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%I #19 Jul 16 2021 21:04:07
%S 1,2,7,25,92,343,1292,4902,18703,71677,275694,1063636,4114131,
%T 15948762,61946290,241013869,939125870,3664299332,14314777054,
%U 55982787136,219158088711,858728875776,3367576480747,13216392846128,51905939548950,203989227456894,802164259099114
%N a(n) = [x^n] 1/(1-x)^n * Product_{k=1..n} 1/(1-x^k).
%C Number of ways to pick n units in all partitions of 2n - _Olivier Gérard_, May 07 2020
%H Alois P. Heinz, <a href="/A292613/b292613.txt">Table of n, a(n) for n = 0..1663</a> (first 301 terms from Vaclav Kotesovec)
%F a(n) ~ c * 4^n / sqrt(Pi*n), where c = 1/(2*QPochhammer[1/2, 1/2]) = 1.7313733097275318057689... - _Vaclav Kotesovec_, Sep 20 2017
%F a(n) = A292508(n,n+1). - _Alois P. Heinz_, Jul 16 2021
%e Illustration of comment for n=3, a(3)=25 :
%e Among the 11 integer partitions of 6, 3 have at least 3 ones.
%e 3,1,1,1 ; 2,1,1,1,1; 1,1,1,1,1,1;
%e There are respectively 1, 4 and 20 ways to pick 3 of these.
%t Table[SeriesCoefficient[1/(1-x)^n*Product[1/(1-x^k), {k, 1, n}], {x, 0, n}], {n, 0, 30}]
%Y Cf. A000041, A001700, A014329, A292463.
%Y Cf. A292508, A322211.
%K nonn
%O 0,2
%A _Vaclav Kotesovec_, Sep 20 2017
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