A295456 a(n) = (30*n)!*(5*n)!*(4*n)!/((15*n)!*(12*n)!*(10*n)!*(2*n)!).

1, 168043980, 126694219977836700, 109504706026534324525391988, 100204869963549181630558779565943580, 94632263448378916462441320194245442445186480, 91178760822805067678742350221792424866578707819103300, 89078657321704573467701346299333700225381618490806719145843000

Offset

0,2

Comments

Conjecture: for n >= 0, both a(n/2) = (15*n)!*(5*n/2)!*(2*n)! / ((15*n/2)!*(6*n)!*(5*n)!*n!) and a(n/3) = (10*n)!*(5*n/3)!*(4*n/3)! / ((5*n)!*(4*n)!*(10*n/3)!*(2*n/3)!) are integers, where fractional factorials are defined using the Gamma function; for example (5*n/2)! := Gamma(5*n/2 + 1). - Peter Bala, Jun 13 2026

Links

Table of n, a(n) for n=0..7.

Formula

G.f.: hypergeom([1/30, 7/30, 11/30, 13/30, 17/30, 19/30, 23/30, 29/30], [1/12, 1/3, 5/12, 1/2, 7/12, 2/3, 11/12], 1054687500*x).

a(n) = binomial(30*n,15*n)*binomial(15*n,3*n)*binomial(3*n,n)/(binomial(10*n,5*n)*binomial(5*n,n)). - Chai Wah Wu, Feb 15 2026

a(n) ~ 2^(2*n-1) * 3^(3*n-1/2) * 5^(10*n) / sqrt(Pi*n). - Amiram Eldar, Feb 22 2026

Mathematica

a[n_] := (30*n)!*(5*n)!*(4*n)!/((15*n)!*(12*n)!*(10*n)!*(2*n)!); Array[a, 8, 0] (* Amiram Eldar, Feb 22 2026 *)

Prog

(Python)
from math import comb
def A295456(n): return comb(30*n, 15*n)*comb(15*n, 3*n)*comb(3*n, n)//(comb(10*n, 5*n)*comb(5*n, n)) # Chai Wah Wu, Feb 15 2026

Crossrefs

Cf. A295431.

Sequence in context: A162839 A251473 A102337 * A172567 A172590 A172693

Adjacent sequences: A295453 A295454 A295455 * A295457 A295458 A295459

Keyword

nonn,easy

Author

Gheorghe Coserea, Nov 27 2017

Status

approved