A391700 a(n) = (8*n)!/(n!*(2*n)!*(5*n)!).

1, 168, 120120, 109830336, 111767455800, 120796711203168, 135699814088494656, 156601962871480526400, 184345148248880810942520, 220333212491971075631534400, 266541722680469446902877305120, 325611887892680108101481230360320, 401008119815310928953801784481793600

Offset

0,2

References

P. Bachmann, "Niedere Zahlen Theorie", Erster Teil, B. G. Teubner (1902), Zweiter Teil (1910).

Links

Paolo Xausa, Table of n, a(n) for n = 0..300

Formula

G.f.: hypergeom([1/8, 1/4, 3/8, 5/8, 3/4, 7/8], [1/5, 2/5, 3/5, 4/5, 1], (4194304*x)/3125).

a(n) = Integral_{x=0..4194304/3125} x^n*W(x) dx, n>=0, where W(x) = W1(x) + W2(x) + W3(x) + W4(x) + W5(x) + W6(x), where

W1(x) = 2^(3/4)*hypergeom([1/8, 1/8, 13/40, 21/40, 29/40, 37/40], [1/4, 3/8, 1/2, 3/4, 7/8], (3125*x)/4194304)/(16*sqrt(Pi)*Gamma(3/4)^2*x^(7/8)),

W2(x) = sec((3*Pi)/8)*csc((3*Pi)/8)*sqrt(2)*hypergeom([1/4, 1/4, 9/20, 13/20, 17/20, 21/20], [3/8, 1/2, 5/8, 7/8, 9/8], (3125*x)/4194304)/(128*sqrt(Pi)*Gamma(3/4)^2*x^(3/4)),

W3(x) = -(7*Gamma(3/4)^2*2^(1/4)*hypergeom([3/8, 3/8, 23/40, 31/40, 39/40, 47/40], [1/2, 5/8, 3/4, 9/8, 5/4], (3125*x)/4194304))/(512*Pi^(5/2)*x^(5/8)),

W4(x) = (51*2^(3/4)*hypergeom([5/8, 5/8, 33/40, 41/40, 49/40, 57/40], [3/4, 7/8, 5/4, 11/8, 3/2], (3125*x)/4194304))/(262144*sqrt(Pi)*Gamma(3/4)^2*x^(3/8)),

W5(x) = -(77*sec((3*Pi)/8)*csc((3*Pi)/8)*Gamma(3/4)^2*sqrt(2)*hypergeom([3/4, 3/4, 19/20, 23/20, 27/20, 31/20], [7/8, 9/8, 11/8, 3/2, 13/8], (3125*x)/4194304))/(327680*Pi^(5/2)*x^(1/4)),

W6(x) = -(5643*Gamma(3/4)^2*2^(1/4)*hypergeom([7/8, 7/8, 43/40, 51/40, 59/40, 67/40], [9/8, 5/4, 3/2, 13/8, 7/4], (3125*x)/4194304))/(41943040*Pi^(5/2)*x^(1/8)).

W(x) is positive on the interval [0, 4194304/3125], it diverges at x=0, monotonically decreases for x>0 and W(4194304/3125) = 0. This integral representation is unique, as W(x) is the solution of the Hausdorff moment problem.

a(n) ~ 2^(22*n)*5^(-5*n-1/2)/(n*Pi). - Stefano Spezia, Dec 18 2025

a(n) = binomial(8*n,3*n)*binomial(3*n,n) = binomial(8*n,2*n)*binomial(6*n,n) = binomial(8*n,n)*binomial(7*n,2*n). - Chai Wah Wu, Mar 31 2026

Mathematica

A391700[n_] := (8*n)!/(n!*(2*n)!*(5*n)!); Array[A391700, 15, 0] (* Paolo Xausa, Dec 25 2025 *)

Prog

(PARI) a(n) = binomial(8*n, 3*n) * binomial(3*n, n); \\ Michael B. Porter, Dec 22 2025
(Python)
from math import comb
def A391700(n): return comb(8*n, 3*n)*comb(3*n, n) # Chai Wah Wu, Feb 17 2026

Crossrefs

Cf. A235536, A234461, A364305.

Sequence in context: A275456 A185404 A146200 * A159394 A008699 A223125

Adjacent sequences: A391697 A391698 A391699 * A391701 A391702 A391703

Keyword

nonn,easy

Author

Karol A. Penson, Dec 17 2025

Status

approved