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Triangle T(n,k) = coefficient of x^n in expansion of [x(1+x^2)/(1-x^2)]^k = sum(n>=k, T(n,k) x^n).
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%I #7 Mar 31 2012 10:23:14

%S 1,0,1,2,0,1,0,4,0,1,2,0,6,0,1,0,8,0,8,0,1,2,0,18,0,10,0,1,0,12,0,32,

%T 0,12,0,1,2,0,38,0,50,0,14,0,1,0,16,0,88,0,72,0,16,0,1,2,0,66,0,170,0,

%U 98,0,18,0,1,0,20,0,192,0,292,0,128,0,20,0,1,2,0,102,0,450,0,462,0,162,0,22,0,1,0,24,0,360,0,912,0,688,0,200,0,24,0,1

%N Triangle T(n,k) = coefficient of x^n in expansion of [x(1+x^2)/(1-x^2)]^k = sum(n>=k, T(n,k) x^n).

%F T(n,k)=((sum(i=0..(n-k)/2, binomial(k,(n-k)/2-i)*binomial(k+i-1,k-1)))*((-1)^(n+k)+1))/2.

%e 1

%e 0, 1,

%e 2, 0, 1,

%e 0, 4, 0, 1,

%e 2, 0, 6, 0, 1,

%e 0, 8, 0, 8, 0, 1,

%e 2, 0, 18, 0, 10, 0, 1

%o (Maxima)

%o T(n,k):=((sum(binomial(k,(n-k)/2-i)*binomial(k+i-1,k-1),i,0,(n-k)/2))*((-1)^(n+k)+1))/2

%K nonn,tabl

%O 1,4

%A _Vladimir Kruchinin_, Dec 17 2011