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%I #20 Dec 15 2021 01:21:37
%S 1,2,10,168,4290,136136,4938024,196125600,8318177010,370784099400,
%T 17184867259560,821870841735840,40334204896057800,2022686389717666848,
%U 103312949950998743200,5360873347802169267840,282015983963437605168210
%N Main diagonal of A132339.
%H G. C. Greubel, <a href="/A132341/b132341.txt">Table of n, a(n) for n = 0..500</a>
%H G. Kreweras, <a href="http://www.numdam.org/numdam-bin/item?id=BURO_1965__6__9_0">Sur une classe de problèmes de dénombrement liés au treillis des partitions des entiers</a>, Cahiers du Bureau Universitaire de Recherche Opérationnelle}, Institut de Statistique, Université de Paris, 6 (1965), see equation (67) circa p. 82.
%F a(n) = T(n, n), where T(n,k) if the array of A132339.
%F a(n) = A(2*n, n), where A(n, k) is the antidiagonal triangle of A132339.
%F a(n) ~ 2^(6*n - 9/2) / (Pi*n^3). - _Vaclav Kotesovec_, Mar 27 2016
%F a(n) = binomial(2*n, n)*binomial(4*n-2, 2*n-1)/((2*n)*(2*n-1)), with a(0) = 1. - _G. C. Greubel_, Dec 14 2021
%t a[n_]:= If[n==0, 1, Binomial[2*n, n]*Binomial[4*n-2, 2*n-1]/(2*Binomial[2*n,2])];
%t Table[a[n], {n, 0, 20}] (* _G. C. Greubel_, Dec 14 2021 *)
%o (PARI) a(n) = if (n, 2*(4*n-3)!/(n!^2*(2*n-1)!), 1); \\ _Michel Marcus_, Mar 27 2016
%o (Sage) b=binomial
%o def a(n): return 1 if (n==0) else b(2*n, n)*b(4*n-2, 2*n-1)/(2*b(2*n,2))
%o [a(n) for n in (0..20)] # _G. C. Greubel_, Dec 14 2021
%Y Cf. A132339.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Nov 08 2007
%E More terms from _Max Alekseyev_, Sep 12 2009
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