A318414 Expansion of Product_{i>=1, j>=1, k>=1} (1 + x^(i*j*k))^(i*j*k).

1, 1, 6, 15, 48, 108, 323, 716, 1868, 4217, 10137, 22311, 51477, 110817, 245260, 519918, 1114914, 2318557, 4854952, 9923533, 20335761, 40941170, 82365742, 163413699, 323589060, 633429923, 1236392498, 2390718266, 4606489839, 8805346615, 16768968317, 31713677061, 59747953446

Offset: 0

Author

Ilya Gutkovskiy, Aug 26 2018

Keywords

• nonn

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..10000
• Lida Ahmadi, Ricardo Gómez Aíza, and Mark Daniel Ward, A unified treatment of families of partition functions, La Matematica (2024). Preprint available as arXiv:2303.02240 [math.CO], 2023.

Crossrefs

Cf. A000005, A007425, A034718, A280473, A280541, A318413.

Programs

Maple

a:=series(mul(mul(mul((1+x^(i*j*k))^(i*j*k),k=1..55),j=1..55),i=1..55),x=0,33): seq(coeff(a,x,n),n=0..32); # Paolo P. Lava, Apr 02 2019

Mathematica

nmax = 32; CoefficientList[Series[Product[Product[Product[(1 + x^(i j k))^(i j k), {i, 1, nmax}], {j, 1, nmax}], {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 32; CoefficientList[Series[Product[(1 + x^k)^(k Sum[DivisorSigma[0, d], {d, Divisors[k]}]), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 32; CoefficientList[Series[Exp[Sum[Sum[(-1)^(k/d + 1) d^2 Sum[DivisorSigma[0, j], {j, Divisors[d]}], {d, Divisors[k]}]  x^k/k, {k, 1, nmax}]], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[(-1)^(k/d + 1) d^2 Sum[DivisorSigma[0, j], {j, Divisors[d]}], {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 32}]
nmax = 32; A034718 = Table[n*Sum[DivisorSigma[0, d], {d, Divisors[n]}], {n, 1, nmax}]; s = 1 + x; Do[s *= Sum[Binomial[A034718[[k]], j]*x^(j*k), {j, 0, nmax/k}]; s = Expand[s]; s = Take[s, Min[nmax + 1, Exponent[s, x] + 1, Length[s]]]; , {k, 2, nmax}]; Take[CoefficientList[s, x], nmax + 1] (* Vaclav Kotesovec, Aug 31 2018 *)

Formula

G.f.: Product_{k>=1} (1 + x^k)^(k*tau_3(k)), where tau_3() = A007425.

G.f.: exp(Sum_{k>=1} ( Sum_{d|k} (-1)^(k/d+1)*d^2 * Sum_{j|d} tau(j) ) * x^k/k), where tau() = A000005.

Conjecture: log(a(n)) ~ 3^(2/3) * Zeta(3)^(1/3) * log(n)^(2/3) * n^(2/3) / 2^(5/3). - Vaclav Kotesovec, Sep 02 2018