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E.g.f.: exp(Sum_{k>=1} k^2 * x^k).
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%I #40 May 26 2019 02:09:24

%S 1,1,9,79,841,10821,162601,2777419,52960209,1112813641,25509407401,

%T 632772511911,16870674740569,480717000225229,14568646143888201,

%U 467640968478534691,15841420612530533281,564519727866573515409,21102817266052772063689,825435163723385398719871

%N E.g.f.: exp(Sum_{k>=1} k^2 * 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="/A255807/b255807.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+x)/(1-x)^3).

%F a(n) ~ 2^(-7/8) * 3^(1/8) * n^(n-1/8) * exp(2^(9/4) * 3^(-3/4) * n^(3/4) - n).

%F a(n) = n!*y(n) where y(0)=1 and y(n)=(Sum_{k=0..n-1} (n-k)^3*y(k))/n for n>=1. - _Benedict W. J. Irwin_, Jun 02 2016

%F a(n) = (4*n-3)*a(n-1) - 2*(n-1)*(3*n-8)*a(n-2) + (n-1)*(n-2)*(4*n-11)*a(n-3) - (n-1)*(n-2)*(n-3)*(n-4)*a(n-4). - _Peter Bala_, Nov 12 2017

%F E.g.f.: Product_{k>=1} 1/(1 - x^k)^(J_3(k)/k), where J_3() is the Jordan function (A059376). - _Ilya Gutkovskiy_, May 25 2019

%t nmax=20; CoefficientList[Series[Exp[Sum[k^2*x^k,{k,1,nmax}]],{x,0,nmax}],x] * Range[0,nmax]!

%t nn = 20; Range[0, nn]! * CoefficientList[Series[Product[Exp[k^2*x^k], {k, 1, nn}], {x, 0, nn}], x] (* _Vaclav Kotesovec_, Mar 21 2016 *)

%Y Cf. A082579, A255819, A000262.

%K nonn,easy

%O 0,3

%A _Vaclav Kotesovec_, Mar 07 2015