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A110371 a(n)=[(n+1)(n+2)(n+3)...(2n)]/(1+2+3+...+n).

%I A110371 #11 Jul 15 2016 17:08:33
%S A110371 2,4,20,168,2016,31680,617760,14414400,392071680,12189864960,
%T A110371 426645273600,16606346803200,711700577280000,33307587016704000,
%U A110371 1690360041097728000,92472637542405120000,5425061402487767040000,339780161524233830400000,22629358757513973104640000
%N A110371 a(n)=[(n+1)(n+2)(n+3)...(2n)]/(1+2+3+...+n).
%C A110371 a(n)=2(n-1)!*Catalan(n). a(n)=2*A065866(n-1). - _Emeric Deutsch_, Aug 05 2005
%H A110371 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Numbers/year2016.html">2016</a>, January 2016.
%F A110371 {2*(2n)!}/{(n+1)!*n}
%e A110371 a(4) = 5*6*7*8/10 = 168.
%e A110371 a(5) = 10*9*8*7*6/(5+4+3+2+1) = 2016.
%p A110371 seq(2*(2*n)!/(n+1)!/n,n=1..20); # _Emeric Deutsch_, Aug 05 2005
%t A110371 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 A110371 (PARI)  A110371(n)=binomial(2*n,n-1)/n*(n-1)!*2 \\ _M. F. Hasler_, Jan 31 2016
%Y A110371 Cf. A065866.
%K A110371 easy,nonn
%O A110371 1,1
%A A110371 _Amarnath Murthy_, Jul 24 2005
%E A110371 More terms from _Emeric Deutsch_, Aug 05 2005
%E A110371 Edited by _M. F. Hasler_, Jan 31 2016