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A365026
a(n) = (5*n)!*(9*n/2)!*(n/2)! / ((2*n)!^2 * (5*n/2)!^2 * n!).
3
1, 126, 79380, 65523780, 60634147860, 59774707082376, 61346313465418800, 64736852770959042240, 69724035322703253191700, 76277370761329867481375100, 84482032811073922526904281880, 94508142285721995026811874069200, 106599928449546340546215262030974000
OFFSET
0,2
COMMENTS
Fractional factorials are defined in terms of the gamma function; for example, (9*n/2)! = Gamma(1 + 9*n/2).
Row 2 of A365025.
LINKS
FORMULA
a(n) = Sum_{j = 0..2*n} binomial(5*n, 2*n-j)^2 * binomial(n+j-1, j).
P-recursive: (5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(2*n)^2*(2*n-1)^2*(2*n-2)^2*(2*n-3)^2*a(n)= 9*(9*n-2)*(9*n-4)*(9*n-6)*(9*n-8)*(9*n-10)*(9*n-12)*(9*n-14)*(9*n-16)*(5*n-1)*(5*n-3)*(5*n-7)*(5*n-9)*a(n-2) with a(0) = 1 and a(1) = 126.
a(n) ~ c^n * 3*sqrt(5)/(20*Pi*n), where c = (3^9)/(2^4).
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all integers n and r.
a(n) = [x^n] G(x)^(9*n), where the power series G(x) = 1 + 14*x + 2744*x^2 + 1130724*x^3 + 615596785*x^4 + 388901411712*x^5 + 269588153179744*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^9, where the power series F(x) = 1 + 14*x + 4508*x^2 + 2489004*x^3 + 1728415009*x^4 + 1362984972918*x^5 + 1165343050808188*x^6 + ... appears to have integer coefficients.
MAPLE
seq( simplify((5*n)!*(9*n/2)!*(n/2)! / ((2*n)!^2 * (5*n/2)!^2 * n!)), n = 0..15);
MATHEMATICA
A365026[n_]:=(5n)!(9n/2)!(n/2)!/((2n)!^2(5n/2)!^2n!); Array[A365026, 15, 0] (* Paolo Xausa, Oct 05 2023 *)
PROG
(Python)
from math import factorial
from sympy import factorial2
def A365026(n): return int(factorial(5*n)*factorial2(9*n)*factorial2(n)//((factorial2(5*n)*factorial(n<<1))**2*factorial(n))) # Chai Wah Wu, Aug 24 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Aug 17 2023
STATUS
approved