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%I #48 May 04 2021 09:01:57
%S 1,2,11,74,602,5452,53559,558602,6106034,69298580,811086718,
%T 9740402476,119550632788,1495039156600,19002275811887,244983878813514,
%U 3198363309664658,42225545561470084,563083734161627910,7576864105285884420,102790882283750139060,1404908982711268821720
%N Row sums of A174158.
%H G. C. Greubel, <a href="/A319743/b319743.txt">Table of n, a(n) for n = 1..830</a>
%H Abderrahim Arabi, Hacène Belbachir, and Jean-Philippe Dubernard, <a href="https://arxiv.org/abs/2105.00971">Enumeration of parallelogram polycubes</a>, arXiv:2105.00971 [cs.DM], 2021.
%F a(n) = Sum_{m=1..n} (binomial(n - 1, m - 1)*binomial(n, m - 1)/m)^2.
%F a(n) = Sum_{m=1..n} A000290(A007318(n - 1, m - 1)*A007318(n, m - 1)/m).
%F a(n) = 4F3([1 - n, 1 - n, - n, - n], [1, 2, 2], 1), where F is the generalized hypergeometric function.
%F From _Vaclav Kotesovec_, Dec 24 2018: (Start)
%F Recurrence: n*(n+1)^3*(5*n^2 - 10*n + 4)*a(n) = 2*n*(2*n - 1)*(15*n^4 - 30*n^3 + 7*n^2 + 8*n - 8)*a(n-1) + 4*(n-2)^2*(4*n - 5)*(4*n - 3)*(5*n^2 - 1)*a(n-2).
%F a(n) ~ 2^(4*n + 1/2) / (Pi^(3/2) * n^(7/2)).
%F (End)
%p a := n -> add(binomial(n-1, m-1)^2*binomial(n, m-1)^2/m^2, m = 1 .. n): seq(a(n), n = 1 .. 20)
%t Table[HypergeometricPFQ[{1-n,1-n,-n,-n},{1,2,2},1],{n,1,20}]
%o (GAP) List([1..20], n->Sum([1..n], m->(Binomial(n-1, m-1)*Binomial(n, m-1)/m)^2));
%o (PARI) a(n) = sum(m=1, n, (binomial(n-1, m-1)*binomial(n, m-1)/m)^2);
%o (Sage) [hypergeometric([-n, -n, -n+1, -n+1], [1, 2, 2], 1).simplify_hypergeometric() for n in (1..25)] # _G. C. Greubel_, Feb 15 2021
%o (Magma) [(&+[(Binomial(n-1,j-1)*Binomial(n,j-1)/j)^2 : j in [1..n]]): n in [1..25]]; // _G. C. Greubel_, Feb 15 2021
%Y Cf. A000290, A007318, A174158, A174696.
%K nonn
%O 1,2
%A _Stefano Spezia_, Dec 23 2018
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