OFFSET
0,3
COMMENTS
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)).
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
The above conjecture is true - see the Bala link. - Peter Bala, Jan 20 2018
LINKS
FORMULA
E.g.f.: exp(x*(1 + 4*x + x^2)/(1-x)^4).
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).
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
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
MATHEMATICA
nmax=20; CoefficientList[Series[Exp[Sum[k^3*x^k, {k, 1, nmax}]], {x, 0, nmax}], x] * Range[0, nmax]!
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vaclav Kotesovec, Mar 07 2015
STATUS
approved