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%I #16 Jun 24 2022 19:58:29
%S 1,2,6,15,40,89,210,435,919,1819,3583,6719,12648,22849,41074,72050,
%T 125411,213620,361845,601945,995074,1622338,2626342,4201367,6681992,
%U 10515756,16449852,25509952,39333476,60172701,91577517,138390481,208096282,310976731,462512831
%N a(n) = Sum_{k=0..n} p(k)^2, where p(k) is the partition function A000041.
%C In general, Sum_{k=0..n} p(k)^m ~ sqrt(6*n)/(m*Pi) * p(n)^m ~ exp(m*Pi*sqrt(2*n/3)) / (m * Pi * 3^((m-1)/2) * 2^(2*m-1/2) * n^(m-1/2)), for m >= 1.
%H Alois P. Heinz, <a href="/A259399/b259399.txt">Table of n, a(n) for n = 0..5000</a>
%F a(n) ~ exp(2*Pi*sqrt(2*n/3)) / (16*sqrt(6)*Pi*n^(3/2)).
%F a(n) = 1 + A209536(n). - _Alois P. Heinz_, Oct 21 2018
%p a:= proc(n) option remember; `if`(n<0, 0,
%p combinat[numbpart](n)^2+a(n-1))
%p end:
%p seq(a(n), n=0..40); # _Alois P. Heinz_, Oct 21 2018
%t Table[Sum[PartitionsP[k]^2,{k,0,n}],{n,0,50}]
%Y Cf. A000041, A000070, A209536, A265093.
%Y Partial sums of A001255.
%K nonn
%O 0,2
%A _Vaclav Kotesovec_, Jun 26 2015
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