 760425734, 2860971568, 10763916406, 40497394188, 152364521390, 573245456832, 2156738015294, 8114358001856, 30528884544598, 114859708062884, 432140012215626, 1625852906053008, 6116993561285206, 23014142354936060, 86586775518891606, 325767937779248176, 1225646164201790634, 4611284124710132760, 17349168055096974798, 65273278345832840520, 245579548972188856786, 923951062391070531508, 3476207889730692435414, 13078637802908965906948, 49206138472038541615246, 185129682449871663049116] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 8 7 6 5 4 3 2 (2 t + 1) (6 t - 7 t + 2 t - 7 t - 5 t - t - 2 t + t + 1) ------------------------------------------------------------------------ 10 9 8 7 6 5 4 3 2 8 t + 8 t + 4 t - 15 t - 2 t + 7 t + 9 t + 9 t - 6 t - 3 t + 1 and in Maple notation (2*t+1)*(6*t^8-7*t^7+2*t^6-7*t^5-5*t^4-t^3-2*t^2+t+1)/(8*t^10+8*t^9+4*t^8-15*t^ 7-2*t^6+7*t^5+9*t^4+9*t^3-6*t^2-3*t+1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 11, : Let C(n) be 2 2 2 n (x y + x y + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 28, 6, 36, 28, 112, 6, 36, 36, 168, 28, 168, 112, 456, 6, 36, 36, 168 , 36, 216, 168, 672, 28, 168, 168, 784, 112, 672, 456, 1816, 6, 36, 36, 168, 36 , 216, 168, 672, 36] Just for kicks C(googol) equals , 396943363464639408218200560699216267233418\ 0368609654505818010397006696022016 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 28, 112, 456, 1816, 7288, 29112, 116536, 465976, 1864248, 7456312, 29826616, 119303736, 477220408, 1908870712, 7635504696, 30541975096, 122167987768, 488671776312, 1954687454776, 7818749120056, 31274997878328, 125099988717112, 500399960460856, 2001599830658616, 8006399345004088, 32025597335277112, 128102389430586936, 512409557543390776, 2049638230531477048, 8198552921410080312, 32794211687071977016, 131176846745424596536, 524707386987425009208, 2098829547938246790712, 8395318191775893655096, 33581272767057761635896, 134325091068322672512568, 537300364273107438112312, 2149201457092796256325176] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 4 t + 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (4*t^2+3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 12, : Let C(n) be 2 2 2 n (x y + x y + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 100, 6, 36, 36, 144, 24, 144, 100, 396, 6, 36, 36, 144 , 36, 216, 144, 600, 24, 144, 144, 576, 100, 600, 396, 1596, 6, 36, 36, 144, 36 , 216, 144, 600, 36] Just for kicks C(googol) equals , 137888714657610667256694059187392359170292799328425980226174976000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 100, 396, 1596, 6364, 25500, 101916, 407836, 1631004, 6524700, 26097436, 104392476, 417564444, 1670268700, 6681052956, 26724255516, 106896934684, 427587913500, 1710351304476, 6841405916956, 27365622269724, 109462491875100, 437849961907996, 1751399858816796, 7005599412897564, 28022397696329500, 112089590695839516, 448358362962315036, 1793433451491346204, 7173733806681212700, 28694935225293195036, 114779740904036091676, 459118963610417743644, 1836475854453124220700, 7345903417789590390556, 29383613671204174546716, 117534454684725072217884, 470137818739083540809500, 1880551274955967659362076] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 13, : Let C(n) be 2 2 2 2 n (x y + x y + x + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 96, 6, 36, 36, 144, 24, 144, 96, 372, 6, 36, 36, 144, 36, 216, 144, 576, 24, 144, 144, 576, 96, 576, 372, 1416, 6, 36, 36, 144, 36, 216, 144, 576, 36] Just for kicks C(googol) equals , 62910545093793295945702128611066193383749523036605062293362318822437027840 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 96, 372, 1416, 5340, 20040, 75012, 280392, 1047324, 3910440, 14597508, 54485736, 203357724, 758969736, 2832570372, 10571410056, 39453266460, 147242049000, 549515715972, 2050822387752, 7653776980764, 28564291826760, 106603402909188, 397849344975816, 1484794027325724, 5541326864990376, 20680513633962372, 77180728073512296, 288042399465393180, 1074988871398673160, 4011913089350524932, 14972663492445877512, 55878740893317887004, 208542300106595474280, 778290459584603617668, 2904619538334898211496, 10840187693961147658524, 40456131237922009283016, 150984337258551523194372] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 3 t + 1 - ------------------------ 2 (t - 4 t + 1) (2 t - 1) and in Maple notation -(4*t^3-3*t^2+1)/(t^2-4*t+1)/(2*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 14, : Let C(n) be 2 2 2 2 n (x y + x y + x + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 96, 6, 36, 36, 144, 24, 144, 96, 384, 6, 36, 36, 144, 36, 216, 144, 576, 24, 144, 144, 576, 96, 576, 384, 1536, 6, 36, 36, 144, 36, 216, 144, 576, 36] Just for kicks C(googol) equals , 90754024259153226481010437249434699795319756525164263594879310578514395136 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 96, 384, 1536, 6144, 24576, 98304, 393216, 1572864, 6291456, 25165824, 100663296, 402653184, 1610612736, 6442450944, 25769803776, 103079215104, 412316860416, 1649267441664, 6597069766656, 26388279066624, 105553116266496, 422212465065984, 1688849860263936, 6755399441055744, 27021597764222976, 108086391056891904, 432345564227567616, 1729382256910270464, 6917529027641081856, 27670116110564327424, 110680464442257309696, 442721857769029238784, 1770887431076116955136, 7083549724304467820544, 28334198897217871282176, 113336795588871485128704, 453347182355485940514816, 1813388729421943762059264] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - ------- 4 t - 1 and in Maple notation -(2*t+1)/(4*t-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 15, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 30, 6, 36, 30, 138, 6, 36, 36, 180, 30, 180, 138, 606, 6, 36, 36, 180 , 36, 216, 180, 828, 30, 180, 180, 900, 138, 828, 606, 2586, 6, 36, 36, 180, 36 , 216, 180, 828, 36] Just for kicks C(googol) equals , 181249926865339492095806851390538971660724\ 679113064023923988889600000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 30, 138, 606, 2586, 10830, 44778, 183486, 747066, 3027630, 12228618, 49268766, 198137946, 795740430, 3192527658, 12798808446, 51281327226, 205383589230, 822309197898, 3291561314526, 13173218826906, 52713796014030, 210917946175338, 843860071059006, 3376005143308986, 13505715150454830, 54027944333475978, 216127028528873886, 864553867700405466, 3458352731556351630, 13833822708489595818, 55336526180750951166, 221349810763381508346, 885410361174659144430, 3541674799062035910858, 14166799259338341642846, 56667497226623960569626, 226670889474307624273230, 906686259600665843077098, 3626753143512969410260926] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (4 t - 1) (3 t - 1) and in Maple notation -(t-1)/(4*t-1)/(3*t-1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 16, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 100, 6, 36, 36, 144, 24, 144, 100, 396, 6, 36, 36, 144 , 36, 216, 144, 600, 24, 144, 144, 576, 100, 600, 396, 1596, 6, 36, 36, 144, 36 , 216, 144, 600, 36] Just for kicks C(googol) equals , 137888714657610667256694059187392359170292799328425980226174976000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 100, 396, 1596, 6364, 25500, 101916, 407836, 1631004, 6524700, 26097436, 104392476, 417564444, 1670268700, 6681052956, 26724255516, 106896934684, 427587913500, 1710351304476, 6841405916956, 27365622269724, 109462491875100, 437849961907996, 1751399858816796, 7005599412897564, 28022397696329500, 112089590695839516, 448358362962315036, 1793433451491346204, 7173733806681212700, 28694935225293195036, 114779740904036091676, 459118963610417743644, 1836475854453124220700, 7345903417789590390556, 29383613671204174546716, 117534454684725072217884, 470137818739083540809500, 1880551274955967659362076] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 17, : Let C(n) be 2 2 2 n (x y + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 26, 6, 36, 26, 110, 6, 36, 36, 156, 26, 156, 110, 450, 6, 36, 36, 156 , 36, 216, 156, 660, 26, 156, 156, 676, 110, 660, 450, 1822, 6, 36, 36, 156, 36 , 216, 156, 660, 36] Just for kicks C(googol) equals , 134039144168938693310529048640833965397591\ 6267283481061109845196800000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 26, 110, 450, 1822, 7330, 29406, 117794, 471518, 1886754, 7548382, 30196258, 120790494, 483172898, 1932713438, 7730897442, 30923677150, 123694883362, 494779882974, 1979120230946, 7916482321886, 31665932083746, 126663733927390, 506654946894370, 2026619809947102, 8106479284527650, 32425917227589086, 129703669089313314, 518814676715167198, 2075258707576496674, 8301034831737642462, 33204139329813881378, 132816557324982148574, 531266229311381840418, 2125064917268433853918, 8500259669119548400162, 34001038676569819569630, 136004154706462530216482, 544016618826216624741854, 2176066475305599506719266] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (2 t + 1) (t - 1) - --------------------------- (4 t - 1) (2 t - 1) (t + 1) and in Maple notation -(2*t+1)*(t-1)/(4*t-1)/(2*t-1)/(t+1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 18, : Let C(n) be 2 2 2 2 n (x y + x + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 82, 6, 36, 36, 132, 22, 132, 82, 306, 6, 36, 36, 132, 36, 216, 132, 492, 22, 132, 132, 484, 82, 492, 306, 1142, 6, 36, 36, 132, 36, 216, 132, 492, 36] Just for kicks C(googol) equals , 2761456586817425003071513463990316124228320685072434853691153738366976000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 82, 306, 1142, 4250, 15806, 58730, 218142, 810074, 3007886, 11167914 , 41463774, 153942330, 571535534, 2121906442, 7877856190, 29247532442, 108585059214, 403135208426, 1496688037406, 5556634000122, 20629669372526, 76590116106186, 284349965403518, 1055683240842074, 3919350217589838, 14551056119534250, 54022534883643358, 200565116973926074, 744621965898163502, 2764498035965846666, 10263529334107729214, 38104579208915103002, 141467804043378683150, 525216128767713836394, 1949927644526031082910, 7239339408300824190586, 26876912697517794694126, 99783750340591693583690] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 5 4 3 2 (t - 1) (4 t - 2 t - 6 t - 3 t + 2 t + 1) - ---------------------------------------------- 5 4 2 (t + 1) (4 t - 2 t - t + 4 t - 1) (2 t - 1) and in Maple notation -(t-1)*(4*t^5-2*t^4-6*t^3-3*t^2+2*t+1)/(t+1)/(4*t^5-2*t^4-t^2+4*t-1)/(2*t-1) This ends this theorem, that took, 0.019, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 19, : Let C(n) be 2 2 2 2 n (x y + x + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 92, 6, 36, 36, 144, 24, 144, 92, 340, 6, 36, 36, 144, 36, 216, 144, 552, 24, 144, 144, 576, 92, 552, 340, 1236, 6, 36, 36, 144, 36, 216, 144, 552, 36] Just for kicks C(googol) equals , 25859451075391164324299350593844150774718594295940830824517630860407603200 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 92, 340, 1236, 4452, 15956, 57028, 203508, 725604, 2585876, 9212932, 32818740, 116898468, 416365652, 1482959428, 5281740660, 18811402980, 66998214548, 238618498180, 849854020788, 3026803253028, 10780126189268, 38394001851076, 136742291486196, 487014945269604, 1734529552998932, 6177618817971460, 22001916096783156, 78360987000034212, 279086795341152596, 993982364318493508, 3540120692227720308, 12608326822500017124, 44905221886315230356, 159932319372665202052, 569607402028065020340, 2028686845104403372068, 7225275339919095970772, 25733199711065606284228] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (2 t + 1) (t - 1) - -------------------------- 2 (2 t + 3 t - 1) (2 t - 1) and in Maple notation -(2*t+1)*(t-1)/(2*t^2+3*t-1)/(2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 20, : Let C(n) be 2 2 2 2 n (x y + x + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 30, 6, 36, 30, 138, 6, 36, 36, 180, 30, 180, 138, 606, 6, 36, 36, 180 , 36, 216, 180, 828, 30, 180, 180, 900, 138, 828, 606, 2586, 6, 36, 36, 180, 36 , 216, 180, 828, 36] Just for kicks C(googol) equals , 181249926865339492095806851390538971660724\ 679113064023923988889600000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 30, 138, 606, 2586, 10830, 44778, 183486, 747066, 3027630, 12228618, 49268766, 198137946, 795740430, 3192527658, 12798808446, 51281327226, 205383589230, 822309197898, 3291561314526, 13173218826906, 52713796014030, 210917946175338, 843860071059006, 3376005143308986, 13505715150454830, 54027944333475978, 216127028528873886, 864553867700405466, 3458352731556351630, 13833822708489595818, 55336526180750951166, 221349810763381508346, 885410361174659144430, 3541674799062035910858, 14166799259338341642846, 56667497226623960569626, 226670889474307624273230, 906686259600665843077098, 3626753143512969410260926] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (4 t - 1) (3 t - 1) and in Maple notation -(t-1)/(4*t-1)/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 21, : Let C(n) be 2 2 2 2 n (x y + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 20, 6, 36, 20, 84, 6, 36, 36, 120, 20, 120, 84, 296, 6, 36, 36, 120, 36, 216, 120, 504, 20, 120, 120, 400, 84, 504, 296, 1124, 6, 36, 36, 120, 36, 216, 120, 504, 36] Just for kicks C(googol) equals , 799605762186889740692889634076881341731181793016126185144320000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 20, 84, 296, 1124, 4056, 15100, 55288, 204740, 753336, 2782588, 10252952 , 37831684, 139470616, 514443580, 1896954168, 6996160196, 25799583352, 95147186940, 350881763288, 1294007162500, 4772059430872, 17598637988284, 64900766401656, 239343741625668, 882659972822840, 3255107458681212, 12004301666699032, 44269913022747396, 163260197361058968, 602077001213815356, 2220361636142223288, 8188331534303637444, 30197229928670023672, 111362456662997249020, 410686563525906396760, 1514545026074115320708, 5585394869142171484504, 20598024753141397814972, 75962153600605611299576] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 8 7 6 5 4 3 2 8 t + 16 t - 20 t + 34 t - 2 t - 15 t + 9 t - t - 1 - ------------------------------------------------------------------- 9 8 7 6 5 4 3 2 16 t + 24 t - 40 t + 56 t + 2 t - 44 t + 25 t + t - 5 t + 1 and in Maple notation -(8*t^8+16*t^7-20*t^6+34*t^5-2*t^4-15*t^3+9*t^2-t-1)/(16*t^9+24*t^8-40*t^7+56*t ^6+2*t^5-44*t^4+25*t^3+t^2-5*t+1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 22, : Let C(n) be 2 2 2 n (x y + x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 100, 6, 36, 36, 144, 24, 144, 100, 396, 6, 36, 36, 144 , 36, 216, 144, 600, 24, 144, 144, 576, 100, 600, 396, 1596, 6, 36, 36, 144, 36 , 216, 144, 600, 36] Just for kicks C(googol) equals , 137888714657610667256694059187392359170292799328425980226174976000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 100, 396, 1596, 6364, 25500, 101916, 407836, 1631004, 6524700, 26097436, 104392476, 417564444, 1670268700, 6681052956, 26724255516, 106896934684, 427587913500, 1710351304476, 6841405916956, 27365622269724, 109462491875100, 437849961907996, 1751399858816796, 7005599412897564, 28022397696329500, 112089590695839516, 448358362962315036, 1793433451491346204, 7173733806681212700, 28694935225293195036, 114779740904036091676, 459118963610417743644, 1836475854453124220700, 7345903417789590390556, 29383613671204174546716, 117534454684725072217884, 470137818739083540809500, 1880551274955967659362076] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 23, : Let C(n) be 2 2 2 2 n (x y + x y + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 90, 6, 36, 36, 132, 22, 132, 90, 358, 6, 36, 36, 132, 36, 216, 132, 540, 22, 132, 132, 484, 90, 540, 358, 1434, 6, 36, 36, 132, 36, 216, 132, 540, 36] Just for kicks C(googol) equals , 14056209326117951659557069010655383121710308703415637491300865933312000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 90, 358, 1434, 5734, 22938, 91750, 367002, 1468006, 5872026, 23488102, 93952410, 375809638, 1503238554, 6012954214, 24051816858, 96207267430 , 384829069722, 1539316278886, 6157265115546, 24629060462182, 98516241848730, 394064967394918, 1576259869579674, 6305039478318694, 25220157913274778, 100880631653099110, 403522526612396442, 1614090106449585766, 6456360425798343066, 25825441703193372262, 103301766812773489050, 413207067251093956198, 1652828269004375824794, 6611313076017503299174, 26445252304070013196698, 105781009216280052786790, 423124036865120211147162, 1692496147460480844588646] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 - ----------------- (t + 1) (4 t - 1) and in Maple notation -(3*t+1)/(t+1)/(4*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 24, : Let C(n) be 2 2 2 2 n (x y + x y + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 82, 6, 36, 36, 132, 22, 132, 82, 302, 6, 36, 36, 132, 36, 216, 132, 492, 22, 132, 132, 484, 82, 492, 302, 1106, 6, 36, 36, 132, 36, 216, 132, 492, 36] Just for kicks C(googol) equals , 2413602479613831581859142085056807815258942153585992997640886289759731712 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 82, 302, 1106, 4066, 14902, 54678, 200578, 735770, 2699182, 9901550, 36323050, 133247570, 488805718, 1793137798, 6577952882, 24130592458, 88520767614, 324729961566, 1191240790586, 4369952806274, 16030753627238, 58807285300086, 215728897446594, 791380812129402, 2903104763095054, 10649761956573134, 39067632408628746, 143315870180817330, 525740552457316726, 1928628896079822694, 7074990508936834578, 25953925508046213930, 95209491578317150878, 349266906996069982270, 1281252218664080409818, 4700151130694245039842, 17242054553786609505670, 63250826828592359807062] Using the found enumerative automaton with, 7, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 5 3 2 (2 t + 1) (2 t + t - t + t + 1) - ------------------------------------------------ 7 6 5 4 3 2 8 t - 8 t + 10 t - 6 t - t + 3 t + 3 t - 1 and in Maple notation -(2*t+1)*(2*t^5+t^3-t^2+t+1)/(8*t^7-8*t^6+10*t^5-6*t^4-t^3+3*t^2+3*t-1) This ends this theorem, that took, 0.011, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 25, : Let C(n) be 2 2 2 2 n (x y + x y + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 94, 6, 36, 36, 144, 24, 144, 94, 356, 6, 36, 36, 144, 36, 216, 144, 564, 24, 144, 144, 576, 94, 564, 356, 1338, 6, 36, 36, 144, 36, 216, 144, 564, 36] Just for kicks C(googol) equals , 42757024268292254411190836399528995947490826964920299882746213199528329216 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 94, 356, 1338, 5036, 18942, 71264, 268114, 1008728, 3795162, 14278632, 53720914, 202115688, 760425734, 2860971568, 10763916406, 40497394188, 152364521390, 573245456832, 2156738015294, 8114358001856, 30528884544598, 114859708062884, 432140012215626, 1625852906053008, 6116993561285206, 23014142354936060, 86586775518891606, 325767937779248176, 1225646164201790634, 4611284124710132760, 17349168055096974798, 65273278345832840520, 245579548972188856786, 923951062391070531508, 3476207889730692435414, 13078637802908965906948, 49206138472038541615246, 185129682449871663049116] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 8 7 6 5 4 3 2 (2 t + 1) (6 t - 7 t + 2 t - 7 t - 5 t - t - 2 t + t + 1) ------------------------------------------------------------------------ 10 9 8 7 6 5 4 3 2 8 t + 8 t + 4 t - 15 t - 2 t + 7 t + 9 t + 9 t - 6 t - 3 t + 1 and in Maple notation (2*t+1)*(6*t^8-7*t^7+2*t^6-7*t^5-5*t^4-t^3-2*t^2+t+1)/(8*t^10+8*t^9+4*t^8-15*t^ 7-2*t^6+7*t^5+9*t^4+9*t^3-6*t^2-3*t+1) This ends this theorem, that took, 0.014, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 26, : Let C(n) be 2 2 2 2 n (x y + x y + x + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 96, 6, 36, 36, 144, 24, 144, 96, 372, 6, 36, 36, 144, 36, 216, 144, 576, 24, 144, 144, 576, 96, 576, 372, 1416, 6, 36, 36, 144, 36, 216, 144, 576, 36] Just for kicks C(googol) equals , 62910545093793295945702128611066193383749523036605062293362318822437027840 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 96, 372, 1416, 5340, 20040, 75012, 280392, 1047324, 3910440, 14597508, 54485736, 203357724, 758969736, 2832570372, 10571410056, 39453266460, 147242049000, 549515715972, 2050822387752, 7653776980764, 28564291826760, 106603402909188, 397849344975816, 1484794027325724, 5541326864990376, 20680513633962372, 77180728073512296, 288042399465393180, 1074988871398673160, 4011913089350524932, 14972663492445877512, 55878740893317887004, 208542300106595474280, 778290459584603617668, 2904619538334898211496, 10840187693961147658524, 40456131237922009283016, 150984337258551523194372] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 3 t + 1 - ------------------------ 2 (t - 4 t + 1) (2 t - 1) and in Maple notation -(4*t^3-3*t^2+1)/(t^2-4*t+1)/(2*t-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 27, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 82, 6, 36, 36, 132, 22, 132, 82, 306, 6, 36, 36, 132, 36, 216, 132, 492, 22, 132, 132, 484, 82, 492, 306, 1142, 6, 36, 36, 132, 36, 216, 132, 492, 36] Just for kicks C(googol) equals , 2761456586817425003071513463990316124228320685072434853691153738366976000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 82, 306, 1142, 4250, 15806, 58730, 218142, 810074, 3007886, 11167914 , 41463774, 153942330, 571535534, 2121906442, 7877856190, 29247532442, 108585059214, 403135208426, 1496688037406, 5556634000122, 20629669372526, 76590116106186, 284349965403518, 1055683240842074, 3919350217589838, 14551056119534250, 54022534883643358, 200565116973926074, 744621965898163502, 2764498035965846666, 10263529334107729214, 38104579208915103002, 141467804043378683150, 525216128767713836394, 1949927644526031082910, 7239339408300824190586, 26876912697517794694126, 99783750340591693583690] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 5 4 3 2 (t - 1) (4 t - 2 t - 6 t - 3 t + 2 t + 1) - ---------------------------------------------- 5 4 2 (t + 1) (4 t - 2 t - t + 4 t - 1) (2 t - 1) and in Maple notation -(t-1)*(4*t^5-2*t^4-6*t^3-3*t^2+2*t+1)/(t+1)/(4*t^5-2*t^4-t^2+4*t-1)/(2*t-1) This ends this theorem, that took, 0.014, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 28, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 20, 6, 36, 20, 88, 6, 36, 36, 120, 20, 120, 88, 336, 6, 36, 36, 120, 36, 216, 120, 528, 20, 120, 120, 400, 88, 528, 336, 1376, 6, 36, 36, 120, 36, 216, 120, 528, 36] Just for kicks C(googol) equals , 2737767495721500889251655914073374844297767225771330581299200000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 20, 88, 336, 1376, 5440, 21888, 87296, 349696, 1397760, 5593088, 22368256, 89481216, 357908480, 1431666688, 5726601216, 22906535936, 91625881600 , 366504050688, 1466015154176, 5864062713856, 23456246661120, 93824995033088, 375299963355136, 1501199886974976, 6004799480791040, 24019198057381888, 96076791961092096, 384307168381239296, 1537228672451215360, 6148914691952345088 , 24595658763514413056, 98382635062647586816, 393530540233410478080, 1574122160968001650688, 6296488643803287126016, 25185954575350587457536, 100743818301127471923200, 402975273205059643506688, 1611901092819139062398976] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 t + 1 - ------------------- (2 t + 1) (4 t - 1) and in Maple notation -(4*t+1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.056, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 29, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 96, 6, 36, 36, 144, 24, 144, 96, 372, 6, 36, 36, 144, 36, 216, 144, 576, 24, 144, 144, 576, 96, 576, 372, 1416, 6, 36, 36, 144, 36, 216, 144, 576, 36] Just for kicks C(googol) equals , 62910545093793295945702128611066193383749523036605062293362318822437027840 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 96, 372, 1416, 5340, 20040, 75012, 280392, 1047324, 3910440, 14597508, 54485736, 203357724, 758969736, 2832570372, 10571410056, 39453266460, 147242049000, 549515715972, 2050822387752, 7653776980764, 28564291826760, 106603402909188, 397849344975816, 1484794027325724, 5541326864990376, 20680513633962372, 77180728073512296, 288042399465393180, 1074988871398673160, 4011913089350524932, 14972663492445877512, 55878740893317887004, 208542300106595474280, 778290459584603617668, 2904619538334898211496, 10840187693961147658524, 40456131237922009283016, 150984337258551523194372] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 3 t + 1 - ------------------------ 2 (t - 4 t + 1) (2 t - 1) and in Maple notation -(4*t^3-3*t^2+1)/(t^2-4*t+1)/(2*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 30, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 30, 6, 36, 30, 138, 6, 36, 36, 180, 30, 180, 138, 606, 6, 36, 36, 180 , 36, 216, 180, 828, 30, 180, 180, 900, 138, 828, 606, 2586, 6, 36, 36, 180, 36 , 216, 180, 828, 36] Just for kicks C(googol) equals , 181249926865339492095806851390538971660724\ 679113064023923988889600000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 30, 138, 606, 2586, 10830, 44778, 183486, 747066, 3027630, 12228618, 49268766, 198137946, 795740430, 3192527658, 12798808446, 51281327226, 205383589230, 822309197898, 3291561314526, 13173218826906, 52713796014030, 210917946175338, 843860071059006, 3376005143308986, 13505715150454830, 54027944333475978, 216127028528873886, 864553867700405466, 3458352731556351630, 13833822708489595818, 55336526180750951166, 221349810763381508346, 885410361174659144430, 3541674799062035910858, 14166799259338341642846, 56667497226623960569626, 226670889474307624273230, 906686259600665843077098, 3626753143512969410260926] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is t - 1 - ------------------- (4 t - 1) (3 t - 1) and in Maple notation -(t-1)/(4*t-1)/(3*t-1) This ends this theorem, that took, 0.002, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 31, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 82, 6, 36, 36, 132, 22, 132, 82, 302, 6, 36, 36, 132, 36, 216, 132, 492, 22, 132, 132, 484, 82, 492, 302, 1106, 6, 36, 36, 132, 36, 216, 132, 492, 36] Just for kicks C(googol) equals , 2413602479613831581859142085056807815258942153585992997640886289759731712 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 82, 302, 1106, 4066, 14902, 54678, 200578, 735770, 2699182, 9901550, 36323050, 133247570, 488805718, 1793137798, 6577952882, 24130592458, 88520767614, 324729961566, 1191240790586, 4369952806274, 16030753627238, 58807285300086, 215728897446594, 791380812129402, 2903104763095054, 10649761956573134, 39067632408628746, 143315870180817330, 525740552457316726, 1928628896079822694, 7074990508936834578, 25953925508046213930, 95209491578317150878, 349266906996069982270, 1281252218664080409818, 4700151130694245039842, 17242054553786609505670, 63250826828592359807062] Using the found enumerative automaton with, 7, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 5 3 2 (2 t + 1) (2 t + t - t + t + 1) - ------------------------------------------------ 7 6 5 4 3 2 8 t - 8 t + 10 t - 6 t - t + 3 t + 3 t - 1 and in Maple notation -(2*t+1)*(2*t^5+t^3-t^2+t+1)/(8*t^7-8*t^6+10*t^5-6*t^4-t^3+3*t^2+3*t-1) This ends this theorem, that took, 0.013, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 32, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 28, 6, 36, 28, 112, 6, 36, 36, 168, 28, 168, 112, 456, 6, 36, 36, 168 , 36, 216, 168, 672, 28, 168, 168, 784, 112, 672, 456, 1816, 6, 36, 36, 168, 36 , 216, 168, 672, 36] Just for kicks C(googol) equals , 396943363464639408218200560699216267233418\ 0368609654505818010397006696022016 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 28, 112, 456, 1816, 7288, 29112, 116536, 465976, 1864248, 7456312, 29826616, 119303736, 477220408, 1908870712, 7635504696, 30541975096, 122167987768, 488671776312, 1954687454776, 7818749120056, 31274997878328, 125099988717112, 500399960460856, 2001599830658616, 8006399345004088, 32025597335277112, 128102389430586936, 512409557543390776, 2049638230531477048, 8198552921410080312, 32794211687071977016, 131176846745424596536, 524707386987425009208, 2098829547938246790712, 8395318191775893655096, 33581272767057761635896, 134325091068322672512568, 537300364273107438112312, 2149201457092796256325176] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 4 t + 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (4*t^2+3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.011, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 33, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 92, 6, 36, 36, 144, 24, 144, 92, 340, 6, 36, 36, 144, 36, 216, 144, 552, 24, 144, 144, 576, 92, 552, 340, 1236, 6, 36, 36, 144, 36, 216, 144, 552, 36] Just for kicks C(googol) equals , 25859451075391164324299350593844150774718594295940830824517630860407603200 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 92, 340, 1236, 4452, 15956, 57028, 203508, 725604, 2585876, 9212932, 32818740, 116898468, 416365652, 1482959428, 5281740660, 18811402980, 66998214548, 238618498180, 849854020788, 3026803253028, 10780126189268, 38394001851076, 136742291486196, 487014945269604, 1734529552998932, 6177618817971460, 22001916096783156, 78360987000034212, 279086795341152596, 993982364318493508, 3540120692227720308, 12608326822500017124, 44905221886315230356, 159932319372665202052, 569607402028065020340, 2028686845104403372068, 7225275339919095970772, 25733199711065606284228] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (2 t + 1) (t - 1) - -------------------------- 2 (2 t + 3 t - 1) (2 t - 1) and in Maple notation -(2*t+1)*(t-1)/(2*t^2+3*t-1)/(2*t-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 34, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + x y + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 26, 6, 36, 26, 110, 6, 36, 36, 156, 26, 156, 110, 450, 6, 36, 36, 156 , 36, 216, 156, 660, 26, 156, 156, 676, 110, 660, 450, 1822, 6, 36, 36, 156, 36 , 216, 156, 660, 36] Just for kicks C(googol) equals , 134039144168938693310529048640833965397591\ 6267283481061109845196800000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 26, 110, 450, 1822, 7330, 29406, 117794, 471518, 1886754, 7548382, 30196258, 120790494, 483172898, 1932713438, 7730897442, 30923677150, 123694883362, 494779882974, 1979120230946, 7916482321886, 31665932083746, 126663733927390, 506654946894370, 2026619809947102, 8106479284527650, 32425917227589086, 129703669089313314, 518814676715167198, 2075258707576496674, 8301034831737642462, 33204139329813881378, 132816557324982148574, 531266229311381840418, 2125064917268433853918, 8500259669119548400162, 34001038676569819569630, 136004154706462530216482, 544016618826216624741854, 2176066475305599506719266] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (2 t + 1) (t - 1) - --------------------------- (4 t - 1) (2 t - 1) (t + 1) and in Maple notation -(2*t+1)*(t-1)/(4*t-1)/(2*t-1)/(t+1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 35, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 82, 6, 36, 36, 132, 22, 132, 82, 302, 6, 36, 36, 132, 36, 216, 132, 492, 22, 132, 132, 484, 82, 492, 302, 1106, 6, 36, 36, 132, 36, 216, 132, 492, 36] Just for kicks C(googol) equals , 2413602479613831581859142085056807815258942153585992997640886289759731712 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 82, 302, 1106, 4066, 14902, 54678, 200578, 735770, 2699182, 9901550, 36323050, 133247570, 488805718, 1793137798, 6577952882, 24130592458, 88520767614, 324729961566, 1191240790586, 4369952806274, 16030753627238, 58807285300086, 215728897446594, 791380812129402, 2903104763095054, 10649761956573134, 39067632408628746, 143315870180817330, 525740552457316726, 1928628896079822694, 7074990508936834578, 25953925508046213930, 95209491578317150878, 349266906996069982270, 1281252218664080409818, 4700151130694245039842, 17242054553786609505670, 63250826828592359807062] Using the found enumerative automaton with, 7, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 5 3 2 (2 t + 1) (2 t + t - t + t + 1) - ------------------------------------------------ 7 6 5 4 3 2 8 t - 8 t + 10 t - 6 t - t + 3 t + 3 t - 1 and in Maple notation -(2*t+1)*(2*t^5+t^3-t^2+t+1)/(8*t^7-8*t^6+10*t^5-6*t^4-t^3+3*t^2+3*t-1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 36, : Let C(n) be 2 2 2 2 n (x y + x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 96, 6, 36, 36, 144, 24, 144, 96, 384, 6, 36, 36, 144, 36, 216, 144, 576, 24, 144, 144, 576, 96, 576, 384, 1536, 6, 36, 36, 144, 36, 216, 144, 576, 36] Just for kicks C(googol) equals , 90754024259153226481010437249434699795319756525164263594879310578514395136 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 96, 384, 1536, 6144, 24576, 98304, 393216, 1572864, 6291456, 25165824, 100663296, 402653184, 1610612736, 6442450944, 25769803776, 103079215104, 412316860416, 1649267441664, 6597069766656, 26388279066624, 105553116266496, 422212465065984, 1688849860263936, 6755399441055744, 27021597764222976, 108086391056891904, 432345564227567616, 1729382256910270464, 6917529027641081856, 27670116110564327424, 110680464442257309696, 442721857769029238784, 1770887431076116955136, 7083549724304467820544, 28334198897217871282176, 113336795588871485128704, 453347182355485940514816, 1813388729421943762059264] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 t + 1 - ------- 4 t - 1 and in Maple notation -(2*t+1)/(4*t-1) This ends this theorem, that took, 0.003, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 37, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 20, 6, 36, 20, 84, 6, 36, 36, 120, 20, 120, 84, 296, 6, 36, 36, 120, 36, 216, 120, 504, 20, 120, 120, 400, 84, 504, 296, 1124, 6, 36, 36, 120, 36, 216, 120, 504, 36] Just for kicks C(googol) equals , 799605762186889740692889634076881341731181793016126185144320000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 20, 84, 296, 1124, 4056, 15100, 55288, 204740, 753336, 2782588, 10252952 , 37831684, 139470616, 514443580, 1896954168, 6996160196, 25799583352, 95147186940, 350881763288, 1294007162500, 4772059430872, 17598637988284, 64900766401656, 239343741625668, 882659972822840, 3255107458681212, 12004301666699032, 44269913022747396, 163260197361058968, 602077001213815356, 2220361636142223288, 8188331534303637444, 30197229928670023672, 111362456662997249020, 410686563525906396760, 1514545026074115320708, 5585394869142171484504, 20598024753141397814972, 75962153600605611299576] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 8 7 6 5 4 3 2 8 t + 16 t - 20 t + 34 t - 2 t - 15 t + 9 t - t - 1 - ------------------------------------------------------------------- 9 8 7 6 5 4 3 2 16 t + 24 t - 40 t + 56 t + 2 t - 44 t + 25 t + t - 5 t + 1 and in Maple notation -(8*t^8+16*t^7-20*t^6+34*t^5-2*t^4-15*t^3+9*t^2-t-1)/(16*t^9+24*t^8-40*t^7+56*t ^6+2*t^5-44*t^4+25*t^3+t^2-5*t+1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 38, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 96, 6, 36, 36, 144, 24, 144, 96, 372, 6, 36, 36, 144, 36, 216, 144, 576, 24, 144, 144, 576, 96, 576, 372, 1416, 6, 36, 36, 144, 36, 216, 144, 576, 36] Just for kicks C(googol) equals , 62910545093793295945702128611066193383749523036605062293362318822437027840 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 96, 372, 1416, 5340, 20040, 75012, 280392, 1047324, 3910440, 14597508, 54485736, 203357724, 758969736, 2832570372, 10571410056, 39453266460, 147242049000, 549515715972, 2050822387752, 7653776980764, 28564291826760, 106603402909188, 397849344975816, 1484794027325724, 5541326864990376, 20680513633962372, 77180728073512296, 288042399465393180, 1074988871398673160, 4011913089350524932, 14972663492445877512, 55878740893317887004, 208542300106595474280, 778290459584603617668, 2904619538334898211496, 10840187693961147658524, 40456131237922009283016, 150984337258551523194372] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 4 t - 3 t + 1 - ------------------------ 2 (t - 4 t + 1) (2 t - 1) and in Maple notation -(4*t^3-3*t^2+1)/(t^2-4*t+1)/(2*t-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 39, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 88, 6, 36, 36, 144, 24, 144, 88, 336, 6, 36, 36, 144, 36, 216, 144, 528, 24, 144, 144, 576, 88, 528, 336, 1280, 6, 36, 36, 144, 36, 216, 144, 528, 36] Just for kicks C(googol) equals , 22961890025021468049088246083148323519502246213219980004482861654815539200 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 88, 336, 1280, 4928, 19072, 74240, 290304, 1139712, 4489216, 17731584, 70197248, 278429696, 1106083840, 4399628288, 17518559232, 69815500800 , 278424715264, 1110989340672, 4435189170176, 17712382214144, 70757707153408, 282733687341056, 1129973180006400, 4516781016219648, 18057054379835392, 72195631334031360, 288677074225332224, 1154367049938501632, 4616363901385179136 , 18461882020951752704, 73835963721153773568, 295306431820952764416, 1181104623705873448960, 4724026595412969259008, 18894838168519077527552, 75575248650168612945920, 302287713700327859421184, 1209107876904987464302592] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 2 16 t + 8 t - 1 - -------------------------- 2 (4 t - 1) (4 t + 2 t - 1) and in Maple notation -(16*t^3+8*t^2-1)/(4*t-1)/(4*t^2+2*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 40, : Let C(n) be 2 2 2 2 n (x y + x y + x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 100, 6, 36, 36, 144, 24, 144, 100, 396, 6, 36, 36, 144 , 36, 216, 144, 600, 24, 144, 144, 576, 100, 600, 396, 1596, 6, 36, 36, 144, 36 , 216, 144, 600, 36] Just for kicks C(googol) equals , 137888714657610667256694059187392359170292799328425980226174976000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 100, 396, 1596, 6364, 25500, 101916, 407836, 1631004, 6524700, 26097436, 104392476, 417564444, 1670268700, 6681052956, 26724255516, 106896934684, 427587913500, 1710351304476, 6841405916956, 27365622269724, 109462491875100, 437849961907996, 1751399858816796, 7005599412897564, 28022397696329500, 112089590695839516, 448358362962315036, 1793433451491346204, 7173733806681212700, 28694935225293195036, 114779740904036091676, 459118963610417743644, 1836475854453124220700, 7345903417789590390556, 29383613671204174546716, 117534454684725072217884, 470137818739083540809500, 1880551274955967659362076] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 41, : Let C(n) be 2 2 2 2 n (x y + x y + x y + x y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 100, 6, 36, 36, 144, 24, 144, 100, 396, 6, 36, 36, 144 , 36, 216, 144, 600, 24, 144, 144, 576, 100, 600, 396, 1596, 6, 36, 36, 144, 36 , 216, 144, 600, 36] Just for kicks C(googol) equals , 137888714657610667256694059187392359170292799328425980226174976000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 100, 396, 1596, 6364, 25500, 101916, 407836, 1631004, 6524700, 26097436, 104392476, 417564444, 1670268700, 6681052956, 26724255516, 106896934684, 427587913500, 1710351304476, 6841405916956, 27365622269724, 109462491875100, 437849961907996, 1751399858816796, 7005599412897564, 28022397696329500, 112089590695839516, 448358362962315036, 1793433451491346204, 7173733806681212700, 28694935225293195036, 114779740904036091676, 459118963610417743644, 1836475854453124220700, 7345903417789590390556, 29383613671204174546716, 117534454684725072217884, 470137818739083540809500, 1880551274955967659362076] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 --------------------------- (t - 1) (2 t + 1) (4 t - 1) and in Maple notation (3*t+1)/(t-1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 42, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 26, 6, 36, 26, 110, 6, 36, 36, 156, 26, 156, 110, 450, 6, 36, 36, 156 , 36, 216, 156, 660, 26, 156, 156, 676, 110, 660, 450, 1822, 6, 36, 36, 156, 36 , 216, 156, 660, 36] Just for kicks C(googol) equals , 134039144168938693310529048640833965397591\ 6267283481061109845196800000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 26, 110, 450, 1822, 7330, 29406, 117794, 471518, 1886754, 7548382, 30196258, 120790494, 483172898, 1932713438, 7730897442, 30923677150, 123694883362, 494779882974, 1979120230946, 7916482321886, 31665932083746, 126663733927390, 506654946894370, 2026619809947102, 8106479284527650, 32425917227589086, 129703669089313314, 518814676715167198, 2075258707576496674, 8301034831737642462, 33204139329813881378, 132816557324982148574, 531266229311381840418, 2125064917268433853918, 8500259669119548400162, 34001038676569819569630, 136004154706462530216482, 544016618826216624741854, 2176066475305599506719266] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (2 t + 1) (t - 1) - --------------------------- (4 t - 1) (2 t - 1) (t + 1) and in Maple notation -(2*t+1)*(t-1)/(4*t-1)/(2*t-1)/(t+1) This ends this theorem, that took, 0.006, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 43, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + x y + y + x) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 24, 6, 36, 24, 94, 6, 36, 36, 144, 24, 144, 94, 356, 6, 36, 36, 144, 36, 216, 144, 564, 24, 144, 144, 576, 94, 564, 356, 1338, 6, 36, 36, 144, 36, 216, 144, 564, 36] Just for kicks C(googol) equals , 42757024268292254411190836399528995947490826964920299882746213199528329216 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 24, 94, 356, 1338, 5036, 18942, 71264, 268114, 1008728, 3795162, 14278632, 53720914, 202115688, 760425734, 2860971568, 10763916406, 40497394188, 152364521390, 573245456832, 2156738015294, 8114358001856, 30528884544598, 114859708062884, 432140012215626, 1625852906053008, 6116993561285206, 23014142354936060, 86586775518891606, 325767937779248176, 1225646164201790634, 4611284124710132760, 17349168055096974798, 65273278345832840520, 245579548972188856786, 923951062391070531508, 3476207889730692435414, 13078637802908965906948, 49206138472038541615246, 185129682449871663049116] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 8 7 6 5 4 3 2 (2 t + 1) (6 t - 7 t + 2 t - 7 t - 5 t - t - 2 t + t + 1) ------------------------------------------------------------------------ 10 9 8 7 6 5 4 3 2 8 t + 8 t + 4 t - 15 t - 2 t + 7 t + 9 t + 9 t - 6 t - 3 t + 1 and in Maple notation (2*t+1)*(6*t^8-7*t^7+2*t^6-7*t^5-5*t^4-t^3-2*t^2+t+1)/(8*t^10+8*t^9+4*t^8-15*t^ 7-2*t^6+7*t^5+9*t^4+9*t^3-6*t^2-3*t+1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 44, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 82, 6, 36, 36, 132, 22, 132, 82, 306, 6, 36, 36, 132, 36, 216, 132, 492, 22, 132, 132, 484, 82, 492, 306, 1142, 6, 36, 36, 132, 36, 216, 132, 492, 36] Just for kicks C(googol) equals , 2761456586817425003071513463990316124228320685072434853691153738366976000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 82, 306, 1142, 4250, 15806, 58730, 218142, 810074, 3007886, 11167914 , 41463774, 153942330, 571535534, 2121906442, 7877856190, 29247532442, 108585059214, 403135208426, 1496688037406, 5556634000122, 20629669372526, 76590116106186, 284349965403518, 1055683240842074, 3919350217589838, 14551056119534250, 54022534883643358, 200565116973926074, 744621965898163502, 2764498035965846666, 10263529334107729214, 38104579208915103002, 141467804043378683150, 525216128767713836394, 1949927644526031082910, 7239339408300824190586, 26876912697517794694126, 99783750340591693583690] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 5 4 3 2 (t - 1) (4 t - 2 t - 6 t - 3 t + 2 t + 1) - ---------------------------------------------- 5 4 2 (t + 1) (4 t - 2 t - t + 4 t - 1) (2 t - 1) and in Maple notation -(t-1)*(4*t^5-2*t^4-6*t^3-3*t^2+2*t+1)/(t+1)/(4*t^5-2*t^4-t^2+4*t-1)/(2*t-1) This ends this theorem, that took, 0.017, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 45, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 20, 6, 36, 20, 84, 6, 36, 36, 120, 20, 120, 84, 296, 6, 36, 36, 120, 36, 216, 120, 504, 20, 120, 120, 400, 84, 504, 296, 1124, 6, 36, 36, 120, 36, 216, 120, 504, 36] Just for kicks C(googol) equals , 799605762186889740692889634076881341731181793016126185144320000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 20, 84, 296, 1124, 4056, 15100, 55288, 204740, 753336, 2782588, 10252952 , 37831684, 139470616, 514443580, 1896954168, 6996160196, 25799583352, 95147186940, 350881763288, 1294007162500, 4772059430872, 17598637988284, 64900766401656, 239343741625668, 882659972822840, 3255107458681212, 12004301666699032, 44269913022747396, 163260197361058968, 602077001213815356, 2220361636142223288, 8188331534303637444, 30197229928670023672, 111362456662997249020, 410686563525906396760, 1514545026074115320708, 5585394869142171484504, 20598024753141397814972, 75962153600605611299576] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 8 7 6 5 4 3 2 8 t + 16 t - 20 t + 34 t - 2 t - 15 t + 9 t - t - 1 - ------------------------------------------------------------------- 9 8 7 6 5 4 3 2 16 t + 24 t - 40 t + 56 t + 2 t - 44 t + 25 t + t - 5 t + 1 and in Maple notation -(8*t^8+16*t^7-20*t^6+34*t^5-2*t^4-15*t^3+9*t^2-t-1)/(16*t^9+24*t^8-40*t^7+56*t ^6+2*t^5-44*t^4+25*t^3+t^2-5*t+1) This ends this theorem, that took, 0.077, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 46, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + x y + y ) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 6, 6, 22, 6, 36, 22, 90, 6, 36, 36, 132, 22, 132, 90, 358, 6, 36, 36, 132, 36, 216, 132, 540, 22, 132, 132, 484, 90, 540, 358, 1434, 6, 36, 36, 132, 36, 216, 132, 540, 36] Just for kicks C(googol) equals , 14056209326117951659557069010655383121710308703415637491300865933312000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 6, 22, 90, 358, 1434, 5734, 22938, 91750, 367002, 1468006, 5872026, 23488102, 93952410, 375809638, 1503238554, 6012954214, 24051816858, 96207267430 , 384829069722, 1539316278886, 6157265115546, 24629060462182, 98516241848730, 394064967394918, 1576259869579674, 6305039478318694, 25220157913274778, 100880631653099110, 403522526612396442, 1614090106449585766, 6456360425798343066, 25825441703193372262, 103301766812773489050, 413207067251093956198, 1652828269004375824794, 6611313076017503299174, 26445252304070013196698, 105781009216280052786790, 423124036865120211147162, 1692496147460480844588646] Using the found enumerative automaton with, 2, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 3 t + 1 - ----------------- (t + 1) (4 t - 1) and in Maple notation -(3*t+1)/(t+1)/(4*t-1) This ends this theorem, that took, 0.002, seconds. ------------------------------------------------------------------ This concludes this webbook, that took, 0.851, seconds. to generate. k is , 7 n On Sequences of the form, P(x, y) , modolu , 2, Evaluated at, {x = 1, y = 1} For all Polynomials that are Sums of Monomials taken from, 2 2 2 2 2 2 {1, x, y, x , y , x y, x y , x y, x y } By Shalosh B. Ekhad In this webbook, we will consider the sequences described in the title, that\ after normalization and weeding out obvious symmetry, concerns the following set of, 21, polynomials 2 2 2 2 2 2 {x y + x + x y + y + x + y + 1, x y + x y + x y + y + x + y + 1, 2 2 2 2 2 2 2 2 x y + x y + x + y + x + y + 1, x y + x y + x + x y + y + y + 1, 2 2 2 2 2 2 2 2 x y + x y + x + x y + y + x + y, x y + x + x y + y + x + y + 1, 2 2 2 2 2 2 2 2 2 x y + x y + x y + y + x + y + 1, x y + x y + x + y + x + y + 1, 2 2 2 2 2 2 2 2 2 x y + x y + x + x y + x + y + 1, x y + x y + x + x y + y + y + 1, 2 2 2 2 2 2 2 2 2 2 x y + x y + x + x y + y + x + 1, x y + x y + x + x y + y + x + y, 2 2 2 2 2 2 2 2 2 x y + x y + x y + y + x + y + 1, x y + x y + x y + x y + x + y + 1, 2 2 2 2 2 x y + x y + x y + x y + y + y + 1, 2 2 2 2 2 x y + x y + x y + x y + y + x + 1, 2 2 2 2 2 x y + x y + x y + x y + y + x + y, 2 2 2 2 2 2 x y + x y + x y + x + y + y + 1, 2 2 2 2 2 2 x y + x y + x y + x + y + x + y, 2 2 2 2 2 2 x y + x y + x y + x + x y + y + 1, 2 2 2 2 2 2 x y + x y + x y + x + x y + y + y} by finding enumerative automata with at most, 200, states . ----------------------------------------------------------------------------\ ---- Theorem Number, 1, : Let C(n) be 2 2 2 n (x y + x + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 27, 7, 49, 27, 113, 7, 49, 49, 189, 27, 189, 113, 447, 7, 49, 49, 189 , 49, 343, 189, 791, 27, 189, 189, 729, 113, 791, 447, 1743, 7, 49, 49, 189, 49 , 343, 189, 791, 49] Just for kicks C(googol) equals , 645420424506004584929011636059207871314243\ 461376867193471738352674661533096929 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 27, 113, 447, 1743, 6789, 26371, 102399, 397425, 1542327, 5984815, 23222957, 90110355, 349647247, 1356699401, 5264252887, 20426289087, 79257818197 , 307535089027, 1193293339871, 4630199918049, 17966035966423, 69711557106543, 270493791315325, 1049566156154579, 4072511647171791, 15802101672278393, 61315089769937015, 237913937730074175, 923150271448256549, 3581994530438293763, 13898804141498109151, 53929941802446858449, 209258191798574502199, 811960654346536991919, 3150558162336227602509, 12224750892949051210387, 47434304238914509746063, 184053911472961467551977, 714163364932527133333911] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 12 t + 12 t + 6 t - 6 t + 3 t - 4 t - 1 - ------------------------------------------------------------ 7 6 5 4 3 2 (12 t - 8 t + 2 t + 2 t + 9 t - 2 t - 4 t + 1) (t + 1) and in Maple notation -(12*t^7+12*t^6+6*t^5-6*t^4+3*t^3-4*t-1)/(12*t^7-8*t^6+2*t^5+2*t^4+9*t^3-2*t^2-\ 4*t+1)/(t+1) This ends this theorem, that took, 0.026, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 2, : Let C(n) be 2 2 2 n (x y + x y + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 25, 7, 49, 25, 107, 7, 49, 49, 175, 25, 175, 107, 413, 7, 49, 49, 175 , 49, 343, 175, 749, 25, 175, 175, 625, 107, 749, 413, 1615, 7, 49, 49, 175, 49 , 343, 175, 749, 49] Just for kicks C(googol) equals , 115252299357716884324756285860116522960740\ 376590900080638490617275238037109375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 25, 107, 413, 1615, 6241, 24227, 93973, 364743, 1415289, 5491643, 21307533, 82673535, 320773713, 1244606995, 4829092037, 18736941431, 72699560681 , 282075165611, 1094454989757, 4246498377391, 16476464213569, 63928877142147, 248044803475829, 962416786448167, 3734188572465241, 14488696052099227, 56216312919224813, 218119962409118431, 846308047088580913, 3283685283337329651, 12740737934703822117, 49434214644262756823, 191805340477042460873, 744207001168233417355, 2887532011414972857117, 11203658530298171425807, 43470328282875858747169, 168665390498178293159011, 654423720170279028952021] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 3 2 16 t + 16 t - 12 t + 7 t - 2 t - 1 - --------------------------------------------- 7 6 4 3 2 16 t + 16 t - 16 t + 9 t + 3 t - 5 t + 1 and in Maple notation -(16*t^6+16*t^5-12*t^3+7*t^2-2*t-1)/(16*t^7+16*t^6-16*t^4+9*t^3+3*t^2-5*t+1) This ends this theorem, that took, 0.018, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 3, : Let C(n) be 2 2 2 2 n (x y + x y + x + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 21, 7, 49, 21, 95, 7, 49, 49, 147, 21, 147, 95, 333, 7, 49, 49, 147, 49, 343, 147, 665, 21, 147, 147, 441, 95, 665, 333, 1319, 7, 49, 49, 147, 49, 343, 147, 665, 49] Just for kicks C(googol) equals , 196686290352557863064275017100411599165471\ 0381658422259464945186318492984375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 21, 95, 333, 1319, 4837, 18447, 68733, 259447, 972565, 3661535, 13756333 , 51754567, 194586181, 731919279, 2752461533, 10352254743, 38932913525, 146424889471, 550683608589, 2071066796007, 7789015542949, 29293584500047, 110169505843517, 414334209685687, 1558259850417109, 5860425507488287, 22040342341502061, 82891039280982407, 311743071539532549, 1172427660162381551, 4409357319499233181, 16583054888865919575, 62366845725293767221, 234554096714279936959, 882129335103095923405, 3317580786194232038183, 12477016493309192526181, 46924536476035659892879, 176477455491828408944125] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 3 2 8 t - 28 t + 16 t - t - 1 - ------------------------------------- 5 4 3 2 8 t - 44 t + 24 t + 5 t - 6 t + 1 and in Maple notation -(8*t^4-28*t^3+16*t^2-t-1)/(8*t^5-44*t^4+24*t^3+5*t^2-6*t+1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 4, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 31, 7, 49, 31, 127, 7, 49, 49, 217, 31, 217, 127, 511, 7, 49, 49, 217 , 49, 343, 217, 889, 31, 217, 217, 961, 127, 889, 511, 2031, 7, 49, 49, 217, 49 , 343, 217, 889, 49] Just for kicks C(googol) equals , 164752321121957837163297617370335113695910\ 20179216238196866738014918684480804019 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 31, 127, 511, 2031, 8043, 31735, 125063, 492367, 1937763, 7624303, 29995559, 118000431, 464192219, 1826013415, 7183010967, 28255752751, 111149170563, 437224979743, 1719900889847, 6765528227247, 26613372893339, 104688284286487, 411809368955175, 1619922922644527, 6372245108777955, 25066320646301935, 98602677307103399, 387870564961339119, 1525755477446319803, 6001821189395112679, 23609194337170247479, 92870820300375594959, 365323320199954434595, 1437062016292606269055, 5652930224902411765943, 22236771805879284928335, 87472160573481473129435, 344086765028636844289367, 1353524379544585488514087] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 9 8 7 5 4 3 2 6 t - 2 t + 9 t - 2 t + 5 t - 7 t - 4 t + 2 t + 1 - ----------------------------------------------------------------------- 10 9 8 7 6 5 4 3 10 t + 22 t - 11 t + 31 t - 24 t + 3 t + 18 t - 21 t + 5 t - 1 and in Maple notation -(6*t^9-2*t^8+9*t^7-2*t^5+5*t^4-7*t^3-4*t^2+2*t+1)/(10*t^10+22*t^9-11*t^8+31*t^ 7-24*t^6+3*t^5+18*t^4-21*t^3+5*t-1) This ends this theorem, that took, 0.018, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 5, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 25, 7, 49, 25, 103, 7, 49, 49, 175, 25, 175, 103, 409, 7, 49, 49, 175 , 49, 343, 175, 721, 25, 175, 175, 625, 103, 721, 409, 1639, 7, 49, 49, 175, 49 , 343, 175, 721, 49] Just for kicks C(googol) equals , 972558273237647608553997316362800310018984\ 94047600366420982778072357177734375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 25, 103, 409, 1639, 6553, 26215, 104857, 419431, 1677721, 6710887, 26843545, 107374183, 429496729, 1717986919, 6871947673, 27487790695, 109951162777, 439804651111, 1759218604441, 7036874417767, 28147497671065, 112589990684263, 450359962737049, 1801439850948199, 7205759403792793, 28823037615171175, 115292150460684697, 461168601842738791, 1844674407370955161, 7378697629483820647, 29514790517935282585, 118059162071741130343, 472236648286964521369, 1888946593147858085479, 7555786372591432341913, 30223145490365729367655, 120892581961462917470617, 483570327845851669882471, 1934281311383406679529881] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 t + 1 - ----------------- (t + 1) (4 t - 1) and in Maple notation -(4*t+1)/(t+1)/(4*t-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 6, : Let C(n) be 2 2 2 2 n (x y + x + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 31, 7, 49, 31, 145, 7, 49, 49, 217, 31, 217, 145, 601, 7, 49, 49, 217 , 49, 343, 217, 1015, 31, 217, 217, 961, 145, 1015, 601, 2551, 7, 49, 49, 217, 49, 343, 217, 1015, 49] Just for kicks C(googol) equals , 102491932901292648485438113781371441636491\ 360038512797908586703720353810714484375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 31, 145, 601, 2551, 10351, 42433, 170761, 690247, 2768191, 11123185, 44543161, 178525591, 714455311, 2860291873, 11443638121, 45791846887, 183184681951, 732859788625, 2931560215321, 11727088287031, 46909200573871, 187642734275713, 750576869083081, 3002349000193927, 12009437524637311, 48038040765580465, 192152453729353081, 768611849586630871, 3074449433015742031, 12297811974747497953, 49191262141674521641, 196765148265489795367, 787060692760750890271, 3148243468934545522705, 12592974573629724052441, 50371903179759689941111, 201487617604279553495791, 805950504613803770102593, 3223802052651900636529801] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (4 t + 1) (t - 1) - -------------------- 2 (4 t - 1) (7 t - 1) and in Maple notation -(4*t+1)*(t-1)/(4*t-1)/(7*t^2-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 7, : Let C(n) be 2 2 2 2 n (x y + x y + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 27, 7, 49, 27, 113, 7, 49, 49, 189, 27, 189, 113, 447, 7, 49, 49, 189 , 49, 343, 189, 791, 27, 189, 189, 729, 113, 791, 447, 1743, 7, 49, 49, 189, 49 , 343, 189, 791, 49] Just for kicks C(googol) equals , 645420424506004584929011636059207871314243\ 461376867193471738352674661533096929 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 27, 113, 447, 1743, 6789, 26371, 102399, 397425, 1542327, 5984815, 23222957, 90110355, 349647247, 1356699401, 5264252887, 20426289087, 79257818197 , 307535089027, 1193293339871, 4630199918049, 17966035966423, 69711557106543, 270493791315325, 1049566156154579, 4072511647171791, 15802101672278393, 61315089769937015, 237913937730074175, 923150271448256549, 3581994530438293763, 13898804141498109151, 53929941802446858449, 209258191798574502199, 811960654346536991919, 3150558162336227602509, 12224750892949051210387, 47434304238914509746063, 184053911472961467551977, 714163364932527133333911] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 12 t + 12 t + 6 t - 6 t + 3 t - 4 t - 1 - ------------------------------------------------------------ 7 6 5 4 3 2 (12 t - 8 t + 2 t + 2 t + 9 t - 2 t - 4 t + 1) (t + 1) and in Maple notation -(12*t^7+12*t^6+6*t^5-6*t^4+3*t^3-4*t-1)/(12*t^7-8*t^6+2*t^5+2*t^4+9*t^3-2*t^2-\ 4*t+1)/(t+1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 8, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 25, 7, 49, 25, 103, 7, 49, 49, 175, 25, 175, 103, 409, 7, 49, 49, 175 , 49, 343, 175, 721, 25, 175, 175, 625, 103, 721, 409, 1639, 7, 49, 49, 175, 49 , 343, 175, 721, 49] Just for kicks C(googol) equals , 972558273237647608553997316362800310018984\ 94047600366420982778072357177734375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 25, 103, 409, 1639, 6553, 26215, 104857, 419431, 1677721, 6710887, 26843545, 107374183, 429496729, 1717986919, 6871947673, 27487790695, 109951162777, 439804651111, 1759218604441, 7036874417767, 28147497671065, 112589990684263, 450359962737049, 1801439850948199, 7205759403792793, 28823037615171175, 115292150460684697, 461168601842738791, 1844674407370955161, 7378697629483820647, 29514790517935282585, 118059162071741130343, 472236648286964521369, 1888946593147858085479, 7555786372591432341913, 30223145490365729367655, 120892581961462917470617, 483570327845851669882471, 1934281311383406679529881] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 t + 1 - ----------------- (t + 1) (4 t - 1) and in Maple notation -(4*t+1)/(t+1)/(4*t-1) This ends this theorem, that took, 0.010, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 9, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 31, 7, 49, 31, 127, 7, 49, 49, 217, 31, 217, 127, 511, 7, 49, 49, 217 , 49, 343, 217, 889, 31, 217, 217, 961, 127, 889, 511, 2031, 7, 49, 49, 217, 49 , 343, 217, 889, 49] Just for kicks C(googol) equals , 164752321121957837163297617370335113695910\ 20179216238196866738014918684480804019 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 31, 127, 511, 2031, 8043, 31735, 125063, 492367, 1937763, 7624303, 29995559, 118000431, 464192219, 1826013415, 7183010967, 28255752751, 111149170563, 437224979743, 1719900889847, 6765528227247, 26613372893339, 104688284286487, 411809368955175, 1619922922644527, 6372245108777955, 25066320646301935, 98602677307103399, 387870564961339119, 1525755477446319803, 6001821189395112679, 23609194337170247479, 92870820300375594959, 365323320199954434595, 1437062016292606269055, 5652930224902411765943, 22236771805879284928335, 87472160573481473129435, 344086765028636844289367, 1353524379544585488514087] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 9 8 7 5 4 3 2 6 t - 2 t + 9 t - 2 t + 5 t - 7 t - 4 t + 2 t + 1 - ----------------------------------------------------------------------- 10 9 8 7 6 5 4 3 10 t + 22 t - 11 t + 31 t - 24 t + 3 t + 18 t - 21 t + 5 t - 1 and in Maple notation -(6*t^9-2*t^8+9*t^7-2*t^5+5*t^4-7*t^3-4*t^2+2*t+1)/(10*t^10+22*t^9-11*t^8+31*t^ 7-24*t^6+3*t^5+18*t^4-21*t^3+5*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 10, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 31, 7, 49, 31, 145, 7, 49, 49, 217, 31, 217, 145, 601, 7, 49, 49, 217 , 49, 343, 217, 1015, 31, 217, 217, 961, 145, 1015, 601, 2551, 7, 49, 49, 217, 49, 343, 217, 1015, 49] Just for kicks C(googol) equals , 102491932901292648485438113781371441636491\ 360038512797908586703720353810714484375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 31, 145, 601, 2551, 10351, 42433, 170761, 690247, 2768191, 11123185, 44543161, 178525591, 714455311, 2860291873, 11443638121, 45791846887, 183184681951, 732859788625, 2931560215321, 11727088287031, 46909200573871, 187642734275713, 750576869083081, 3002349000193927, 12009437524637311, 48038040765580465, 192152453729353081, 768611849586630871, 3074449433015742031, 12297811974747497953, 49191262141674521641, 196765148265489795367, 787060692760750890271, 3148243468934545522705, 12592974573629724052441, 50371903179759689941111, 201487617604279553495791, 805950504613803770102593, 3223802052651900636529801] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (4 t + 1) (t - 1) - -------------------- 2 (4 t - 1) (7 t - 1) and in Maple notation -(4*t+1)*(t-1)/(4*t-1)/(7*t^2-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 11, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 29, 7, 49, 29, 103, 7, 49, 49, 203, 29, 203, 103, 373, 7, 49, 49, 203 , 49, 343, 203, 721, 29, 203, 203, 841, 103, 721, 373, 1407, 7, 49, 49, 203, 49 , 343, 203, 721, 49] Just for kicks C(googol) equals , 330535228988519304590102283146694122428133\ 409876308146495742847941505181368039 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 29, 103, 373, 1407, 5277, 19639, 73157, 272943, 1018157, 3796839, 14159317, 52806751, 196940221, 734469911, 2739138277, 10215390607, 38097452877, 142081224135, 529879903477, 1976142458303, 7369856791005, 27485259393911, 102503957075973, 382279865222383, 1425680525146477, 5316955307198503, 19829136499982357, 73951092622897951, 275794364527453693, 1028551828039807959, 3835897317102466341, 14305655608510212175, 53351736366068729741, 198970802259344383623, 742044830182934578997, 2767393626345435272319, 10320761188039610723869, 38490408623657063872567, 143546733523198150428229] Using the found enumerative automaton with, 7, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 (2 t + 1) (4 t + 2 t + 1) - -------------------------- 4 3 8 t + 8 t + 3 t - 1 and in Maple notation -(2*t+1)*(4*t^2+2*t+1)/(8*t^4+8*t^3+3*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 12, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 31, 7, 49, 31, 127, 7, 49, 49, 217, 31, 217, 127, 511, 7, 49, 49, 217 , 49, 343, 217, 889, 31, 217, 217, 961, 127, 889, 511, 2031, 7, 49, 49, 217, 49 , 343, 217, 889, 49] Just for kicks C(googol) equals , 164752321121957837163297617370335113695910\ 20179216238196866738014918684480804019 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 31, 127, 511, 2031, 8043, 31735, 125063, 492367, 1937763, 7624303, 29995559, 118000431, 464192219, 1826013415, 7183010967, 28255752751, 111149170563, 437224979743, 1719900889847, 6765528227247, 26613372893339, 104688284286487, 411809368955175, 1619922922644527, 6372245108777955, 25066320646301935, 98602677307103399, 387870564961339119, 1525755477446319803, 6001821189395112679, 23609194337170247479, 92870820300375594959, 365323320199954434595, 1437062016292606269055, 5652930224902411765943, 22236771805879284928335, 87472160573481473129435, 344086765028636844289367, 1353524379544585488514087] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 9 8 7 5 4 3 2 6 t - 2 t + 9 t - 2 t + 5 t - 7 t - 4 t + 2 t + 1 - ----------------------------------------------------------------------- 10 9 8 7 6 5 4 3 10 t + 22 t - 11 t + 31 t - 24 t + 3 t + 18 t - 21 t + 5 t - 1 and in Maple notation -(6*t^9-2*t^8+9*t^7-2*t^5+5*t^4-7*t^3-4*t^2+2*t+1)/(10*t^10+22*t^9-11*t^8+31*t^ 7-24*t^6+3*t^5+18*t^4-21*t^3+5*t-1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 13, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 21, 7, 49, 21, 95, 7, 49, 49, 147, 21, 147, 95, 333, 7, 49, 49, 147, 49, 343, 147, 665, 21, 147, 147, 441, 95, 665, 333, 1319, 7, 49, 49, 147, 49, 343, 147, 665, 49] Just for kicks C(googol) equals , 196686290352557863064275017100411599165471\ 0381658422259464945186318492984375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 21, 95, 333, 1319, 4837, 18447, 68733, 259447, 972565, 3661535, 13756333 , 51754567, 194586181, 731919279, 2752461533, 10352254743, 38932913525, 146424889471, 550683608589, 2071066796007, 7789015542949, 29293584500047, 110169505843517, 414334209685687, 1558259850417109, 5860425507488287, 22040342341502061, 82891039280982407, 311743071539532549, 1172427660162381551, 4409357319499233181, 16583054888865919575, 62366845725293767221, 234554096714279936959, 882129335103095923405, 3317580786194232038183, 12477016493309192526181, 46924536476035659892879, 176477455491828408944125] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 3 2 8 t - 28 t + 16 t - t - 1 - ------------------------------------- 5 4 3 2 8 t - 44 t + 24 t + 5 t - 6 t + 1 and in Maple notation -(8*t^4-28*t^3+16*t^2-t-1)/(8*t^5-44*t^4+24*t^3+5*t^2-6*t+1) This ends this theorem, that took, 0.013, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 14, : Let C(n) be 2 2 2 2 n (x y + x y + x y + x y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 25, 7, 49, 25, 103, 7, 49, 49, 175, 25, 175, 103, 409, 7, 49, 49, 175 , 49, 343, 175, 721, 25, 175, 175, 625, 103, 721, 409, 1639, 7, 49, 49, 175, 49 , 343, 175, 721, 49] Just for kicks C(googol) equals , 972558273237647608553997316362800310018984\ 94047600366420982778072357177734375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 25, 103, 409, 1639, 6553, 26215, 104857, 419431, 1677721, 6710887, 26843545, 107374183, 429496729, 1717986919, 6871947673, 27487790695, 109951162777, 439804651111, 1759218604441, 7036874417767, 28147497671065, 112589990684263, 450359962737049, 1801439850948199, 7205759403792793, 28823037615171175, 115292150460684697, 461168601842738791, 1844674407370955161, 7378697629483820647, 29514790517935282585, 118059162071741130343, 472236648286964521369, 1888946593147858085479, 7555786372591432341913, 30223145490365729367655, 120892581961462917470617, 483570327845851669882471, 1934281311383406679529881] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 t + 1 - ----------------- (t + 1) (4 t - 1) and in Maple notation -(4*t+1)/(t+1)/(4*t-1) This ends this theorem, that took, 0.008, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 15, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 27, 7, 49, 27, 113, 7, 49, 49, 189, 27, 189, 113, 447, 7, 49, 49, 189 , 49, 343, 189, 791, 27, 189, 189, 729, 113, 791, 447, 1743, 7, 49, 49, 189, 49 , 343, 189, 791, 49] Just for kicks C(googol) equals , 645420424506004584929011636059207871314243\ 461376867193471738352674661533096929 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 27, 113, 447, 1743, 6789, 26371, 102399, 397425, 1542327, 5984815, 23222957, 90110355, 349647247, 1356699401, 5264252887, 20426289087, 79257818197 , 307535089027, 1193293339871, 4630199918049, 17966035966423, 69711557106543, 270493791315325, 1049566156154579, 4072511647171791, 15802101672278393, 61315089769937015, 237913937730074175, 923150271448256549, 3581994530438293763, 13898804141498109151, 53929941802446858449, 209258191798574502199, 811960654346536991919, 3150558162336227602509, 12224750892949051210387, 47434304238914509746063, 184053911472961467551977, 714163364932527133333911] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 12 t + 12 t + 6 t - 6 t + 3 t - 4 t - 1 - ------------------------------------------------------------ 7 6 5 4 3 2 (12 t - 8 t + 2 t + 2 t + 9 t - 2 t - 4 t + 1) (t + 1) and in Maple notation -(12*t^7+12*t^6+6*t^5-6*t^4+3*t^3-4*t-1)/(12*t^7-8*t^6+2*t^5+2*t^4+9*t^3-2*t^2-\ 4*t+1)/(t+1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 16, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + x y + y + x + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 31, 7, 49, 31, 127, 7, 49, 49, 217, 31, 217, 127, 511, 7, 49, 49, 217 , 49, 343, 217, 889, 31, 217, 217, 961, 127, 889, 511, 2031, 7, 49, 49, 217, 49 , 343, 217, 889, 49] Just for kicks C(googol) equals , 164752321121957837163297617370335113695910\ 20179216238196866738014918684480804019 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 31, 127, 511, 2031, 8043, 31735, 125063, 492367, 1937763, 7624303, 29995559, 118000431, 464192219, 1826013415, 7183010967, 28255752751, 111149170563, 437224979743, 1719900889847, 6765528227247, 26613372893339, 104688284286487, 411809368955175, 1619922922644527, 6372245108777955, 25066320646301935, 98602677307103399, 387870564961339119, 1525755477446319803, 6001821189395112679, 23609194337170247479, 92870820300375594959, 365323320199954434595, 1437062016292606269055, 5652930224902411765943, 22236771805879284928335, 87472160573481473129435, 344086765028636844289367, 1353524379544585488514087] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 9 8 7 5 4 3 2 6 t - 2 t + 9 t - 2 t + 5 t - 7 t - 4 t + 2 t + 1 - ----------------------------------------------------------------------- 10 9 8 7 6 5 4 3 10 t + 22 t - 11 t + 31 t - 24 t + 3 t + 18 t - 21 t + 5 t - 1 and in Maple notation -(6*t^9-2*t^8+9*t^7-2*t^5+5*t^4-7*t^3-4*t^2+2*t+1)/(10*t^10+22*t^9-11*t^8+31*t^ 7-24*t^6+3*t^5+18*t^4-21*t^3+5*t-1) This ends this theorem, that took, 0.014, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 17, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 25, 7, 49, 25, 107, 7, 49, 49, 175, 25, 175, 107, 413, 7, 49, 49, 175 , 49, 343, 175, 749, 25, 175, 175, 625, 107, 749, 413, 1615, 7, 49, 49, 175, 49 , 343, 175, 749, 49] Just for kicks C(googol) equals , 115252299357716884324756285860116522960740\ 376590900080638490617275238037109375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 25, 107, 413, 1615, 6241, 24227, 93973, 364743, 1415289, 5491643, 21307533, 82673535, 320773713, 1244606995, 4829092037, 18736941431, 72699560681 , 282075165611, 1094454989757, 4246498377391, 16476464213569, 63928877142147, 248044803475829, 962416786448167, 3734188572465241, 14488696052099227, 56216312919224813, 218119962409118431, 846308047088580913, 3283685283337329651, 12740737934703822117, 49434214644262756823, 191805340477042460873, 744207001168233417355, 2887532011414972857117, 11203658530298171425807, 43470328282875858747169, 168665390498178293159011, 654423720170279028952021] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 3 2 16 t + 16 t - 12 t + 7 t - 2 t - 1 - --------------------------------------------- 7 6 4 3 2 16 t + 16 t - 16 t + 9 t + 3 t - 5 t + 1 and in Maple notation -(16*t^6+16*t^5-12*t^3+7*t^2-2*t-1)/(16*t^7+16*t^6-16*t^4+9*t^3+3*t^2-5*t+1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 18, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 25, 7, 49, 25, 103, 7, 49, 49, 175, 25, 175, 103, 409, 7, 49, 49, 175 , 49, 343, 175, 721, 25, 175, 175, 625, 103, 721, 409, 1639, 7, 49, 49, 175, 49 , 343, 175, 721, 49] Just for kicks C(googol) equals , 972558273237647608553997316362800310018984\ 94047600366420982778072357177734375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 25, 103, 409, 1639, 6553, 26215, 104857, 419431, 1677721, 6710887, 26843545, 107374183, 429496729, 1717986919, 6871947673, 27487790695, 109951162777, 439804651111, 1759218604441, 7036874417767, 28147497671065, 112589990684263, 450359962737049, 1801439850948199, 7205759403792793, 28823037615171175, 115292150460684697, 461168601842738791, 1844674407370955161, 7378697629483820647, 29514790517935282585, 118059162071741130343, 472236648286964521369, 1888946593147858085479, 7555786372591432341913, 30223145490365729367655, 120892581961462917470617, 483570327845851669882471, 1934281311383406679529881] Using the found enumerative automaton with, 5, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 t + 1 - ----------------- (t + 1) (4 t - 1) and in Maple notation -(4*t+1)/(t+1)/(4*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 19, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 21, 7, 49, 21, 95, 7, 49, 49, 147, 21, 147, 95, 333, 7, 49, 49, 147, 49, 343, 147, 665, 21, 147, 147, 441, 95, 665, 333, 1319, 7, 49, 49, 147, 49, 343, 147, 665, 49] Just for kicks C(googol) equals , 196686290352557863064275017100411599165471\ 0381658422259464945186318492984375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 21, 95, 333, 1319, 4837, 18447, 68733, 259447, 972565, 3661535, 13756333 , 51754567, 194586181, 731919279, 2752461533, 10352254743, 38932913525, 146424889471, 550683608589, 2071066796007, 7789015542949, 29293584500047, 110169505843517, 414334209685687, 1558259850417109, 5860425507488287, 22040342341502061, 82891039280982407, 311743071539532549, 1172427660162381551, 4409357319499233181, 16583054888865919575, 62366845725293767221, 234554096714279936959, 882129335103095923405, 3317580786194232038183, 12477016493309192526181, 46924536476035659892879, 176477455491828408944125] Using the found enumerative automaton with, 9, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 4 3 2 8 t - 28 t + 16 t - t - 1 - ------------------------------------- 5 4 3 2 8 t - 44 t + 24 t + 5 t - 6 t + 1 and in Maple notation -(8*t^4-28*t^3+16*t^2-t-1)/(8*t^5-44*t^4+24*t^3+5*t^2-6*t+1) This ends this theorem, that took, 0.015, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 20, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + x y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 31, 7, 49, 31, 145, 7, 49, 49, 217, 31, 217, 145, 601, 7, 49, 49, 217 , 49, 343, 217, 1015, 31, 217, 217, 961, 145, 1015, 601, 2551, 7, 49, 49, 217, 49, 343, 217, 1015, 49] Just for kicks C(googol) equals , 102491932901292648485438113781371441636491\ 360038512797908586703720353810714484375 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 31, 145, 601, 2551, 10351, 42433, 170761, 690247, 2768191, 11123185, 44543161, 178525591, 714455311, 2860291873, 11443638121, 45791846887, 183184681951, 732859788625, 2931560215321, 11727088287031, 46909200573871, 187642734275713, 750576869083081, 3002349000193927, 12009437524637311, 48038040765580465, 192152453729353081, 768611849586630871, 3074449433015742031, 12297811974747497953, 49191262141674521641, 196765148265489795367, 787060692760750890271, 3148243468934545522705, 12592974573629724052441, 50371903179759689941111, 201487617604279553495791, 805950504613803770102593, 3223802052651900636529801] Using the found enumerative automaton with, 3, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is (4 t + 1) (t - 1) - -------------------- 2 (4 t - 1) (7 t - 1) and in Maple notation -(4*t+1)*(t-1)/(4*t-1)/(7*t^2-1) This ends this theorem, that took, 0.007, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 21, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + x y + y + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 7, 7, 27, 7, 49, 27, 113, 7, 49, 49, 189, 27, 189, 113, 447, 7, 49, 49, 189 , 49, 343, 189, 791, 27, 189, 189, 729, 113, 791, 447, 1743, 7, 49, 49, 189, 49 , 343, 189, 791, 49] Just for kicks C(googol) equals , 645420424506004584929011636059207871314243\ 461376867193471738352674661533096929 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 7, 27, 113, 447, 1743, 6789, 26371, 102399, 397425, 1542327, 5984815, 23222957, 90110355, 349647247, 1356699401, 5264252887, 20426289087, 79257818197 , 307535089027, 1193293339871, 4630199918049, 17966035966423, 69711557106543, 270493791315325, 1049566156154579, 4072511647171791, 15802101672278393, 61315089769937015, 237913937730074175, 923150271448256549, 3581994530438293763, 13898804141498109151, 53929941802446858449, 209258191798574502199, 811960654346536991919, 3150558162336227602509, 12224750892949051210387, 47434304238914509746063, 184053911472961467551977, 714163364932527133333911] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 12 t + 12 t + 6 t - 6 t + 3 t - 4 t - 1 - ------------------------------------------------------------ 7 6 5 4 3 2 (12 t - 8 t + 2 t + 2 t + 9 t - 2 t - 4 t + 1) (t + 1) and in Maple notation -(12*t^7+12*t^6+6*t^5-6*t^4+3*t^3-4*t-1)/(12*t^7-8*t^6+2*t^5+2*t^4+9*t^3-2*t^2-\ 4*t+1)/(t+1) This ends this theorem, that took, 0.016, seconds. ------------------------------------------------------------------ This concludes this webbook, that took, 0.447, seconds. to generate. k is , 8 n On Sequences of the form, P(x, y) , modolu , 2, Evaluated at, {x = 1, y = 1} For all Polynomials that are Sums of Monomials taken from, 2 2 2 2 2 2 {1, x, y, x , y , x y, x y , x y, x y } By Shalosh B. Ekhad In this webbook, we will consider the sequences described in the title, that\ after normalization and weeding out obvious symmetry, concerns the following set of, 6, polynomials 2 2 2 2 {x y + x y + x + x y + y + x + y + 1, 2 2 2 2 2 x y + x y + x + x y + y + x + y + 1, 2 2 2 2 2 x y + x y + x y + x y + y + x + y + 1, 2 2 2 2 2 2 x y + x y + x y + x + y + x + y + 1, 2 2 2 2 2 2 x y + x y + x y + x + x y + y + y + 1, 2 2 2 2 2 2 x y + x y + x y + x + x y + y + x + y} by finding enumerative automata with at most, 200, states . ----------------------------------------------------------------------------\ ---- Theorem Number, 1, : Let C(n) be 2 2 2 2 n (x y + x y + x + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 8, 8, 28, 8, 64, 28, 128, 8, 64, 64, 224, 28, 224, 128, 480, 8, 64, 64, 224 , 64, 512, 224, 1024, 28, 224, 224, 784, 128, 1024, 480, 2008, 8, 64, 64, 224, 64, 512, 224, 1024, 64] Just for kicks C(googol) equals , 536302581750246321379201703534136557383317\ 961865997292621289158903830654065049600 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 8, 28, 128, 480, 2008, 7776, 31672, 124528, 500792, 1982144, 7926808, 31474832, 125567832, 499339552, 1989964024, 7918843056, 31542845816, 125560568192, 500031498968, 1990726198480, 7927069429912, 31561303322592, 125671455827384, 500370494068976, 1992345405195448, 7932781395312704, 31585987282333080, 125764465113945360, 500754799313981912, 1993837728988083872, 7938823894798691960, 31609774757232483632, 125859902486851922936, 501132847948360507136, 1995348198660019909720, 7944824072535027774800, 31633702928519138207512, 125955075944750743176544, 501512066699785452333880, 1996857400443241977061744] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 2 128 t - 128 t + 36 t + 49 t - 56 t + 20 t - 1 - ------------------------------------------------------------------ 8 7 6 5 4 3 2 32 t + 24 t - 164 t + 236 t - 145 t + 24 t + 16 t - 8 t + 1 and in Maple notation -(128*t^7-128*t^6+36*t^5+49*t^4-56*t^3+20*t^2-1)/(32*t^8+24*t^7-164*t^6+236*t^5 -145*t^4+24*t^3+16*t^2-8*t+1) This ends this theorem, that took, 0.093, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 2, : Let C(n) be 2 2 2 2 2 n (x y + x y + x + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 8, 8, 32, 8, 64, 32, 140, 8, 64, 64, 256, 32, 256, 140, 580, 8, 64, 64, 256 , 64, 512, 256, 1120, 32, 256, 256, 1024, 140, 1120, 580, 2300, 8, 64, 64, 256, 64, 512, 256, 1120, 64] Just for kicks C(googol) equals , 118226011774895411565724707929119896330068\ 96220794327852752523138957312000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 8, 32, 140, 580, 2300, 9212, 36708, 145684, 579132, 2300748, 9136868, 36293220, 144150716, 572523772, 2273961796, 9031663444, 35871633244, 142474173772, 565874928900, 2247525801188, 8926662348828, 35454669141436, 140817868278436, 559296518443284, 2221398396161020, 8822888544377292, 35042504230172580, 139180846736483300, 552794628537942780, 2195574381701116156, 8720321462102628996, 34635131036227958100, 137562853293842350620, 546368327161158607052, 2170050575295851005636, 8618946716466384276324, 34232493632389008179548, 135963669209776649987772, 540016732158320044870372, 2144823486344577826664468] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 3 4 t - 8 t - t - 6 t - 4 t - 1 - -------------------------------------------- 7 6 5 4 3 8 t - 8 t - 12 t + 29 t - 6 t - 4 t + 1 and in Maple notation -(4*t^6-8*t^5-t^4-6*t^3-4*t-1)/(8*t^7-8*t^6-12*t^5+29*t^4-6*t^3-4*t+1) This ends this theorem, that took, 0.019, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 3, : Let C(n) be 2 2 2 2 2 n (x y + x y + x y + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 8, 8, 28, 8, 64, 28, 128, 8, 64, 64, 224, 28, 224, 128, 480, 8, 64, 64, 224 , 64, 512, 224, 1024, 28, 224, 224, 784, 128, 1024, 480, 2008, 8, 64, 64, 224, 64, 512, 224, 1024, 64] Just for kicks C(googol) equals , 536302581750246321379201703534136557383317\ 961865997292621289158903830654065049600 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 8, 28, 128, 480, 2008, 7776, 31672, 124528, 500792, 1982144, 7926808, 31474832, 125567832, 499339552, 1989964024, 7918843056, 31542845816, 125560568192, 500031498968, 1990726198480, 7927069429912, 31561303322592, 125671455827384, 500370494068976, 1992345405195448, 7932781395312704, 31585987282333080, 125764465113945360, 500754799313981912, 1993837728988083872, 7938823894798691960, 31609774757232483632, 125859902486851922936, 501132847948360507136, 1995348198660019909720, 7944824072535027774800, 31633702928519138207512, 125955075944750743176544, 501512066699785452333880, 1996857400443241977061744] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 2 128 t - 128 t + 36 t + 49 t - 56 t + 20 t - 1 - ------------------------------------------------------------------ 8 7 6 5 4 3 2 32 t + 24 t - 164 t + 236 t - 145 t + 24 t + 16 t - 8 t + 1 and in Maple notation -(128*t^7-128*t^6+36*t^5+49*t^4-56*t^3+20*t^2-1)/(32*t^8+24*t^7-164*t^6+236*t^5 -145*t^4+24*t^3+16*t^2-8*t+1) This ends this theorem, that took, 0.017, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 4, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 8, 8, 24, 8, 64, 24, 112, 8, 64, 64, 192, 24, 192, 112, 416, 8, 64, 64, 192 , 64, 512, 192, 896, 24, 192, 192, 576, 112, 896, 416, 1728, 8, 64, 64, 192, 64 , 512, 192, 896, 64] Just for kicks C(googol) equals , 157450401983196230042883642107102147671851\ 82750941749742799889226113884197224448 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 8, 24, 112, 416, 1728, 6784, 27392, 109056, 437248, 1746944, 6991872, 27959296, 111853568, 447381504, 1789591552, 7158235136, 28633202688, 114532286464, 458130194432, 1832518680576, 7330078916608, 29320307277824, 117281245888512, 469124949999616, 1876499867107328, 7505999334211584, 30023997605281792, 120095989884256256, 480383960610766848, 1921535840295583744, 7686143365477302272, 30744573453319274496, 122978293830456967168, 491913175287468130304, 1967652701218591997952, 7870610804736929038336, 31482443219222594060288, 125929772876340620427264, 503719091506461993336832, 2014876366023648950091776] Using the found enumerative automaton with, 6, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 t + 1 - ------------------- (2 t + 1) (4 t - 1) and in Maple notation -(6*t+1)/(2*t+1)/(4*t-1) This ends this theorem, that took, 0.009, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 5, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + x y + y + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 8, 8, 32, 8, 64, 32, 140, 8, 64, 64, 256, 32, 256, 140, 580, 8, 64, 64, 256 , 64, 512, 256, 1120, 32, 256, 256, 1024, 140, 1120, 580, 2300, 8, 64, 64, 256, 64, 512, 256, 1120, 64] Just for kicks C(googol) equals , 118226011774895411565724707929119896330068\ 96220794327852752523138957312000000000000 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 8, 32, 140, 580, 2300, 9212, 36708, 145684, 579132, 2300748, 9136868, 36293220, 144150716, 572523772, 2273961796, 9031663444, 35871633244, 142474173772, 565874928900, 2247525801188, 8926662348828, 35454669141436, 140817868278436, 559296518443284, 2221398396161020, 8822888544377292, 35042504230172580, 139180846736483300, 552794628537942780, 2195574381701116156, 8720321462102628996, 34635131036227958100, 137562853293842350620, 546368327161158607052, 2170050575295851005636, 8618946716466384276324, 34232493632389008179548, 135963669209776649987772, 540016732158320044870372, 2144823486344577826664468] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 6 5 4 3 4 t - 8 t - t - 6 t - 4 t - 1 - -------------------------------------------- 7 6 5 4 3 8 t - 8 t - 12 t + 29 t - 6 t - 4 t + 1 and in Maple notation -(4*t^6-8*t^5-t^4-6*t^3-4*t-1)/(8*t^7-8*t^6-12*t^5+29*t^4-6*t^3-4*t+1) This ends this theorem, that took, 0.016, seconds. ----------------------------------------------------------------------------\ ---- Theorem Number, 6, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + x y + y + x + y) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 8, 8, 28, 8, 64, 28, 128, 8, 64, 64, 224, 28, 224, 128, 480, 8, 64, 64, 224 , 64, 512, 224, 1024, 28, 224, 224, 784, 128, 1024, 480, 2008, 8, 64, 64, 224, 64, 512, 224, 1024, 64] Just for kicks C(googol) equals , 536302581750246321379201703534136557383317\ 961865997292621289158903830654065049600 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 8, 28, 128, 480, 2008, 7776, 31672, 124528, 500792, 1982144, 7926808, 31474832, 125567832, 499339552, 1989964024, 7918843056, 31542845816, 125560568192, 500031498968, 1990726198480, 7927069429912, 31561303322592, 125671455827384, 500370494068976, 1992345405195448, 7932781395312704, 31585987282333080, 125764465113945360, 500754799313981912, 1993837728988083872, 7938823894798691960, 31609774757232483632, 125859902486851922936, 501132847948360507136, 1995348198660019909720, 7944824072535027774800, 31633702928519138207512, 125955075944750743176544, 501512066699785452333880, 1996857400443241977061744] Using the found enumerative automaton with, 10, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 7 6 5 4 3 2 128 t - 128 t + 36 t + 49 t - 56 t + 20 t - 1 - ------------------------------------------------------------------ 8 7 6 5 4 3 2 32 t + 24 t - 164 t + 236 t - 145 t + 24 t + 16 t - 8 t + 1 and in Maple notation -(128*t^7-128*t^6+36*t^5+49*t^4-56*t^3+20*t^2-1)/(32*t^8+24*t^7-164*t^6+236*t^5 -145*t^4+24*t^3+16*t^2-8*t+1) This ends this theorem, that took, 0.018, seconds. ------------------------------------------------------------------ This concludes this webbook, that took, 0.229, seconds. to generate. k is , 9 n On Sequences of the form, P(x, y) , modolu , 2, Evaluated at, {x = 1, y = 1} For all Polynomials that are Sums of Monomials taken from, 2 2 2 2 2 2 {1, x, y, x , y , x y, x y , x y, x y } By Shalosh B. Ekhad In this webbook, we will consider the sequences described in the title, that\ after normalization and weeding out obvious symmetry, concerns the following set of, 1, polynomials 2 2 2 2 2 2 {x y + x y + x y + x + x y + y + x + y + 1} by finding enumerative automata with at most, 200, states . ----------------------------------------------------------------------------\ ---- Theorem Number, 1, : Let C(n) be 2 2 2 2 2 2 n (x y + x y + x y + x + x y + y + x + y + 1) , modulo , 2 evaluated at , {x = 1, y = 1} The first, 41, terms of C(n) staring at n=0 are [1, 9, 9, 25, 9, 81, 25, 121, 9, 81, 81, 225, 25, 225, 121, 441, 9, 81, 81, 225 , 81, 729, 225, 1089, 25, 225, 225, 625, 121, 1089, 441, 1849, 9, 81, 81, 225, 81, 729, 225, 1089, 81] Just for kicks C(googol) equals , 455505931434869056749202906195805635147762\ 1208254704911743061281740665435791015625 i The first few terms of the sparse subsequence, , C(2 - 1), for i from 0 to, 40, are : [1, 9, 25, 121, 441, 1849, 7225, 29241, 116281, 466489, 1863225, 7458361, 29822521, 119311929, 477204025, 1908903481, 7635439161, 30542106169, 122167725625, 488672300601, 1954686406201, 7818751217209, 31274993684025, 125099997105721, 500399943683641, 2001599864213049, 8006399277895225, 32025597469494841, 128102389162151481, 512409558080261689, 2049638229457735225, 8198552923557563961, 32794211682777009721, 131176846754014531129, 524707386970245140025, 2098829547972606529081, 8395318191707174178361, 33581272767195200589369, 134325091068047794605625, 537300364273657193926201, 2149201457091696744697401] Using the found enumerative automaton with, 4, states, that we omit, it follows that the (rigorously) PROVED rational generating function for that sparse subsequ\ ence is 2 8 t - 6 t - 1 - --------------------------- (2 t + 1) (4 t - 1) (t - 1) and in Maple notation -(8*t^2-6*t-1)/(2*t+1)/(4*t-1)/(t-1) This ends this theorem, that took, 0.009, seconds. ------------------------------------------------------------------ This concludes this webbook, that took, 0.014, seconds. to generate.