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Search: id:A092879
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| A092879 |
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Triangle of coefficients of the product of two consecutive Fibonacci polynomials. |
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1
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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
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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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 (rlbagulatftn(AT)yahoo.com), Apr 09 2008
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EXAMPLE
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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; ...
P(4) = 1 + 3x + x^2, P(3) = 1 + 2x, so (1 + 3x + x^2)(1 + 2x) = 1 + 5x + 7x^2 + 2x3 is T(3,k).
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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 (rlbagulatftn(AT)yahoo.com), Apr 09 2008
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PROGRAM
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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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Row sums are A001654(n+1).
Cf. A109954, A136674.
Sequence in context: A110712 A138483 A065366 this_sequence A073370 A129675 A081277
Adjacent sequences: A092876 A092877 A092878 this_sequence A092880 A092881 A092882
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KEYWORD
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nonn,tabl
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AUTHOR
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Michael Somos, Mar 10 2004
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