A364517 a(n) = (5/7) * (9*n)!*(7*n/2)!^2/((9*n/2)!*(7*n)!*(5*n/2)!*n!^2) for n >= 1, with a(0) = 1.

1, 40, 7650, 1847560, 494944450, 140625140040, 41500392000480, 12576565436409000, 3886690320522202050, 1219380045859742166400, 387154587452271772676400, 124120231850529022319265600, 40113527971798583517288018400, 13052024252899352166622940568000

Offset: 0

Author

Peter Bala, Aug 03 2023

Keywords

• nonn
• easy

Comments

Fractional factorials are defined in terms of the gamma function; for example, (7*n/2)! = gamma(7*n/2 + 1).

Row 7 of A364513.

Links

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

Crossrefs

Cf. A364513.

Programs

Maple

seq( (5/7) * (9*n)!*(7*n/2)!^2/((9*n/2)!*(7*n)!*(5*n/2)!*n!^2), n = 1..15);

Mathematica

A364517[n_]:=If[n==0,1,(5/7)(9n)!(7n/2)!^2/((9n/2)!(7n)!(5n/2)!n!^2)];Array[A364517,15,0] (* Paolo Xausa, Oct 05 2023 *)

Python

from math import factorial
from sympy import factorial2
def A364517(n): return int(5*factorial(9*n)*factorial2(7*n)**2//factorial2(9*n)//factorial(7*n)//factorial2(5*n)//factorial(n)**2//7) if n else 1 # Chai Wah Wu, Aug 08 2023

Formula

a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(7*n-1, (1 + x)/(1 - x)) for n >= 1.

a(n) = Sum_{k = 0..n} binomial(7*n - 1, n - k)^2 * binomial(5*n + k - 2, k).

a(n) = (5/7) * binomial(9*n,2*n)*binomial(9*n/2,2*n)*binomial(2*n,n)^2 / binomial(9*n/2,n)^2 for n >= 1.

a(n) = (7*n-1)! * ((9*n-1)/2)! * ((5*n-1)/2)!/( (5*n-1)! * ((7*n-1)/2)!^2 * n!^2 ) for n >= 1.

a(n) ~ c^n * sqrt(35)/(14*Pi*n), where c = (3^9)/(5^3) * sqrt(5) = 352.1002080....

Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

For n > 0, a(n) = 5*(9*n)!*(7*n)!!^2/(7*(9*n)!!*(7*n)!*(5*n)!!*n!^2). - Chai Wah Wu, Aug 08 2023