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%I #21 Dec 18 2017 03:07:22
%S 1,1,2,7,135,53174,6620830889,39020148000237259665,
%T 133523474368721196662101633251149823925,
%U 14042421942608880253531745690954970851431472263832971258973477309202081861
%N Number of partitions of the Catalan number binomial(2n,n)/(n+1).
%H Michael De Vlieger, <a href="/A120379/b120379.txt">Table of n, a(n) for n = 0..13</a>
%H Henry Bottomley, <a href="http://www.se16.info/js/partitions.htm">Partition and composition calculator using a Java applet</a>
%H G. P. Michon, <a href="http://numericana.com/data/partition.htm">Partition Function</a>
%e a(3)=7 because binomial(6,3)/4 = 5 and the number of partitions of 5 is 7.
%p with(combinat): seq(numbpart(binomial(2*n,n)/(n+1)),n=0..8); # _Emeric Deutsch_, Jul 23 2006
%t Array[PartitionsP@ CatalanNumber@ # &, 10, 0] (* _Michael De Vlieger_, Dec 17 2017 *)
%o (MuPAD) combinat::partitions::count(binomial(2*n,n)/(n+1)) $n=0..10 // _Zerinvary Lajos_, Apr 16 2007
%Y Cf. A003107, A000108.
%K nonn
%O 0,3
%A _Zerinvary Lajos_, Jun 28 2006
%E Edited by _Emeric Deutsch_, Jul 23 2006