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A378805 a(n) = n^2 * 2^n * binomial(4*n, n).

%I A378805 #12 Apr 26 2025 06:00:41
%S A378805 0,8,448,15840,465920,12403200,310109184,7426298880,172331827200,
%T A378805 3904310108160,86800438067200,1900523591622656,41092173674053632,
%U A378805 879143252561100800,18640526785000243200,392189380800086016000,8196121945202522849280,170275478835299912515584,3519020291600950115696640
%N A378805 a(n) = n^2 * 2^n * binomial(4*n, n).
%H A378805 Amiram Eldar, <a href="/A378805/b378805.txt">Table of n, a(n) for n = 0..500</a>
%H A378805 Necdet Batir and Anthony Sofo, <a href="http://dx.doi.org/10.1016/j.amc.2013.05.053">On some series involving reciprocals of binomial coefficients</a>, Appl. Math. Comp., Vol. 220 (2013), pp. 331-338.
%F A378805 a(n) = A007758(n) * A005810(n).
%F A378805 a(n) = 2^n * A378803(n).
%F A378805 a(n) = n * A378804(n).
%F A378805 a(n) == 0 (mod 8).
%F A378805 Sum_{n>=1} 1/a(n) = -(3/2)*log((c-1)/(c+1))^2 + (3/4) * arctan(2*sqrt(c^2+2*c)/(c^2+2*c-1))^2 + (3/4) * arctan(2*sqrt(c^2-2*c)/(c^2-2*c-1))^2 = 0.12729750445123620540..., where c = sqrt(1 + (16*sqrt(2/3))*cos(arctan(sqrt(485/27))/3)) (Batir and Sofo, 2013, p. 336, Example 1).
%t A378805 a[n_] := n^2 * 2^n * Binomial[4*n, n]; Array[a, 20, 0]
%o A378805 (PARI) a(n) = n^2 * 2^n * binomial(4*n, n);
%Y A378805 Cf. A005810, A007758, A378803, A378804.
%K A378805 nonn,easy
%O A378805 0,2
%A A378805 _Amiram Eldar_, Dec 07 2024