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A108350
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Number triangle T(n,k) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*((j+1) mod 2).
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3
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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, 65, 65, 31, 7, 1, 1, 8, 43, 116, 161, 116, 43, 8, 1, 1, 9, 57, 189, 341, 341, 189, 57, 9, 1, 1, 10, 73, 288, 645, 842, 645, 288, 73, 10, 1, 1, 11, 91, 417, 1121, 1827, 1827, 1121, 417, 91
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OFFSET
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0,5
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COMMENTS
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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).
A symmetric number triangle based on 1/(1-x^2).
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.
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LINKS
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FORMULA
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Row k (and column k) has g.f. (1+C(k,2)x^2)/(1-x)^(k+1).
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EXAMPLE
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Triangle rows begin
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;
As a square array read by antidiagonals, rows begin
1, 1, 1, 1, 1, 1, 1, ...
1, 2, 3, 4, 5, 6, 7, ...
1, 3, 7, 13, 21, 31, 43, ...
1, 4, 13, 32, 65, 116, 189, ...
1, 5, 21, 65, 161, 341, 645, ...
1, 6, 31, 116, 341, 842, 1827, ...
1, 7, 43, 189, 645, 1827, 4495, ...
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PROG
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(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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