cp's OEIS Frontend

A275654 a(n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2.

%I A275654 #23 Aug 08 2023 14:47:19
%S A275654 1,28,2646,316540,42031990,5921058528,866486466720,130220534668224,
%T A275654 19958454291525750,3105489721784166640,489023391870111994896,
%U A275654 77758775451291032116200,12464212878673327376454304,2011515147856766922731424000
%N A275654 a(n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2.
%C A275654 Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(5*n + k,5*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2.
%C A275654 We also have Sum_{k = 0..5*n} (-1)^k*binomial(5*n + k,5*n - k)* binomial(2*k,k) *binomial(2*n - k,n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2.
%C A275654 Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n). See also A275652, A275653 and A275655.
%F A275654 a(n) = (3*n/2)!*(5*n)!*(7*n/2)!/(n!^2*(3*n)!*(5*n/2)!^2).
%F A275654 Recurrence: 3*a(n)*n^2*(n - 1)^2*(3*n - 1)*(3*n - 5)*(5*n - 2)*(5*n - 4)*(5*n - 6)*(5*n - 8) = 7*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)*(7*n - 2)*(7*n - 4)*(7*n - 6)*(7*n - 8)*(7*n - 10)*(7*n - 12)*a(n-2).
%F A275654 a(n) = [x^n] G(x)^(7*n) where G(x) = 1 + 4*x + 85*x^2+ 4220*x^3 + 283285*x^4 + 22308156*x^5 + 1939419083*x^6 + ... appears to have integer coefficients.
%F A275654 exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^7, where F(x) = 1 + 4*x + 197*x^2 + 15840*x^3 + 1580819*x^4 + 178220584*x^5 + 21729476664*x^6 + ... appears to have integer coefficients.
%F A275654 a(n) ~ 7^(7*n/2+1/2)/(2*sqrt(5)*Pi*3^(3*n/2)*n). - _Ilya Gutkovskiy_, Aug 07 2016
%F A275654 From _Peter Bala_, Mar 23 2022: (Start)
%F A275654 a(n) = Sum_{k = 0..n} binomial(4*n-k-1,n-k)*binomial(5*n,k)^2.
%F A275654 For n >= 1, a(n) = (7/5)*binomial(m*n,2*n)*binomial(m*n/2,2*n)* binomial(2*n,n)^2/binomial(m*n/2,n)^2 at m = -3. Se A352651 for the case m = 1.
%F A275654 a(n) = [x^n] (1 - x)^(2*n) * P(5*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
%F A275654 a(p) == a(1) (mod p^3) for primes p >= 5.
%F A275654 Conjecture: The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)
%p A275654 seq(simplify(factorial(3*n/2)*factorial(5*n)*factorial(7*n/2)/(factorial(n)^2*factorial(3*n)*factorial(5*n/2)^2)), n = 0 .. 20);
%t A275654 Table[(5 n)!/((3 n)! n!^2) ((3 n/2)! (7 n/2)!)/(5 n/2)!^2, {n, 0, 13}] (* _Michael De Vlieger_, Aug 07 2016 *)
%o A275654 (PARI) a(n) = sum(k = 0, n, binomial(4*n-k-1,n-k)*binomial(5*n,k)^2); \\ _Michel Marcus_, Apr 21 2022
%o A275654 (Python)
%o A275654 from math import factorial
%o A275654 from sympy import factorial2
%o A275654 def A275654(n): return int(factorial(5*n)*factorial2(3*n)*factorial2(7*n)//factorial(3*n)//factorial(n)**2//factorial2(5*n)**2) # _Chai Wah Wu_, Aug 08 2023
%Y A275654 Cf. A275652, A275653, A275655, A276098, A276100, A276101, A276102, A352651, A352652.
%K A275654 nonn,easy
%O A275654 0,2
%A A275654 _Peter Bala_, Aug 04 2016