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a(n) = floor(binomial(2n,2k)/binomial(n,k)).
1

%I #6 Jul 31 2015 21:12:06

%S 1,1,1,1,3,1,1,5,5,1,1,7,11,7,1,1,9,21,21,9,1,1,11,33,46,33,11,1,1,13,

%T 47,85,85,47,13,1,1,15,65,143,183,143,65,15,1,1,17,85,221,347,347,221,

%U 85,17,1,1,19,107,323,599,733,599,323,107,19,1,1,21,133,452,969,1399

%N a(n) = floor(binomial(2n,2k)/binomial(n,k)).

%C Row sums of inverse appear to be A000007.

%F Number triangle T(n, k)=if(k<=n, floor(C(2n, 2k)/C(n, k)), 0).

%e Triangle begins

%e 1;

%e 1, 1;

%e 1, 3, 1;

%e 1, 5, 5, 1;

%e 1, 7, 11, 7, 1;

%e 1, 9, 21, 21, 9, 1;

%e 1, 11, 33, 46, 33, 11, 1;

%t Flatten[ Table[ Floor[ Binomial[2n, 2k]/Binomial[n, k]], {n, 0, 11}, {k, 0, n}]] (* _Robert G. Wilson v_, Oct 21 2005 *)

%Y Even terms are in A111304.

%K easy,nonn,tabl

%O 0,5

%A _Paul Barry_, Oct 20 2005