A295432 a(n) = (12*n)!*(3*n)!*(2*n)!/((6*n)!*(6*n)!*(4*n)!*n!).

1, 462, 579462, 825012300, 1240292449350, 1922257130698212, 3037843525927941276, 4866407206411278522456, 7873830606510488097067590, 12837724391783995395083457780, 21058175422921386792478888300212, 34712605923460405627194955063564200

Offset

0,2

Links

Table of n, a(n) for n=0..11.

Formula

G.f.: hypergeom([1/12, 5/12, 7/12, 11/12], [1/6, 1/2, 5/6], 1728*x).

a(n) = a(n-1)*6*(12*n - 1)*(12*n - 5)*(12*n - 7)*(12*n - 11)/(n*(2*n - 1)*(6*n - 1)*(6*n - 5)). - Neven Sajko, Jul 22 2023

a(n) ~ 2^(6*n - 1) * 3^(3*n) / sqrt(Pi*n). - Vaclav Kotesovec, Jul 11 2025

a(n) = binomial(12*n,6*n)*binomial(2*n,n)/binomial(4*n,n) = binomial(12*n,6*n)*binomial(5*n,n)/binomial(5*n,2*n). - Chai Wah Wu, Feb 15 2026

From Peter Bala, Jun 10 2026: (Start)

For n >= 1, the 2-adic valuation v_2(a(n)) = s_2(n) = A000120(n), where s_2(n) is the number of 1's in the binary expansion of n. Hence a(n) is even and a(n)/2 is odd iff n is a power of 2.

For n >= 1, the 3-adic valuation v_3(a(n)) = (1/2) * s_3(2*n) >= 1, where s_3(n) is the sum of the digits in the base 3 expansion of n. Hence a(n) is divisible by 3 and v_3(a(n)) = 1 iff n is an element of A073216.

Conjecture: for n >= 1, 6*(6*n + 1)*(12*n - 1) divides a(n).(End)

Mathematica

Array[(12 #)!*(3 #)!*(2 #)!/((6 #)!*(6 #)!*(4 #)!*#!) &, 12, 0] (* Michael De Vlieger, Nov 23 2017 *)
(* Alternative: *)
CoefficientList[ Series[ HypergeometricPFQ[{1/12, 5/12, 7/12, 11/12}, {1/6, 1/2, 5/6}, 1728 x], {x, 0, 11}], x] (* Robert G. Wilson v, Nov 23 2017 *)

Prog

(Python)
from math import comb
def A295432(n): return comb(12*n, 6*n)*comb(2*n, n)//comb(4*n, n) # Chai Wah Wu, Feb 15 2026

Crossrefs

Cf. A000120, A073216, A295431.

Sequence in context: A208619 A318266 A383796 * A213406 A294853 A025038

Adjacent sequences: A295429 A295430 A295431 * A295433 A295434 A295435

Keyword

nonn,easy,changed

Author

Gheorghe Coserea, Nov 23 2017

Status

approved