login
a(n)=[(n+1)(n+2)(n+3)...(2n)]/(1+2+3+...+n).
1

%I #11 Jul 15 2016 17:08:33

%S 2,4,20,168,2016,31680,617760,14414400,392071680,12189864960,

%T 426645273600,16606346803200,711700577280000,33307587016704000,

%U 1690360041097728000,92472637542405120000,5425061402487767040000,339780161524233830400000,22629358757513973104640000

%N a(n)=[(n+1)(n+2)(n+3)...(2n)]/(1+2+3+...+n).

%C a(n)=2(n-1)!*Catalan(n). a(n)=2*A065866(n-1). - _Emeric Deutsch_, Aug 05 2005

%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Numbers/year2016.html">2016</a>, January 2016.

%F {2*(2n)!}/{(n+1)!*n}

%e a(4) = 5*6*7*8/10 = 168.

%e a(5) = 10*9*8*7*6/(5+4+3+2+1) = 2016.

%p seq(2*(2*n)!/(n+1)!/n,n=1..20); # _Emeric Deutsch_, Aug 05 2005

%t Table[(Times@@Range[n+1,2n])/((n(n+1))/2),{n,20}] (* or *) Table[ 2(n-1)! CatalanNumber[n],{n,20}] (* _Harvey P. Dale_, Jul 15 2016 *)

%o (PARI) A110371(n)=binomial(2*n,n-1)/n*(n-1)!*2 \\ _M. F. Hasler_, Jan 31 2016

%Y Cf. A065866.

%K easy,nonn

%O 1,1

%A _Amarnath Murthy_, Jul 24 2005

%E More terms from _Emeric Deutsch_, Aug 05 2005

%E Edited by _M. F. Hasler_, Jan 31 2016