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A092879
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Triangle of coefficients of the product of two consecutive Fibonacci polynomials.
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2
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1, 1, 1, 1, 3, 2, 1, 5, 7, 2, 1, 7, 16, 13, 3, 1, 9, 29, 40, 22, 3, 1, 11, 46, 91, 86, 34, 4, 1, 13, 67, 174, 239, 166, 50, 4, 1, 15, 92, 297, 541, 553, 296, 70, 5, 1, 17, 121, 468, 1068, 1461, 1163, 496, 95, 5, 1, 19, 154, 695, 1912, 3300, 3544, 2269, 791, 125, 6, 1, 21, 191
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OFFSET
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0,5
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COMMENTS
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The Fibonacci polynomials are defined by F(0,x) = 1, F(1,x) = 1 and F(n, x) = F(n-1, x) + x*F(n-2, x).
This is also the reflected triangle of coefficients of the polynomials defined by the recursion: c0=-1; p(x, n) = (2 + c0 - x)*p(x, n - 1) + (-1 - c0*(2 - x))*p(x, n - 2) + c0*p(x, n - 3). - Roger L. Bagula, Apr 09 2008
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LINKS
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FORMULA
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T(n, k) = Sum_{i=0..k} (-1)^(i+k)*binomial(i+2*n-2*k+1, i).
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2) - T(n-3,k-3), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 12 2013
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EXAMPLE
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Triangle begins;
1;
1,1;
1,3,2;
1,5,7,2;
1,7,16,13,3;
1,9,29,40,22,3;
...
F(3,x) = 1 + 2*x and F(4,x) = 1 + 3*x + x^2 so F(3,x)*F(4,x)=(1 + 3*x + x^2)*(1 + 2*x) = 1 + 5*x + 7*x^2 + 2*x^3 leads to T(3,k) = [1,5,7,2].
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MAPLE
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T:=proc(n, k): add((-1)^(i+k)*binomial(i+2*n-2*k+1, i), i=0..k) end: seq(seq(T(n, k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
T:=proc(n, k): coeff(F(n, x)*F(n+1, x), x, k) end: F:=proc(n, x) option remember: if n=0 then 1 elif n=1 then 1 else procname(n-1, x) + x*procname(n-2, x) fi: end: seq(seq(T(n, k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
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MATHEMATICA
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c0 = -1; p[x, -1] = 0; p[x, 0] = 1; p[x, 1] = 2 - x + c0; p[x_, n_] :=p[x, n] = (2 + c0 -x)*p[x, n - 1] + (-1 - c0 (2 - x))*p[x, n - 2] + c0*p[x, n - 3]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[Reverse[CoefficientList[p[x, n], x]], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 09 2008 *)
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PROG
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(PARI) T(n, k)=local(m); if(k<0 || k>n, 0, n++; m=contfracpnqn(matrix(2, n, i, j, x)); polcoeff(m[1, 1]*m[2, 1]/x^n, n-k))
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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