|
%I #38 Mar 26 2023 02:13:02
%S 1,12,240,5376,126720,3075072,76038144,1905131520,48199827456,
%T 1228623052800,31504481648640,811751838842880,20999667135283200,
%U 545086744471535616,14189559697354260480,370298578584748425216,9684502341534993088512,253765034617761850982400
%N a(n) = 4^n*(3*n)!/((2*n)!*n!).
%D W. A. Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 35.
%D The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972; Eq. 3.115, page 35.
%H Vincenzo Librandi, <a href="/A006588/b006588.txt">Table of n, a(n) for n = 0..710</a>
%H M. Le Brun, <a href="/A006577/a006577.pdf">Email to N. J. A. Sloane, Jul 1991</a>
%F a(n) = Sum_{k=0..n} C(4*n+1, 2*n-2*k)*C(n+k, k) = 4^n*C(3*n, n).
%F a(n) ~ (1/2)*3^(1/2)*Pi^(-1/2)*n^(-1/2)*3^(3*n)*{1 - (7/72)*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
%F a(n) = A013609(3*n, 2*n). - _Johannes W. Meijer_, Aug 22 2013
%F a(n) = [x^n] hypergeom([1/3, 2/3], [1/2], 27*x). - _Peter Luschny_, Sep 30 2018
%F a(n) = Sum_{k = n..3*n} binomial(3*n,k)*binomial(k,n). - _Peter Bala_, Mar 25 2023
%p A006588 := n->add( binomial(4*n+1,2*n-2*k)*binomial(n+k,k),k=0..n): seq(A006588(n), n=0..15);
%p h := proc(x) hypergeom([1/3, 2/3], [1/2], 27*x) end: ser := series(h(x), x, 20): seq(coeff(ser, x, n), n=0..15); # _Peter Luschny_, Sep 30 2018
%t Table[4^n*(3*n)!/((2*n)!*n!), {n, 0, 20}] (* _Erich Friedman_, Mar 22 2008 *)
%o (PARI) a(n) = 4^n*(3*n)!/((2*n)!*n!) \\ P L Patodia (pannalal(AT)usa.net), Feb 24 2007
%o (PARI) a(n) = sum(k=0,n,binomial(4*n+1,2*n-2*k)*binomial(n+k,k)) \\ P L Patodia (pannalal(AT)usa.net), Feb 24 2007
%o (Magma) [4^n*Factorial(3*n)/(Factorial(2*n)* Factorial(n)): n in [0..20]]; // _Vincenzo Librandi_, Oct 01 2018
%K nonn,easy,nice
%O 0,2
%A _N. J. A. Sloane_
|
