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a(n) = n^2 * 2^n * binomial(3*n, n).
2

%I #7 Dec 07 2024 05:42:34

%S 0,6,240,6048,126720,2402400,42771456,729308160,12049956864,

%T 194372006400,3076609536000,47959947509760,738269547724800,

%U 11245075661094912,169748150676357120,2542638555345715200,37830087271621066752,559525260959878348800,8232406073859904634880,120560661522092497305600

%N a(n) = n^2 * 2^n * binomial(3*n, n).

%D Jonathan Borwein, David Bailey, and Roland Girgensohn, Experimentation in Mathematics: Computational Paths to Discovery, A K Peters, Natick, MA, 2004. See p. 26.

%H Amiram Eldar, <a href="/A378778/b378778.txt">Table of n, a(n) for n = 0..500</a>

%H Necdet Batir, <a href="https://doi.org/10.1007/BF02829799">On the series Sum_{k=1..oo} binomial(3k,k)^{-1} k^{-n} x^k</a>, Proc. Indian Acad. Sci. (Math. Sci.), Vol. 115, No. 4 (2005), pp. 371-381; <a href="https://arxiv.org/abs/math/0512310">arXiv preprint</a>, arXiv:math/0512310 [math.AC], 2005. See p. 379, eq. (3.6).

%H Jonathan M. Borwein and Roland Girgensohn, <a href="https://doi.org/10.1007/s00010-005-2774-x">Evaluations of binomial series</a>, aequationes mathematicae, Vol. 70, No. 1 (2005), pp. 25-36. See p. 32, eq. (43).

%F a(n) = A007758(n) * A005809(n).

%F a(n) = n^2 * A228484(n).

%F a(n) = n * A378780(n).

%F a(n) == 0 (mod 6).

%F Sum_{n>=1) 1/a(n) = Pi^2/24 - log(2)^2/2 (Borwein et al., 2004; Borwein and Girgensohn, 2005; Batir, 2005).

%t a[n_] := n^2 * 2^n * Binomial[3*n, n]; Array[a, 25, 0]

%o (PARI) a(n) = n^2 * 2^n * binomial(3*n, n);

%Y Cf. A005809, A007758, A228484, A378780.

%K nonn,easy,new

%O 0,2

%A _Amiram Eldar_, Dec 07 2024