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%I #12 Nov 15 2017 09:40:44
%S 1,-1,-6,-12,6,65,179,202,-137,-1392,-3492,-5135,-1325,15437,52934,
%T 101787,116827,-16945,-462603,-1350732,-2475989,-2889620,-343236,
%U 8559858,26972213,53099230,72521956,47535918,-86985043,-409729146,-952305325,-1577038736
%N Expansion of Product_{k>=1} 1/(1 + x^k)^(k*(5*k-3)/2).
%C This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = n*(5*n-3)/2, g(n) = -1.
%H Seiichi Manyama, <a href="/A295122/b295122.txt">Table of n, a(n) for n = 0..10000</a>
%F Convolution inverse of A294837.
%F G.f.: Product_{k>=1} 1/(1 + x^k)^A000566(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*(5*d-3)*(-1)^(n/d).
%o (PARI) N=66; x='x+O('x^N); Vec(1/prod(k=1, N, (1+x^k)^(k*(5*k-3)/2)))
%Y Cf. A294846 (b=3), A284896 (b=4), A295086 (b=5), A295121 (b=6), this sequence (b=7), A295123 (b=8).
%K sign
%O 0,3
%A _Seiichi Manyama_, Nov 15 2017
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