A296164 a(n) = [x^n] Product_{k>=1} ((1 + x^k)/(1 + x^(3*k)))^n.

1, 1, 3, 10, 35, 131, 498, 1919, 7459, 29170, 114653, 452552, 1792754, 7124040, 28386081, 113372690, 453743907, 1819317153, 7306575042, 29386858821, 118348662525, 477188876405, 1926137365804, 7782398551661, 31472648050930, 127384123318906, 515978637418884

Offset: 0

Author

Ilya Gutkovskiy, Dec 06 2017

Keywords

• nonn

Links

• G. C. Greubel, Table of n, a(n) for n = 0..500
• Eric Weisstein's World of Mathematics, Schur's Partition Theorem

Crossrefs

Cf. A003105, A058484, A058539, A103262, A255526, A296163.

Programs

Mathematica

Table[SeriesCoefficient[Product[((1 + x^k)/(1 + x^(3 k)))^n, {k, 1, n}], {x, 0, n}], {n, 0, 26}]
Table[SeriesCoefficient[Product[1/((1 - x^(6 k - 1)) (1 - x^(6 k - 5)))^n, {k, 1, n}], {x, 0, n}], {n, 0, 26}]
(* Calculation of constants {d,c}: *) With[{k = 3}, {1/r, Sqrt[QPochhammer[-1, (r*s)^k] / (2*Pi*(r^2*s*Derivative[0, 2][QPochhammer][-1, r*s] - k^2*(r*s)^(2*k) * Derivative[0, 2][QPochhammer][-1, (r*s)^k] - k*(1 + k)*(r*s)^k * Derivative[0, 1][QPochhammer][-1, (r*s)^k]))]} /. FindRoot[{s == QPochhammer[-1, r*s]/QPochhammer[-1, (r*s)^k], QPochhammer[-1, (r*s)^k] + k*(r*s)^k*Derivative[0, 1][QPochhammer][-1, (r*s)^k] == r*Derivative[0, 1][QPochhammer][-1, r*s]}, {r, 1/4}, {s, 2}, WorkingPrecision -> 70]] (* Vaclav Kotesovec, Jan 17 2024 *)

Formula

a(n) = [x^n] Product_{k>=1} 1/((1 - x^(6*k-1))*(1 - x^(6*k-5)))^n.

a(n) ~ c * d^n / sqrt(n), where d = 4.129321588075726742506... and c = 0.25764349816429874321... - Vaclav Kotesovec, May 18 2018