A378780 a(n) = n * 2^n * binomial(3*n, n).

0, 6, 120, 2016, 31680, 480480, 7128576, 104186880, 1506244608, 21596889600, 307660953600, 4359995228160, 61522462310400, 865005820084224, 12124867905454080, 169509237023047680, 2364380454476316672, 32913250644698726400, 457355892992216924160, 6345297974846973542400

Offset: 0

Author

Amiram Eldar, Dec 07 2024

Keywords

• nonn
• easy

References

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

Links

• Amiram Eldar, Table of n, a(n) for n = 0..500
• Necdet Batir, On the series Sum_{k=1..oo} binomial(3k,k)^{-1} k^{-n} x^k, Proc. Indian Acad. Sci. (Math. Sci.), Vol. 115, No. 4 (2005), pp. 371-381; arXiv preprint, arXiv:math/0512310 [math.AC], 2005. See p. 379, eq. (3.9).
• Jonathan M. Borwein and Roland Girgensohn, Evaluations of binomial series, aequationes mathematicae, Vol. 70, No. 1 (2005), pp. 25-36. See p. 32, eq. (43).

Crossrefs

Cf. A005809, A036289, A228484.

Programs

Mathematica

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

PARI

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

Formula

a(n) = A036289(n) * A005809(n).

a(n) = n * A228484(n).

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

Sum_{n>=1} 1/a(n) = Pi/10 - log(2)/5 (Borwein et al., 2004; Borwein and Girgensohn, 2005; Batir, 2005).