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 A275654 a(n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2. 7
 1, 28, 2646, 316540, 42031990, 5921058528, 866486466720, 130220534668224, 19958454291525750, 3105489721784166640, 489023391870111994896, 77758775451291032116200, 12464212878673327376454304, 2011515147856766922731424000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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. 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. 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. LINKS Table of n, a(n) for n=0..13. FORMULA a(n) = (3*n/2)!*(5*n)!*(7*n/2)!/(n!^2*(3*n)!*(5*n/2)!^2). 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). 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. 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. a(n) ~ 7^(7*n/2+1/2)/(2*sqrt(5)*Pi*3^(3*n/2)*n). - Ilya Gutkovskiy, Aug 07 2016 From Peter Bala, Mar 23 2022: (Start) a(n) = Sum_{k = 0..n} binomial(4*n-k-1,n-k)*binomial(5*n,k)^2. 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. 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. a(p) == a(1) (mod p^3) for primes p >= 5. 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) MAPLE 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); MATHEMATICA 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 *) PROG (PARI) a(n) = sum(k = 0, n, binomial(4*n-k-1, n-k)*binomial(5*n, k)^2); \\ Michel Marcus, Apr 21 2022 (Python) from math import factorial from sympy import factorial2 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 CROSSREFS Cf. A275652, A275653, A275655, A276098, A276100, A276101, A276102, A352651, A352652. Sequence in context: A100548 A103660 A317799 * A283519 A284770 A107444 Adjacent sequences: A275651 A275652 A275653 * A275655 A275656 A275657 KEYWORD nonn,easy AUTHOR Peter Bala, Aug 04 2016 STATUS approved

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Last modified July 19 10:10 EDT 2024. Contains 374392 sequences. (Running on oeis4.)