%I #34 May 26 2019 02:09:06
%S 1,1,17,211,3049,54221,1131601,26714647,700868561,20208794329,
%T 634445325361,21512122643771,782497124407417,30364699568650981,
%U 1251108918727992689,54512805637285532671,2502891521610396838561,120718449425308259052977,6099522639316776103853521
%N E.g.f.: exp(Sum_{k>=1} k^3 * x^k).
%C In general, if e.g.f. = exp(Sum_{k>=1} k^m * x^k) and m>0, then a(n) ~ (m+2)^(-1/2) * Gamma(m+2)^(1/(2*m+4)) * exp((m+2)/(m+1) * Gamma(m+2)^(1/(m+2)) * n^((m+1)/(m+2)) + zeta(-m) - n) * n^(n - 1/(2*m+4)).
%C It appears that the sequence a(n) taken modulo 10 is periodic with period 5. More generally, we conjecture that for k = 2,3,4,... the difference a(n+k) - a(n) is divisible by k: if true, then the sequence a(n) taken modulo k would be periodic with period dividing k. - _Peter Bala_, Nov 14 2017
%C The above conjecture is true - see the Bala link. - _Peter Bala_, Jan 20 2018
%H G. C. Greubel, <a href="/A255819/b255819.txt">Table of n, a(n) for n = 0..250</a>
%H P. Bala, <a href="/A047974/a047974_1.pdf">Integer sequences that become periodic on reduction modulo k for all k</a>
%F E.g.f.: exp(x*(1 + 4*x + x^2)/(1-x)^4).
%F a(n) ~ 2^(3/10) * 3^(1/10) * 5^(-1/2) * n^(n-1/10) * exp(1/120 + 5 * 2^(-7/5) * 3^(1/5) * n^(4/5) - n).
%F a(n) = y(n)*n! where y(0)=1 and y(n)=(Sum_{k=0..n-1} (n-k)^4*y(k))/n for n>=1. - _Benedict W. J. Irwin_, Jun 02 2016
%F E.g.f.: Product_{k>=1} 1/(1 - x^k)^(J_4(k)/k), where J_4(k) is the Jordan function (A059377). - _Ilya Gutkovskiy_, May 25 2019
%t nmax=20; CoefficientList[Series[Exp[Sum[k^3*x^k,{k,1,nmax}]],{x,0,nmax}],x] * Range[0,nmax]!
%t nn = 20; Range[0, nn]! * CoefficientList[Series[Product[Exp[k^3*x^k], {k, 1, nn}], {x, 0, nn}], x] (* _Vaclav Kotesovec_, Mar 21 2016 *)
%Y Cf. A082579, A255807, A000262.
%K nonn,easy
%O 0,3
%A _Vaclav Kotesovec_, Mar 07 2015