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A279219 Expansion of Product_{k>=1} 1/(1 - x^k)^(k*(k+1)*(2*k-1)/2). 7

%I #11 May 08 2017 00:22:14

%S 1,1,10,40,155,560,2051,7080,24064,79370,257067,815593,2545201,

%T 7812699,23639459,70551216,207932549,605611061,1744513262,4973116444,

%U 14038641287,39263308551,108849552289,299248060986,816159923366,2209102273109,5936069692320,15840122529455,41987363787469,110584436073149

%N Expansion of Product_{k>=1} 1/(1 - x^k)^(k*(k+1)*(2*k-1)/2).

%C Euler transform of the octagonal pyramidal numbers (A002414).

%H Vaclav Kotesovec, <a href="/A279219/b279219.txt">Table of n, a(n) for n = 0..1000</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 <a href="/index/Ps#pyramidal_numbers">Index to sequences related to pyramidal numbers</a>

%F G.f.: Product_{k>=1} 1/(1 - x^k)^(k*(k+1)*(2*k-1)/2).

%F a(n) ~ exp(-Zeta'(-1)/2 - Zeta(3)/(8*Pi^2) - Pi^16/(671846400000*Zeta(5)^3) - Pi^8*Zeta(3)/(5184000*Zeta(5)^2) - Zeta(3)^2/(240*Zeta(5)) + Zeta'(-3) + (Pi^12/(388800000*2^(3/5)*3^(1/5)*Zeta(5)^(11/5)) + Pi^4*Zeta(3)/(3600*2^(3/5) * 3^(1/5)*Zeta(5)^(6/5))) * n^(1/5) + (-Pi^8/(432000*2^(1/5)*3^(2/5)*Zeta(5)^(7/5)) - Zeta(3)/(2^(11/5)*(3*Zeta(5))^(2/5))) * n^(2/5) + (Pi^4/(180*2^(4/5)*(3*Zeta(5))^(3/5))) * n^(3/5) + ((5*(3*Zeta(5))^(1/5))/(2^(7/5))) * n^(4/5)) * (3*Zeta(5))^(9/100) / (2^(23/100) * sqrt(5*Pi) * n^(59/100)). - _Vaclav Kotesovec_, Dec 08 2016

%t nmax=29; CoefficientList[Series[Product[1/(1 - x^k)^(k (k + 1) (2 k - 1)/2), {k, 1, nmax}], {x, 0, nmax}], x]

%Y Cf. A000335, A002414, A279215, A279216, A279217, A279218.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Dec 08 2016

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