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A114197
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A Pascal-Fibonacci triangle.
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14
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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, 31, 21, 6, 1, 1, 7, 31, 61, 61, 31, 7, 1, 1, 8, 43, 106, 142, 106, 43, 8, 1, 1, 9, 57, 169, 286, 286, 169, 57, 9, 1, 1, 10, 73, 253, 520, 659, 520, 253, 73, 10, 1
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OFFSET
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0,5
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COMMENTS
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T(2n,n) is A114198. Row sums are A114199. Row sums of inverse are 0^n.
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LINKS
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FORMULA
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As a number triangle, T(n,k) = Sum_{j=0..n-k} C(n-k, j)C(k, j)F(j);
As a number triangle, T(n,k) = Sum_{j=0..n} C(n-k, n-j)C(k, j-k)F(j-k);
As a number triangle, T(n,k) = Sum_{j=0..n} C(k, j)C(n-k, n-j)F(k-j) if k <= n, 0 otherwise.
As a square array, T(n,k) = Sum_{j=0..n} C(n, j)C(k, j)F(j);
As a square array, T(n,k) = Sum_{j=0..n+k} C(n, n+k-j)C(k, j-k)F(j-k);
Column k has g.f.: (Sum_{j=0..k} C(k, j)F(j+1)(x/(1-x))^j)*x^k/(1-x);
G.f.: -((x^2-x)*y-x+1)/((x^4+x^3-x^2)*y^2+(x^3-3*x^2+2*x)*y-x^2+2*x-1). - Vladimir Kruchinin, Jan 15 2018
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EXAMPLE
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Triangle begins
1;
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 4, 7, 4, 1;
1, 5, 13, 13, 5, 1;
1, 6, 21, 31, 21, 6, 1;
1, 7, 31, 61, 61, 31, 7, 1;
1, 8, 43, 106, 142, 106, 43, 8, 1;
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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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