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%I #15 Nov 17 2017 00:48:44
%S 1,1,6,21,58,178,494,1365,3640,9533,24401,61384,151958,370335,890565,
%T 2113913,4959199,11505799,26420628,60082005,135386341,302448477,
%U 670148898,1473387787,3215519032,6968266907,14999453058,32079714584,68187859040,144083404856,302727633735,632579826174
%N Expansion of Product_{k>=1} (1 + x^k)^(k*(2*k-1)).
%C Weigh transform of the hexagonal numbers (A000384).
%C This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = -n*(2*n-1), g(n) = -1. - _Seiichi Manyama_, Nov 14 2017
%H Seiichi Manyama, <a href="/A294836/b294836.txt">Table of n, a(n) for n = 0..9392</a>
%H M. Bernstein and N. J. A. Sloane, <a href="http://arXiv.org/abs/math.CO/0205301">Some canonical sequences of integers</a>, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to arXiv version]
%H M. Bernstein and N. J. A. Sloane, <a href="/A003633/a003633_1.pdf">Some canonical sequences of integers</a>, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HexagonalNumber.html">Hexagonal Number</a>
%F G.f.: Product_{k>=1} (1 + x^k)^A000384(k).
%F a(n) ~ 7^(1/8) * exp(Pi*2^(3/2) * (7/15)^(1/4) * n^(3/4)/3 - 3*Zeta(3)*sqrt(15*n/7) / (2*Pi^2) - 135*Zeta(3)^2 * (15*n/7)^(1/4) / (28*sqrt(2)*Pi^5) - 2025*Zeta(3)^3 / (196*Pi^8)) / (2^(5/3) * 15^(1/8) * n^(5/8)). - _Vaclav Kotesovec_, Nov 10 2017
%F a(0) = 1 and a(n) = (1/n) * Sum_{k=1..n} b(k)*a(n-k) where b(n) = Sum_{d|n} d^2*(2*d-1)*(-1)^(1+n/d). - _Seiichi Manyama_, Nov 14 2017
%t nmax = 31; CoefficientList[Series[Product[(1 + x^k)^(k (2 k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
%t a[n_] := a[n] = If[n == 0, 1, Sum[Sum[(-1)^(k/d + 1) d^2 (2 d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 31}]
%Y Cf. A000384, A027998, A028377, A294102, A294837, A294838.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Nov 09 2017