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Number triangle T(n,k)=sum{j=0..n-k, C(k,2j)C(n-k,2j)*2^j}.
2

%I #2 Mar 30 2012 18:59:15

%S 1,1,1,1,1,1,1,1,1,1,1,1,3,1,1,1,1,7,7,1,1,1,1,13,19,13,1,1,1,1,21,37,

%T 37,21,1,1,1,1,31,61,77,61,31,1,1,1,1,43,91,141,141,91,43,1,1,1,1,57,

%U 127,241,301,241,127,57,1,1

%N Number triangle T(n,k)=sum{j=0..n-k, C(k,2j)C(n-k,2j)*2^j}.

%C Row sums are A119330. Product of Pascal's triangle A007318 and A119331. T(n,k)=T(n,n-k).

%F Column k has g.f. (x^k/(1-x))*sum{j=0..k, C(k,2j)*2^j*(x/(1-x))^(2j)}

%e Triangle begins

%e 1,

%e 1, 1,

%e 1, 1, 1,

%e 1, 1, 1, 1,

%e 1, 1, 3, 1, 1,

%e 1, 1, 7, 7, 1, 1,

%e 1, 1, 13, 19, 13, 1, 1,

%e 1, 1, 21, 37, 37, 21, 1, 1

%Y Cf. A119326.

%K easy,nonn,tabl

%O 0,13

%A _Paul Barry_, May 14 2006