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%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
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