A028377 Expansion of Product_{m>0} (1+q^m)^(m(m+1)/2).
1, 1, 3, 9, 19, 46, 100, 218, 460, 965, 1975, 3993, 7975, 15712, 30650, 59150, 113093, 214300, 402812, 751165, 1390714, 2557004, 4670770, 8479232, 15302657, 27462424, 49021252, 87057783, 153850769, 270614429, 473850031, 826125184, 1434286323, 2480145226
Offset: 0
Keywords
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Alois P. Heinz)
Crossrefs
Programs
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Maple
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(binomial(i*(i+1)/2, j)*b(n-i*j, i-1), j=0..n/i))) end: a:= n-> b(n$2): seq(a(n), n=0..50); # Alois P. Heinz, Aug 03 2013
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Mathematica
b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[Binomial[i*(i+1)/2, j]*b[n-i*j, i-1], {j, 0, n/i}]]]; a[n_] := b[n, n]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Oct 13 2014, after Alois P. Heinz *)
Formula
a(n) ~ 7^(1/8) * exp(2 * 7^(1/4) * Pi * n^(3/4) / (3^(5/4) * 5^(1/4)) + 3^(3/2) * 5^(1/2) * Zeta(3) * n^(1/2) / (2 * 7^(1/2) * Pi^2) - 3^(13/4) * 5^(5/4) * Zeta(3)^2 * n^(1/4) / (4 * 7^(5/4) * Pi^5) + 2025 * Zeta(3)^3 / (98*Pi^8)) / (2^(49/24) * 15^(1/8) * n^(5/8)), where Zeta(3) = A002117. - Vaclav Kotesovec, Mar 11 2015
a(0) = 1 and a(n) = (1/(2*n)) * Sum_{k=1..n} b(k)*a(n-k) where b(n) = Sum_{d|n} d^2*(d+1)*(-1)^(1+n/d). - Seiichi Manyama, Nov 14 2017
G.f.: exp(Sum_{k>=1} (-1)^(k+1)*x^k/(k*(1 - x^k)^3)). - Ilya Gutkovskiy, May 28 2018
Comments