A365771 a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n) for n >= 0.

1, 2, 20, 280, 4620, 84084, 1633632, 33256080, 701149020, 15191562100, 336424047960, 7584833081280, 173575987821600, 4022766574898400, 94247674040476800, 2228957491057276320, 53150802525726081660, 1276661433215969608500, 30863850087221160009000, 750478881068219785482000

Offset

0,2

Comments

Equals the central terms of triangle A365770.

Conjectures: given A033042 is the sums of distinct powers of 5, then

(1) a(5*A033042(n)) == 4 (mod 5) for n > 0,

(2) a(5*A033042(n) + 1) == 2 (mod 5) for n > 0,

(3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5*A033042(k) for some k >= 0.

Links

Paolo Xausa, Table of n, a(n) for n = 0..500

Formula

a(n) = A365770(2*n,n) for n >= 0.

a(n) = A000108(n) * A165817(n) for n >= 0.

a(n) = 2*A319578(n) = (2/3) * A007004(n) for n >= 1. - Peter Bala, Aug 25 2025

a(n) ~ 3^(3*n-1/2) / (n^2 * Pi). - Amiram Eldar, Sep 23 2025

Mathematica

A365771[n_] := Binomial[2*n + 1, n]/(2*n + 1)*Binomial[3*n - 1, n];
Array[A365771, 20, 0] (* Paolo Xausa, Oct 12 2024 *)

Prog

(PARI) {a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n)}
for(n=0, 30, print1(a(n), ", "))
(Python)
from math import comb
def A365771(n): return comb(m:=(n<<1)+1, n)*comb(m+n-2, n)//m if n else 1 # Chai Wah Wu, Oct 11 2023

Crossrefs

Cf. A007004, A000108, A165817, A033042, A319578, A356770.

Sequence in context: A303616 A231499 A217364 * A381820 A326010 A363380

Adjacent sequences: A365768 A365769 A365770 * A365772 A365773 A365774

Keyword

nonn,easy

Author

Paul D. Hanna, Oct 10 2023

Status

approved