Linear Recurrences For the Sums of the , 1, Power over The Characters of the Symmetric Group on n elements over shapes with n cells and , 3, rows, where mu is mostly 1's, where the non-one part of mu is a pa\ rtition with at most , 7, cells. By Shalosh B. Ekhad ----------------------------------------------------- Theorem number, 1, : Let , A[2](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 1, cells and at most , 3, rows with mu equal to, [2], followed by , n - 1, ones. A[2](n), satisfies the following homogeneous linear recurrence of order, 2, with polynomial coefficients. 3 (n + 1) n A[2](n) (2 n + 3) (n + 1) A[2](n + 1) - ------------------- - ----------------------------- + A[2](n + 2) = 0 (n + 5) (n - 1) (n + 5) (n - 1) and in Maple format: -3*(n+1)*n/(n+5)/(n-1)*A[2](n)-(2*n+3)*(n+1)/(n+5)/(n-1)*A[2](n+1)+A[2](n+2) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [2] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 1, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 2 `` [0, 0, 1, 3, 9, 25, 69, 189, 518, 1422, 3915, 10813, 29964, 83304, 232323, 649845, 1822824, 5126520, 14453451, 40843521, 115668105, 328233969, 933206967, 2657946907, 7583013474, 21668135850, 62007732605, 177696228411, 509899901553, 1464990733969, 4214045993925] `` ----------------------------------------------------- Theorem number, 2, : Let , A[3](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 2, cells and at most , 3, rows with mu equal to, [3], followed by , n - 1, ones. A[3](n), satisfies the following homogeneous linear recurrence of order, 2, with polynomial coefficients. 2 3 n (n - 1) (n - 9) A[3](n) n (2 n - 17 n - 39) A[3](n + 1) - --------------------------- - -------------------------------- + A[3](n + 2) (n - 10) (n - 3) (n + 6) (n - 10) (n - 3) (n + 6) = 0 and in Maple format: -3*n*(n-1)*(n-9)/(n-10)/(n-3)/(n+6)*A[3](n)-n*(2*n^2-17*n-39)/(n-10)/(n-3)/(n+6 )*A[3](n+1)+A[3](n+2) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [3] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 2, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 3 `` [1, 0, 0, 0, 1, 5, 19, 64, 203, 622, 1866, 5523, 16203, 47256, 137280, 397761, 1150539, 3324504, 9600648, 27718398, 80026803, 231089229, 667508779, 1928892218 , 5576500765, 16130167306, 46682405630, 135180435130, 391675097385, 1135512672021, 3293913149595] `` ----------------------------------------------------- Theorem number, 3, : Let , A[4](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 3, cells and at most , 3, rows with mu equal to, [4], followed by , n - 1, ones. A[4](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 3 n (n + 1) A[4](n) (n + 1) (n - 3) A[4](n + 1) ------------------- - --------------------------- (n + 8) (n - 4) (n + 8) (n - 4) 2 (3 n + 9 n - 29) A[4](n + 2) - ----------------------------- + A[4](n + 3) = 0 (n + 8) (n - 4) and in Maple format: 3*n*(n+1)/(n+8)/(n-4)*A[4](n)-(n+1)*(n-3)/(n+8)/(n-4)*A[4](n+1)-(3*n^2+9*n-29)/ (n+8)/(n-4)*A[4](n+2)+A[4](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [4] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 3, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 4 `` [1, 1, 1, 1, 1, 1, 2, 8, 35, 139, 508, 1750, 5787, 18591, 58488, 181182, 554763 , 1683579, 5074161, 15210717, 45402819, 135065035, 400705726, 1186215088, 3505433957, 10344491501, 30491979521, 89798474041, 264264722881, 777252626881, 2285032818916] `` ----------------------------------------------------- Theorem number, 4, : Let , A[2, 2](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 3, cells and at most , 3, rows with mu equal to, [2, 2], followed by , n - 1, ones. A[2, 2](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 6 (n + 2) A[2, 2](n) (n - 1) A[2, 2](n + 1) (4 n + 23) A[2, 2](n + 2) -------------------- + ---------------------- - ------------------------- n + 8 n + 8 n + 8 + A[2, 2](n + 3) = 0 and in Maple format: 6*(n+2)/(n+8)*A[2,2](n)+(n-1)/(n+8)*A[2,2](n+1)-(4*n+23)/(n+8)*A[2,2](n+2)+A[2, 2](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [2, 2] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 3, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 4 `` [1, 1, 3, 7, 19, 51, 140, 386, 1071, 2983, 8338, 23376, 65715, 185199, 523134, 1480872, 4200411, 11936619, 33981063, 96897759, 276739029, 791532973, 2267119660, 6502108902, 18671460905, 53680763201, 154507444731, 445190930863, 1284064525987, 3707234094819, 10713124806766] `` ----------------------------------------------------- Theorem number, 5, : Let , A[5](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 4, cells and at most , 3, rows with mu equal to, [5], followed by , n - 1, ones. A[5](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 2 (n + 1) (8 n - 173 n - 150) A[5](n) - ------------------------------------ (n + 9) (19 n - 44) 3 2 (16 n - 135 n - 1111 n - 2010) A[5](n + 1) - 1/3 -------------------------------------------- (n + 9) (19 n - 44) 3 2 (8 n - 255 n - 2009 n + 816) A[5](n + 2) + 1/3 ------------------------------------------ + A[5](n + 3) = 0 (n + 9) (19 n - 44) and in Maple format: -(n+1)*(8*n^2-173*n-150)/(n+9)/(19*n-44)*A[5](n)-1/3*(16*n^3-135*n^2-1111*n-\ 2010)/(n+9)/(19*n-44)*A[5](n+1)+1/3*(8*n^3-255*n^2-2009*n+816)/(n+9)/(19*n-44)* A[5](n+2)+A[5](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [5] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 4, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 5 `` [1, 1, 2, 3, 5, 8, 13, 21, 35, 64, 142, 397, 1299, 4512, 15744, 54069, 181899, 600033, 1945770, 6219537, 19643911, 61430916, 190529837, 586880099, 1797337669, 5477709741, 16625771516, 50285853907, 151639856413, 456110403290, 1368899080473 ] `` ----------------------------------------------------- Theorem number, 6, : Let , A[3, 2](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 4, cells and at most , 3, rows with mu equal to, [3, 2], followed by , n - 1, ones. A[3, 2](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 9 (n + 1) (n - 6) A[3, 2](n) 3 (n - 2) (n + 1) A[3, 2](n + 1) ---------------------------- + -------------------------------- (n + 9) (n - 12) (n + 9) (n - 12) 2 5 ((n - 2) + n - 68) A[3, 2](n + 2) - ------------------------------------ + A[3, 2](n + 3) = 0 (n + 9) (n - 12) and in Maple format: 9*(n+1)*(n-6)/(n+9)/(n-12)*A[3,2](n)+3*(n-2)*(n+1)/(n+9)/(n-12)*A[3,2](n+1)-5*( (n-2)^2+n-68)/(n+9)/(n-12)*A[3,2](n+2)+A[3,2](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [3, 2] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 4, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 5 `` [0, 0, 1, 3, 9, 26, 75, 216, 622, 1791, 5157, 14850, 42768, 123201, 355017, 1023426, 2951640, 8517102, 24590007, 71035623, 205330321, 593874660, 1718716329 , 4977165776, 14422071018, 41815623870, 121314227125, 352162477251, 1022887805553, 2972770926004, 8644463486751] `` ----------------------------------------------------- For the case where mu is, [6], with , n - 6, ones, and the summation is over the, 1, power of all shapes with n cells and at most , 3, rows , there is no linear recurrence of ORDER+DEGREE <= , 16 For the sake of the OEIS, here are the first, 31, terms, starting with , n = 6 `` [1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 513, 1035, 2124, 4512, 10167, 24897, 66993, 196113, 609468, 1959490, 6392684, 20913340, 68173790, 220795930, 709719005, 2263959301, 7169785496, 22555568662, 70532873794, 219378933578, 679079371306] `` ----------------------------------------------------- Theorem number, 8, : Let , A[4, 2](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 5, cells and at most , 3, rows with mu equal to, [4, 2], followed by , n - 1, ones. A[4, 2](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 2 (n + 2) (3 n - 405 n - 140) A[4, 2](n) -3/2 --------------------------------------- (n + 10) (99 n - 134) 3 2 (6 n - 195 n + 257 n + 3790) A[4, 2](n + 1) - 1/2 --------------------------------------------- (n + 10) (99 n - 134) 3 2 (3 n - 786 n - 5097 n + 5630) A[4, 2](n + 2) + 1/2 ---------------------------------------------- + A[4, 2](n + 3) = 0 (n + 10) (99 n - 134) and in Maple format: -3/2*(n+2)*(3*n^2-405*n-140)/(n+10)/(99*n-134)*A[4,2](n)-1/2*(6*n^3-195*n^2+257 *n+3790)/(n+10)/(99*n-134)*A[4,2](n+1)+1/2*(3*n^3-786*n^2-5097*n+5630)/(n+10)/( 99*n-134)*A[4,2](n+2)+A[4,2](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [4, 2] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 5, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 6 `` [-1, -1, -1, -1, 0, 4, 19, 69, 230, 734, 2287, 7017, 21306, 64206, 192399, 574053, 1707003, 5062395, 14981385, 44259397, 130575656, 384803636, 1133003781, 3333623587, 9802996519, 28814514999, 84667774799, 248723181119, 730527565154, 2145384673910, 6300047764439] `` ----------------------------------------------------- Theorem number, 9, : Let , A[3, 3](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 5, cells and at most , 3, rows with mu equal to, [3, 3], followed by , n - 1, ones. A[3, 3](n), satisfies the following homogeneous linear recurrence of order, 5, with polynomial coefficients. 9 n A[3, 3](n) 3 (2 n + 3) A[3, 3](n + 1) 3 (4 n + 21) A[3, 3](n + 2) - -------------- - -------------------------- + --------------------------- n + 12 n + 12 n + 12 3 (n + 6) A[3, 3](n + 3) (5 n + 48) A[3, 3](4 + n) + ------------------------ - ------------------------- + A[3, 3](n + 5) = n + 12 n + 12 0 and in Maple format: -9*n/(n+12)*A[3,3](n)-3*(2*n+3)/(n+12)*A[3,3](n+1)+3*(4*n+21)/(n+12)*A[3,3](n+2 )+3*(n+6)/(n+12)*A[3,3](n+3)-(5*n+48)/(n+12)*A[3,3](4+n)+A[3,3](n+5) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [3, 3] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 5, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 6 `` [3, 1, 2, 4, 10, 26, 70, 192, 534, 1500, 4245, 12081, 34530, 99024, 284727, 820419, 2368071, 6845121, 19810764, 57396286, 166446290, 483091798, 1403191348, 4078580366, 11862720535, 34524293913, 100534596756, 292916438922, 853883068564, 2490404277620, 7266905670496] `` ----------------------------------------------------- Theorem number, 10, : Let , A[2, 2, 2](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 5, cells and at most , 3, rows with mu equal to, [2, 2, 2], followed by , n - 1, ones. A[2, 2, 2](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 6 (n + 3) A[2, 2, 2](n) (n - 3) A[2, 2, 2](n + 1) ----------------------- + ------------------------- n + 10 n + 10 (4 n + 29) A[2, 2, 2](n + 2) - ---------------------------- + A[2, 2, 2](n + 3) = 0 n + 10 and in Maple format: 6*(n+3)/(n+10)*A[2,2,2](n)+(n-3)/(n+10)*A[2,2,2](n+1)-(4*n+29)/(n+10)*A[2,2,2]( n+2)+A[2,2,2](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [2, 2, 2] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 5, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 6 `` [1, 1, 5, 13, 38, 106, 299, 841, 2372, 6700, 18963, 53769, 152736, 434604, 1238667, 3535797, 10107825, 28935633, 82943511, 238054915, 684053714, 1967869582, 5667243101, 16337841391, 47145918329, 136176041401, 393682664261, 1139105042845, 3298656617128, 9559852863548, 27726297377971] `` ----------------------------------------------------- For the case where mu is, [7], with , n - 7, ones, and the summation is over the, 1, power of all shapes with n cells and at most , 3, rows , there is no linear recurrence of ORDER+DEGREE <= , 16 For the sake of the OEIS, here are the first, 31, terms, starting with , n = 7 `` [1, 1, 2, 4, 9, 20, 45, 101, 227, 510, 1146, 2575, 5787, 13014, 29316, 66276, 150843, 347277, 814114, 1959490, 4885243, 12708988, 34621557, 98596424, 291683517, 888505481, 2761694276, 8691491422, 27533253169, 87432917714, 277570161985] `` ----------------------------------------------------- Theorem number, 12, : Let , A[5, 2](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 6, cells and at most , 3, rows with mu equal to, [5, 2], followed by , n - 1, ones. A[5, 2](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 2 3 (n + 2) (3241 (n - 1) - 90903 n - 62297) A[5, 2](n) ------------------------------------------------------ 2 (n + 11) (2101 (n - 1) - 51042 n + 107023) 3 2 (179 (n - 1) - 56238 (n - 1) - 528751 n - 1592927) A[5, 2](n + 1) + ------------------------------------------------------------------- 2 (n + 11) (2101 (n - 1) - 51042 n + 107023) 3 2 (7443 (n - 1) - 129420 (n - 1) - 1761025 n + 982891) A[5, 2](n + 2) - --------------------------------------------------------------------- 2 (n + 11) (2101 (n - 1) - 51042 n + 107023) + A[5, 2](n + 3) = 0 and in Maple format: 3*(n+2)*(3241*(n-1)^2-90903*n-62297)/(n+11)/(2101*(n-1)^2-51042*n+107023)*A[5,2 ](n)+(179*(n-1)^3-56238*(n-1)^2-528751*n-1592927)/(n+11)/(2101*(n-1)^2-51042*n+ 107023)*A[5,2](n+1)-(7443*(n-1)^3-129420*(n-1)^2-1761025*n+982891)/(n+11)/(2101 *(n-1)^2-51042*n+107023)*A[5,2](n+2)+A[5,2](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [5, 2] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 6, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 7 `` [0, -1, -1, -2, -3, -5, -7, -6, 14, 113, 505, 1914, 6720, 22581, 73761, 236235, 745704, 2327997, 7204837, 22143094, 67668005, 205820425, 623577471, 1883034403, 5670352034, 17034310875, 51068148599, 152830690464, 456678273893, 1362818372507 , 4062271972345] `` ----------------------------------------------------- Theorem number, 13, : Let , A[4, 3](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 6, cells and at most , 3, rows with mu equal to, [4, 3], followed by , n - 1, ones. A[4, 3](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 3 2 3 (n + 1) (154323 n - 11650819 n + 75396568 n - 344166084) A[4, 3](n) - ----------------------------------------------------------------------- - 2 2 (n + 11) (4672813 n - 38579405 n + 226658458) 4 3 2 (154323 n - 4255792 n + 42622293 n + 106223758 n - 673053510) / 2 A[4, 3](n + 1) / ((n + 11) (4672813 n - 38579405 n + 226658458)) + / 4 3 2 (154323 n - 20379153 n + 8526567 n + 616782617 n - 7969343066) / 2 A[4, 3](n + 2) / ((n + 11) (4672813 n - 38579405 n + 226658458)) / + A[4, 3](n + 3) = 0 and in Maple format: -3*(n+1)*(154323*n^3-11650819*n^2+75396568*n-344166084)/(n+11)/(4672813*n^2-\ 38579405*n+226658458)*A[4,3](n)-2*(154323*n^4-4255792*n^3+42622293*n^2+ 106223758*n-673053510)/(n+11)/(4672813*n^2-38579405*n+226658458)*A[4,3](n+1)+( 154323*n^4-20379153*n^3+8526567*n^2+616782617*n-7969343066)/(n+11)/(4672813*n^2 -38579405*n+226658458)*A[4,3](n+2)+A[4,3](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [4, 3] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 6, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 7 `` [1, 1, 1, 2, 5, 14, 40, 115, 331, 954, 2754, 7965, 23079, 66990, 194754, 566970 , 1652523, 4821405, 14079061, 41142772, 120306367, 351983798, 1030306808, 3017150282, 8838837349, 25902775261, 75934396801, 222670360396, 653146023637, 1916355333340, 5624121393254] `` ----------------------------------------------------- Theorem number, 14, : Let , A[3, 2, 2](n), be the sum of the, 1, powers of the values of the symmetric group characters over all shapes with , n + 6, cells and at most , 3, rows with mu equal to, [3, 2, 2], followed by , n - 1, ones. A[3, 2, 2](n), satisfies the following homogeneous linear recurrence of order, 3, with polynomial coefficients. 2 3 (n + 2) (485 n + 563 n - 16932) A[3, 2, 2](n) ------------------------------------------------ 2 (n + 11) (217 n - 446 n - 11264) 3 2 (319 n + 651 n - 9934 n + 1464) A[3, 2, 2](n + 1) + --------------------------------------------------- 2 (n + 11) (217 n - 446 n - 11264) 3 2 (919 n + 6219 n - 54818 n - 370248) A[3, 2, 2](n + 2) - ------------------------------------------------------- 2 (n + 11) (217 n - 446 n - 11264) + A[3, 2, 2](n + 3) = 0 and in Maple format: 3*(n+2)*(485*n^2+563*n-16932)/(n+11)/(217*n^2-446*n-11264)*A[3,2,2](n)+(319*n^3 +651*n^2-9934*n+1464)/(n+11)/(217*n^2-446*n-11264)*A[3,2,2](n+1)-(919*n^3+6219* n^2-54818*n-370248)/(n+11)/(217*n^2-446*n-11264)*A[3,2,2](n+2)+A[3,2,2](n+3) = 0 Sketch of (semi-rigorous) proof: By using the constant term expression for the character with shape, [a[1], a[2], a[3]], and mu being, [3, 2, 2] with as many as needed ones appended to make it a partition of, a[1] + a[2] + a[3] It is (fully rigorously!) derived that that number is given by a certain c\ losed-form expression that we spare you. It follows by the fundamental theorem of WZ theory that the sum of the, 1, power of these over all shapes, [a[1], a[2], a[3]], that sum to , n + 6, satisfies SOME linear recurrence with polynomial coefficients whose order can be easily bound. This justifies looking for such a recurrenc\ e by guessing, that nevertheless, can be easily justified, if desired. For the sake of the OEIS, here are the first, 31, terms, starting with , n = 7 `` [1, 1, 3, 8, 23, 66, 190, 547, 1575, 4536, 13068, 37665, 108615, 313392, 904788 , 2613822, 7555803, 21855609, 63259075, 183214018, 530967009, 1539733118, 4467739466, 12971481834, 37682979385, 109534023001, 318562851051, 926995314898, 2698921634743, 7861897185348, 22913048366756] `` ----------------------------------------------------- This concludes this article, that took, 173.086, seconds. to generate.