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Search: id:A108350
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| A108350 |
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Number triangle T(n,k)=sum{j=0..n-k,C(k,j)C(n-j,k)*mod(j+1,2)}, as a square array read by anti-diagonals, T(n,k)=sum{j=0..n,C(k,j)C(n+k-j,k)*mod(j+1,2)}. Row k (and column k) has g.f. (1+C(k,2)x^2)/(1-x)^(k+1) |
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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
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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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,C(k,j)C(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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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,...
1,2,3,4,5,6,...
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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CROSSREFS
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Sequence in context: A094525 A130671 A114197 this_sequence A086617 A094526 A088699
Adjacent sequences: A108347 A108348 A108349 this_sequence A108351 A108352 A108353
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KEYWORD
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easy,nonn,tabl
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AUTHOR
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Paul Barry (pbarry(AT)wit.ie), May 31 2005
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