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%I #9 Dec 05 2016 02:40:41
%S 1,1,1,1,2,1,1,3,3,1,1,4,7,4,1,1,5,13,13,5,1,1,6,21,32,21,6,1,1,7,31,
%T 65,65,31,7,1,1,8,43,116,161,116,43,8,1,1,9,57,189,341,341,189,57,9,1,
%U 1,10,73,288,645,842,645,288,73,10,1,1,11,91,417,1121,1827,1827,1121,417,91
%N Number triangle T(n,k) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*((j+1) mod 2).
%C Or as a square array read by antidiagonals, T(n,k) = Sum_{j=0..n} binomial(k,j)*binomial(n+k-j,k)*((j+1) mod 2).
%C A symmetric number triangle based on 1/(1-x^2).
%C The construction of a symmetric triangle in this example is general. Let f(n) be a sequence, preferably with f(0)=1. Then T(n,k) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*f(j) yields a symmetric triangle. When f(n)=1^n, we get Pascal's triangle. When f(n)=2^n, we get the Delannoy triangle (see A008288). In general, f(n)=k^n yields a (1,k,1)-Pascal triangle (see A081577, A081578). Row sums of triangle are A100131. Diagonal sums of the triangle are A108351. Triangle mod 2 is A106465.
%F Row k (and column k) has g.f. (1+C(k,2)x^2)/(1-x)^(k+1).
%e Triangle rows begin
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 3, 3, 1;
%e 1, 4, 7, 4, 1;
%e 1, 5, 13, 13, 5, 1;
%e 1, 6, 21, 32, 21, 6, 1;
%e As a square array read by antidiagonals, rows begin
%e 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 2, 3, 4, 5, 6, 7, ...
%e 1, 3, 7, 13, 21, 31, 43, ...
%e 1, 4, 13, 32, 65, 116, 189, ...
%e 1, 5, 21, 65, 161, 341, 645, ...
%e 1, 6, 31, 116, 341, 842, 1827, ...
%e 1, 7, 43, 189, 645, 1827, 4495, ...
%o (PARI) trgn(nn) = {for (n= 0, nn, for (k = 0, n, print1(sum(j=0, n-k, binomial(k,j)*binomial(n-j,k)*((j+1) % 2)), ", ");); print(););} \\ _Michel Marcus_, Sep 11 2013
%K easy,nonn,tabl
%O 0,5
%A _Paul Barry_, May 31 2005
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