# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a001459 Showing 1-1 of 1 %I A001459 #36 Feb 03 2022 00:51:09 %S A001459 1,30,3150,420420,62355150,9816086280,1605660228900,269764879032000, %T A001459 46225898052627150,8042050347997165500,1415997888807961859400, %U A001459 251762943910387780962000,45125969443194371927422500,8143514687130622653091029120,1478138194032735032800001630400 %N A001459 a(n) = (5*n)!/((2*n)!*(2*n)!*n!). %H A001459 James Spahlinger, Table of n, a(n) for n = 0..400 %H A001459 Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011. %F A001459 From _Ilya Gutkovskiy_, Feb 07 2017: (Start) %F A001459 O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1/2,1/2,1; 3125*x/16). %F A001459 E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1/2,1/2,1,1; 3125*x/16). %F A001459 a(n) ~ 5^(5*n+1/2)/(4*Pi*16^n*n). (End) %F A001459 From _Peter Bala_, Sep 20 2021: (Start) %F A001459 a(n) = 5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/(4*n^2*(2*n - 1)^2)*a(n-1). %F A001459 a(n) = Sum_{k = n..3*n} binomial(3*n,k)^2*binomial(k,n). Cf. A006480. %F A001459 Congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(5*n,2*n)*C(3*n,n) and apply Mestrovic, equation 39, p. 12). (End) %p A001459 f := n->(5*n)!/((2*n)!*(2*n)!*n!); %t A001459 Table[(5 n)!/((2 n)! (2 n)!*n!), {n, 0, 12}] (* _Michael De Vlieger_, Feb 07 2017 *) %Y A001459 Cf. A006480. %K A001459 nonn %O A001459 0,2 %A A001459 _N. J. A. Sloane_ # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE