%I #9 Nov 14 2017 17:41:15
%S 1,-1,-2,-4,0,3,17,24,40,9,-24,-149,-250,-435,-395,-281,514,1528,3542,
%T 5127,6920,5416,1368,-11136,-28533,-57051,-82846,-107315,-95655,
%U -43646,107826,345877,727771,1150968,1601729,1766547,1495154,183944,-2339567,-6770991,-12701854
%N Expansion of Product_{k>=1} 1/(1 + x^k)^(k*(k+1)/2).
%C Convolution inverse of A028377.
%C This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = n*(n+1)/2, g(n) = -1. - _Seiichi Manyama_, Nov 14 2017
%H Seiichi Manyama, <a href="/A294846/b294846.txt">Table of n, a(n) for n = 0..10000</a>
%F G.f.: Product_{k>=1} 1/(1 + x^k)^A000217(k).
%F 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)^(n/d). - _Seiichi Manyama_, Nov 14 2017
%t nmax = 40; CoefficientList[Series[Product[1/(1 + x^k)^(k (k + 1)/2), {k, 1, nmax}], {x, 0, nmax}], x]
%t a[n_] := a[n] = If[n == 0, 1, -Sum[Sum[(-1)^(k/d + 1) d^2 (d + 1)/2, {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 40}]
%Y Cf. A000217, A000294, A028377, A255528, A284896, A292386.
%K sign
%O 0,3
%A _Ilya Gutkovskiy_, Nov 09 2017