A391700 - OEIS

%I #35 Mar 31 2026 20:28:43

%S 1,168,120120,109830336,111767455800,120796711203168,

%T 135699814088494656,156601962871480526400,184345148248880810942520,

%U 220333212491971075631534400,266541722680469446902877305120,325611887892680108101481230360320,401008119815310928953801784481793600

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

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

%H Paolo Xausa, <a href="/A391700/b391700.txt">Table of n, a(n) for n = 0..300</a>

%F 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).

%F 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

%F   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)),

%F   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)),

%F   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)),

%F   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)),

%F   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)),

%F   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)).

%F 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.

%F a(n) ~ 2^(22*n)*5^(-5*n-1/2)/(n*Pi). - _Stefano Spezia_, Dec 18 2025

%F 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

%t A391700[n_] := (8*n)!/(n!*(2*n)!*(5*n)!); Array[A391700, 15, 0] (* _Paolo Xausa_, Dec 25 2025 *)

%o (PARI) a(n) = binomial(8*n, 3*n) * binomial(3*n, n); \\ _Michael B. Porter_, Dec 22 2025

%o (Python)

%o from math import comb

%o def A391700(n): return comb(8*n,3*n)*comb(3*n,n) # _Chai Wah Wu_, Feb 17 2026

%Y Cf. A235536, A234461, A364305.

%K nonn,easy

%O 0,2

%A _Karol A. Penson_, Dec 17 2025