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%I #34 Feb 07 2024 11:53:28
%S 1,0,2,4,12,16,48,60,148,220,438,618,1302,1740,3216,4788,8170,11512,
%T 19862,27570,45448,64600,100808,141724,223080,307512,465736,652518,
%U 968180,1334030,1972164,2691132,3902432,5347176,7611484,10358426,14697028,19790508,27691500
%N Number of ordered pairs of partitions of n with no common parts.
%H Reinhard Zumkeller, <a href="/A054440/b054440.txt">Table of n, a(n) for n = 0..5000</a>
%H Sylvie Corteel, Carla D. Savage, Herbert S. Wilf, Doron Zeilberger, <a href="https://doi.org/10.1006/jcta.1997.2846">A pentagonal number sieve</a>, J. Combin. Theory Ser. A 82 (1998), no. 2, 186-192.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PentagonalNumberTheorem.html">Pentagonal Number Theorem</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Pentagonal_number_theorem">Pentagonal number theorem</a>
%F G.f.: Sum[p(n)^2*x^n]/Sum[p(n)*x^n], with p(n)=number of partitions of n.
%F a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n)) / (64 * 2^(1/4) * n^(7/4)). - _Vaclav Kotesovec_, May 20 2018
%e a(3)=4 because of the 4 pairs of partitions of 3: (3,21),(3,111),(21,3),(111,3).
%p with(combinat): p1 := sum(numbpart(n)^2*x^n, n=0..500): it := p1*product((1-x^i), i=1..500): s := series(it, x, 500): for i from 0 to 100 do printf(`%d,`,coeff(s,x,i)) od:
%t nmax = 50; CoefficientList[Series[Sum[PartitionsP[k]^2*x^k, {k, 0, nmax}]/Sum[PartitionsP[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Jul 04 2016 *)
%o (Haskell)
%o a054440 = sum . zipWith (*) a087960_list . map a001255 . a260672_row
%o -- _Reinhard Zumkeller_, Nov 15 2015
%Y Cf. A000041, A001255, A001318, A087960, A260672, A260664, A260669, A304873, A304877.
%Y Main diagonal of A284592.
%K easy,nonn
%O 0,3
%A Herbert S. Wilf, May 13 2000
%E Corrected and extended by _James A. Sellers_, May 23 2000
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