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G.f.: Product_{k>=1} (1+x^k)^(3*k+2).
3

%I #10 Mar 08 2015 04:21:24

%S 1,5,18,61,182,506,1338,3369,8172,19197,43833,97636,212748,454461,

%T 953505,1968095,4001627,8024295,15885484,31074351,60111277,115071431,

%U 218126868,409662895,762679151,1408172844,2579599582,4690277001,8467363674,15182486586

%N G.f.: Product_{k>=1} (1+x^k)^(3*k+2).

%C In general, if g.f. = Product_{k>=1} (1+x^k)^(m*k+c), m > 0, then a(n) ~ (m*Zeta(3))^(1/6) * exp(-c^2 * Pi^4 / (1296*m*Zeta(3)) + (c * Pi^2 * n^(1/3)) / (2^(5/3) * 3^(4/3) * (m*Zeta(3))^(1/3)) + 3^(4/3) * (m*Zeta(3))^(1/3) * n^(2/3) / 2^(4/3)) / (2^(m/12 + c/2 + 2/3) * 3^(1/3) * sqrt(Pi) * n^(2/3)). - _Vaclav Kotesovec_, Mar 08 2015

%H Vaclav Kotesovec, <a href="/A255837/b255837.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) ~ Zeta(3)^(1/6) * exp(-Pi^4/(972*Zeta(3)) + Pi^2 * n^(1/3) / (2^(2/3) * 3^(5/3) * Zeta(3)^(1/3)) + 3^(5/3)/2^(4/3) * Zeta(3)^(1/3) * n^(2/3)) / (2^(23/12) * 3^(1/6) * sqrt(Pi) * n^(2/3)), where Zeta(3) = A002117.

%t nmax=50; CoefficientList[Series[Product[(1+x^k)^(3*k+2),{k,1,nmax}],{x,0,nmax}],x]

%Y Cf. A026007 (k), A219555 (k+1), A052812 (k-1), A255834 (2*k+1), A255835 (2*k-1), A255836 (3*k+1).

%Y Cf. A255803.

%K nonn

%O 0,2

%A _Vaclav Kotesovec_, Mar 07 2015