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A061177
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Coefficients of polynomials ( (1 -x +sqrt(x))^(n+1) - (1 -x -sqrt(x))^(n+1) )/(2*sqrt(x)).
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7
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1, 2, -2, 3, -5, 3, 4, -8, 8, -4, 5, -10, 11, -10, 5, 6, -10, 6, -6, 10, -6, 7, -7, -14, 29, -14, -7, 7, 8, 0, -56, 120, -120, 56, 0, -8, 9, 12, -126, 288, -365, 288, -126, 12, 9, 10, 30, -228, 540, -770, 770, -540, 228, -30, -10, 11, 55, -363, 858
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OFFSET
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0,2
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COMMENTS
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The row polynomial pFo(m,x) = Sum_{j=0..m} T(m, j)*x^j is the numerator of the g.f. for the m-th column sequence of A060921, the odd part of the bisected Fibonacci triangle.
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LINKS
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FORMULA
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T(n, k) = coefficient of x^k of ( (1 -x +sqrt(x))^(n+1) - (1 -x -sqrt(x))^(n+1) )/(2*sqrt(x)).
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1, 2*j+1)*binomial(n-2*j, k-j), if 0 <= k <= floor(n/2), T(n, k) = (-1)^n*T(n, n-k) if floor(n/2) < k <= n else 0.
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EXAMPLE
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The first few polynomials are:
pFo(0, x) = 1.
pFo(1, x) = 2 - 2*x.
pFo(2, x) = 3 - 5*x + 3*x^2.
pFo(3, x) = 4 - 8*x + 8*x^2 - 4*x^3.
pFo(4, x) = 5 - 10*x + 11*x^2 - 10*x^3 + 5*x^4.
pFo(5, x) = 6 - 10*x + 6*x^2 - 6*x^3 + 10*x^4 - 6*x^5.
Number triangle begins as:
1;
2, -2;
3, -5, 3;
4, -8, 8, -4;
5, -10, 11, -10, 5;
6, -10, 6, -6, 10, -6;
7, -7, -14, 29, -14, -7, 7;
8, 0, -56, 120, -120, 56, 0, -8;
9, 12, -126, 288, -365, 288, -126, 12, 9;
10, 30, -228, 540, -770, 770, -540, 228, -30, -10;
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MATHEMATICA
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T[n_, k_]:= Sum[(-1)^(k-j)*Binomial[n+1, 2*j+1]*Binomial[n-2*j, k-j], {j, 0, k}];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Apr 06 2021 *)
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PROG
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(Magma)
A061177:= func< n, k | (&+[(-1)^(k+j)*Binomial(n+1, 2*j+1)*Binomial(n-2*j, k-j): j in [0..k]]) >;
(Sage)
def A061177(n, k): return sum((-1)^(k+j)*binomial(n+1, 2*j+1)*binomial(n-2*j, k-j) for j in (0..k))
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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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