On computing the Mod, 2, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (1/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 33, : Let A(n) be the constant term, in x, of n (3/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (3/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 0 all the congruences classes mod, 2, show up Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 2 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 2, equals , 1 The congruence classes mod, 2, in the following set , {0}, never show up! ------------------------------------------ This ends this fascinating book that took, 0.079, to generate. ----------------------------------------- ----------------------------------------------------- On computing the Mod, 4, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {2, 3}, never show up! Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (3 + 1/x + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {2, 3}, never show up! Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {2, 3}, never show up! Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {2, 3}, never show up! Theorem Number, 33, : Let A(n) be the constant term, in x, of n (1 + 3/x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (1 + 3/x + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 4, equals , 0 The congruence classes mod, 4, in the following set , {3}, never show up! Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 4 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 4, equals , 1 The congruence classes mod, 4, in the following set , {0, 2}, never show up! ------------------------------------------ This ends this fascinating book that took, 0.093, to generate. ----------------------------------------- ----------------------------------------------------- On computing the Mod, 8, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 14, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 5, 6, 7}, never show up! Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 5, 6, 7}, never show up! Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 25, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {2, 3, 5, 6, 7}, never show up! Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {2, 3, 5, 7}, never show up! Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 32, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 6, 7}, never show up! Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 50, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 5, 6, 7}, never show up! Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 9, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 5, 7}, never show up! Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 4, 5, 6, 7}, never show up! Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 11, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 5, 7}, never show up! Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 25, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (3 + 1/x + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 49, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 54, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 50, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {2, 3, 5, 6, 7}, never show up! Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {2, 3, 5, 6, 7}, never show up! Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 32, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 6, 7}, never show up! Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 32, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 6, 7}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 4, 5, 6, 7}, never show up! Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 4, 5, 6, 7}, never show up! Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 54, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7}, never show up! Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 54, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {2, 3, 5, 7}, never show up! Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {2, 3, 5, 6, 7}, never show up! Theorem Number, 33, : Let A(n) be the constant term, in x, of n (1 + 3/x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 14, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 50, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 32, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 6, 7}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (1 + 3/x + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 49, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 5, 6, 7}, never show up! Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 11, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 5, 7}, never show up! Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 4, 5, 6, 7}, never show up! Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 9, states . For example, A(100000), mudolo , 8, equals , 0 The congruence classes mod, 8, in the following set , {3, 5, 7}, never show up! Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 50, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 8 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 54, states . For example, A(100000), mudolo , 8, equals , 1 The congruence classes mod, 8, in the following set , {0, 2, 4, 6}, never show up! ------------------------------------------ This ends this fascinating book that took, 0.232, to generate. ----------------------------------------- ----------------------------------------------------- On computing the Mod, 16, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 11, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 52, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 21, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 21, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 181, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 17, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {2, 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 29, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {2, 3, 5, 7, 9, 10, 11, 13, 14, 15}, never show up! Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 155, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15}, never show up! Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 366, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 6, 8, 10, 12, 14}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 21, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 34, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 5, 7, 9, 10, 11, 13, 14, 15}, never show up! Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 19, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 41, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 5, 7, 9, 11, 13, 14, 15}, never show up! Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 181, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (3 + 1/x + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 357, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 6, 8, 10, 12, 14}, never show up! Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 198, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 366, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 6, 8, 10, 12, 14}, never show up! Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {2, 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15}, never show up! Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 11, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 155, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15}, never show up! Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 12, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 155, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 19, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 17, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 17, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 198, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 15, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 198, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 6, 8, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 29, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {2, 3, 5, 7, 9, 10, 11, 13, 14, 15}, never show up! Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15}, never show up! Theorem Number, 33, : Let A(n) be the constant term, in x, of n (1 + 3/x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 52, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 366, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 6, 8, 10, 12, 14}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 155, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (1 + 3/x + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 357, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 6, 8, 10, 12, 14}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 21, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 41, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 5, 7, 9, 11, 13, 14, 15}, never show up! Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 17, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up! Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 36, states . For example, A(100000), mudolo , 16, equals , 0 The congruence classes mod, 16, in the following set , {3, 5, 7, 9, 10, 11, 13, 14, 15}, never show up! Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 366, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 6, 8, 10, 12, 14}, never show up! Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 16 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 198, states . For example, A(100000), mudolo , 16, equals , 1 The congruence classes mod, 16, in the following set , {0, 2, 4, 6, 8, 9, 10, 11, 12, 13, 14, 15}, never show up! ------------------------------------------ This ends this fascinating book that took, 1.920, to generate. ----------------------------------------- ----------------------------------------------------- On computing the Mod, 3, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 2 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 0 all the congruences classes mod, 3, show up Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 2 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (1/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 0 all the congruences classes mod, 3, show up Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 2 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 2 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 0 all the congruences classes mod, 3, show up Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 33, : Let A(n) be the constant term, in x, of n (3/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (3/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0, 2}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 3, equals , 1 The congruence classes mod, 3, in the following set , {0}, never show up! Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 3 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 3, equals , 0 The congruence classes mod, 3, in the following set , {2}, never show up! ------------------------------------------ This ends this fascinating book that took, 0.087, to generate. ----------------------------------------- ----------------------------------------------------- On computing the Mod, 9, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 31, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 14, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 3, 5, 8}, never show up! Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 3, 4, 5, 7, 8}, never show up! Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 33, states . For example, A(100000), mudolo , 9, equals , 2 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 13, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 8}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 31, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 23, states . For example, A(100000), mudolo , 9, equals , 0 all the congruences classes mod, 9, show up Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 33, states . For example, A(100000), mudolo , 9, equals , 5 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 23, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 4, 5, 6}, never show up! Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (1/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 27, states . For example, A(100000), mudolo , 9, equals , 0 all the congruences classes mod, 9, show up Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 18, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 5, 8}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 7, 8}, never show up! Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 14, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 3, 5, 8}, never show up! Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8}, never show up! Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 8, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 33, states . For example, A(100000), mudolo , 9, equals , 2 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 16, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 5, 8}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 13, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 5, 6, 7, 8}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 33, states . For example, A(100000), mudolo , 9, equals , 5 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 23, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 4, 5, 6}, never show up! Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 18, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 5, 8}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 27, states . For example, A(100000), mudolo , 9, equals , 0 all the congruences classes mod, 9, show up Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8}, never show up! Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 3, 4, 5, 7, 8}, never show up! Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8}, never show up! Theorem Number, 33, : Let A(n) be the constant term, in x, of n (3/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 13, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 8}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 13, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 5, 6, 7, 8}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (3/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 9, equals , 1 The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 23, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 4, 5, 6}, never show up! Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 23, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 4, 5, 6}, never show up! Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 9, equals , 7 The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up! Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 7, 8}, never show up! Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 9 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 3, states . For example, A(100000), mudolo , 9, equals , 0 The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8}, never show up! ------------------------------------------ This ends this fascinating book that took, 0.198, to generate. ----------------------------------------- ----------------------------------------------------- On computing the Mod, 27, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 62, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 691, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 22, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 173, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 3, 5, 8, 11, 12, 14, 17, 20, 21, 23, 26}, never show up! Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 714, states . For example, A(100000), mudolo , 27, equals , 20 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 170, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 691, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 280, states . For example, A(100000), mudolo , 27, equals , 0 all the congruences classes mod, 27, show up Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 714, states . For example, A(100000), mudolo , 27, equals , 14 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 322, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 4, 5, 6, 7, 9, 11, 12, 13, 14, 15, 16, 18, 20, 21, 22, 23, 24}, never show up! Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (1/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 351, states . For example, A(100000), mudolo , 27, equals , 0 all the congruences classes mod, 27, show up Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 240, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 5, 8, 11, 14, 17, 20, 23, 26}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 173, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 3, 5, 8, 11, 12, 14, 17, 20, 21, 23, 26}, never show up! Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 62, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 714, states . For example, A(100000), mudolo , 27, equals , 20 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 180, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 5, 8, 11, 14, 17, 20, 23, 26}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 164, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 714, states . For example, A(100000), mudolo , 27, equals , 14 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 322, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 21, 22, 23, 24}, never show up! Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 240, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 5, 8, 11, 14, 17, 20, 23, 26}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 354, states . For example, A(100000), mudolo , 27, equals , 0 all the congruences classes mod, 27, show up Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26}, never show up! Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 33, : Let A(n) be the constant term, in x, of n (3/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 5, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 170, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 164, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (3/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 6, states . For example, A(100000), mudolo , 27, equals , 1 The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 22, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 322, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 4, 5, 6, 7, 9, 11, 12, 13, 14, 15, 16, 18, 20, 21, 22, 23, 24}, never show up! Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 322, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 21, 22, 23, 24}, never show up! Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 22, states . For example, A(100000), mudolo , 27, equals , 25 The congruence classes mod, 27, in the following set , {0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up! Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up! Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 27 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 7, states . For example, A(100000), mudolo , 27, equals , 0 The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26}, never show up! ------------------------------------------ This ends this fascinating book that took, 8.294, to generate. ----------------------------------------- ----------------------------------------------------- On computing the Mod, 5, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0, 2, 3, 4}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0, 2, 3, 4}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 5, equals , 0 The congruence classes mod, 5, in the following set , {2, 3}, never show up! Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 5, equals , 0 The congruence classes mod, 5, in the following set , {2, 3}, never show up! Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 0 all the congruences classes mod, 5, show up Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 2 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 4 all the congruences classes mod, 5, show up Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 2 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 0 all the congruences classes mod, 5, show up Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (1/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 4 all the congruences classes mod, 5, show up Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 2 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 0 all the congruences classes mod, 5, show up Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 5, equals , 0 The congruence classes mod, 5, in the following set , {2, 3}, never show up! Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 0 all the congruences classes mod, 5, show up Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0, 2, 3, 4}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 0 all the congruences classes mod, 5, show up Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 5, equals , 4 The congruence classes mod, 5, in the following set , {2, 3}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 3 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 2 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 4 all the congruences classes mod, 5, show up Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 2 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 3 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 4 all the congruences classes mod, 5, show up Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 5, equals , 0 The congruence classes mod, 5, in the following set , {2, 3}, never show up! Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 0 all the congruences classes mod, 5, show up Theorem Number, 33, : Let A(n) be the constant term, in x, of n (3/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 1, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0, 2, 3, 4}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 2 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 3 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (3/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 2, states . For example, A(100000), mudolo , 5, equals , 4 The congruence classes mod, 5, in the following set , {2, 3}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 1 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 0 all the congruences classes mod, 5, show up Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 4 all the congruences classes mod, 5, show up Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 3 The congruence classes mod, 5, in the following set , {0}, never show up! Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 0 all the congruences classes mod, 5, show up Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 5 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 4, states . For example, A(100000), mudolo , 5, equals , 4 all the congruences classes mod, 5, show up ------------------------------------------ This ends this fascinating book that took, 0.105, to generate. ----------------------------------------- ----------------------------------------------------- On computing the Mod, 25, of Many Interesting sequences by Shalosh B. Ekhad Theorem Number, 1, : Let A(n) be the constant term, in x, of n (1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 18, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24}, never show up! Theorem Number, 2, : Let A(n) be the constant term, in x, of n (1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 14, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24}, never show up! Theorem Number, 3, : Let A(n) be the constant term, in x, of n (2 + x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 4, : Let A(n) be the constant term, in x, of n (2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 5, : Let A(n) be the constant term, in x, of n (3 + x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 261, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 6, : Let A(n) be the constant term, in x, of n (3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 7, : Let A(n) be the constant term, in x, of n (1/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 67, states . For example, A(100000), mudolo , 25, equals , 15 The congruence classes mod, 25, in the following set , {2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up! Theorem Number, 8, : Let A(n) be the constant term, in x, of n (1/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 67, states . For example, A(100000), mudolo , 25, equals , 15 The congruence classes mod, 25, in the following set , {2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up! Theorem Number, 9, : Let A(n) be the constant term, in x, of n (1/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 224, states . For example, A(100000), mudolo , 25, equals , 0 all the congruences classes mod, 25, show up Theorem Number, 10, : Let A(n) be the constant term, in x, of n (1/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 17 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 11, : Let A(n) be the constant term, in x, of n (1/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 12, : Let A(n) be the constant term, in x, of n (1/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 113, states . For example, A(100000), mudolo , 25, equals , 9 all the congruences classes mod, 25, show up Theorem Number, 13, : Let A(n) be the constant term, in x, of n (1/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 17 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 14, : Let A(n) be the constant term, in x, of n (1/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 221, states . For example, A(100000), mudolo , 25, equals , 0 all the congruences classes mod, 25, show up Theorem Number, 15, : Let A(n) be the constant term, in x, of n (1/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 261, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 16, : Let A(n) be the constant term, in x, of n (1/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261, 212601015, 1024642875, 4950790605 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 117, states . For example, A(100000), mudolo , 25, equals , 9 all the congruences classes mod, 25, show up Theorem Number, 17, : Let A(n) be the constant term, in x, of n (1/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 17 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 18, : Let A(n) be the constant term, in x, of n (1/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 221, states . For example, A(100000), mudolo , 25, equals , 0 all the congruences classes mod, 25, show up Theorem Number, 19, : Let A(n) be the constant term, in x, of n (2/x + x) For the record, the first 15 terms of the sequence are: 0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 67, states . For example, A(100000), mudolo , 25, equals , 15 The congruence classes mod, 25, in the following set , {2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up! Theorem Number, 20, : Let A(n) be the constant term, in x, of n (2/x + 3 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 117, states . For example, A(100000), mudolo , 25, equals , 15 all the congruences classes mod, 25, show up Theorem Number, 21, : Let A(n) be the constant term, in x, of n (2/x + 1) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 18, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24}, never show up! Theorem Number, 22, : Let A(n) be the constant term, in x, of n (2/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545, 17760965, 65753693 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 224, states . For example, A(100000), mudolo , 25, equals , 0 all the congruences classes mod, 25, show up Theorem Number, 23, : Let A(n) be the constant term, in x, of n (2/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 67, states . For example, A(100000), mudolo , 25, equals , 9 The congruence classes mod, 25, in the following set , {2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up! Theorem Number, 24, : Let A(n) be the constant term, in x, of n (2/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 13 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 25, : Let A(n) be the constant term, in x, of n (2/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 17 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 26, : Let A(n) be the constant term, in x, of n (2/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 117, states . For example, A(100000), mudolo , 25, equals , 9 all the congruences classes mod, 25, show up Theorem Number, 27, : Let A(n) be the constant term, in x, of n (2/x + 3) For the record, the first 15 terms of the sequence are: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 28, : Let A(n) be the constant term, in x, of n (2/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 17 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 29, : Let A(n) be the constant term, in x, of n (2/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491, 2081258529, 14011445763, 94594402353, 640188979299 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 261, states . For example, A(100000), mudolo , 25, equals , 13 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 30, : Let A(n) be the constant term, in x, of n (2/x + 3 + 3 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 117, states . For example, A(100000), mudolo , 25, equals , 9 all the congruences classes mod, 25, show up Theorem Number, 31, : Let A(n) be the constant term, in x, of n (3/x + x) For the record, the first 15 terms of the sequence are: 0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 67, states . For example, A(100000), mudolo , 25, equals , 15 The congruence classes mod, 25, in the following set , {2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up! Theorem Number, 32, : Let A(n) be the constant term, in x, of n (3/x + 2 x) For the record, the first 15 terms of the sequence are: 0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 117, states . For example, A(100000), mudolo , 25, equals , 15 all the congruences classes mod, 25, show up Theorem Number, 33, : Let A(n) be the constant term, in x, of n (1 + 3/x) For the record, the first 15 terms of the sequence are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 14, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24}, never show up! Theorem Number, 34, : Let A(n) be the constant term, in x, of n (3/x + 1 + x) For the record, the first 15 terms of the sequence are: 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 17 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 35, : Let A(n) be the constant term, in x, of n (3/x + 1 + 2 x) For the record, the first 15 terms of the sequence are: 1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653, 221446369, 1245188881, 7130897437, 40516456357 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 13 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 36, : Let A(n) be the constant term, in x, of n (3/x + 1 + 3 x) For the record, the first 15 terms of the sequence are: 1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709, 1714206781, 11255897485, 77419522675, 517370395015 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 67, states . For example, A(100000), mudolo , 25, equals , 9 The congruence classes mod, 25, in the following set , {2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up! Theorem Number, 37, : Let A(n) be the constant term, in x, of n (3/x + 2) For the record, the first 15 terms of the sequence are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 1 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 38, : Let A(n) be the constant term, in x, of n (3/x + 2 + x) For the record, the first 15 terms of the sequence are: 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 221, states . For example, A(100000), mudolo , 25, equals , 0 all the congruences classes mod, 25, show up Theorem Number, 39, : Let A(n) be the constant term, in x, of n (3/x + 2 + 2 x) For the record, the first 15 terms of the sequence are: 2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232, 1562588416, 10370408960, 69019648000, 460456939520 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 117, states . For example, A(100000), mudolo , 25, equals , 9 all the congruences classes mod, 25, show up Theorem Number, 40, : Let A(n) be the constant term, in x, of n (3/x + 2 + 3 x) For the record, the first 15 terms of the sequence are: 2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344, 8981891164, 69111416792, 533463087928, 4126851588176 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 265, states . For example, A(100000), mudolo , 25, equals , 13 The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20}, never show up! Theorem Number, 41, : Let A(n) be the constant term, in x, of n (3/x + 3 + x) For the record, the first 15 terms of the sequence are: 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 221, states . For example, A(100000), mudolo , 25, equals , 0 all the congruences classes mod, 25, show up Theorem Number, 42, : Let A(n) be the constant term, in x, of n (3/x + 3 + 2 x) For the record, the first 15 terms of the sequence are: 3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223, 8503237521, 64604559555, 492221474325, 3759348384855 We are interested in A(n) modulo , 25 Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\ size, i.e. log-time in n) with , 117, states . For example, A(100000), mudolo , 25, equals , 9 all the congruences classes mod, 25, show up ------------------------------------------ This ends this fascinating book that took, 8.567, to generate. ----------------------------------------- The whole thing took, 19.611, seconds.