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Search: id:A140575
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| A140575 |
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A signed binomial triangular sequence with end canceling or doubling: torally reversed to get the zero degrees to show: p(x,n)=x^n*(1/x^n-(1-1/x^n)). |
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+0
1
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| 1, -1, 1, 0, 2, -1, 2, -3, 3, -1, 0, 4, -6, 4, -1, 2, -5, 10, -10, 5, -1, 0, 6, -15, 20, -15, 6, -1, 2, -7, 21, -35, 35, -21, 7, -1, 0, 8, -28, 56, -70, 56, -28, 8, -1, 2, -9, 36, -84, 126, -126, 84, -36, 9, -1, 0, 10, -45, 120, -210, 252, -210, 120, -45, 10, -1
(list; table; graph; listen)
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OFFSET
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1,5
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COMMENT
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Row sums are:
{1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1};
This function was inspired by the Binet Fibonacci sequence with golden mean "digits":
Clear[binet]
binet[x, 0] = 0; binet[x, 1] = Sqrt[5];
binet[x_, n_] := binet[x, n] = (x^n - (binet[x, 1]/Sqrt[5] - x )^(n))/Sqrt[5];
Table[Sqrt[5]*CoefficientList[binet[x, n], x], {n, 0, 10}].
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FORMULA
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p(x,n)=Coefficients(x^n*(1/x^n-(1-1/x^n))) t(n,m) = -(1-x)^n coefficients such that
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EXAMPLE
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{1},
{-1, 1},
{0, 2, -1},
{2, -3, 3, -1},
{0, 4, -6, 4, -1},
{2, -5, 10, -10, 5, -1},
{0, 6, -15, 20, -15, 6, -1},
{2, -7, 21, -35, 35, -21, 7, -1},
{0, 8, -28,56, -70, 56, -28, 8, -1},
{2, -9, 36, -84, 126, -126, 84, -36, 9, -1},
{0, 10, -45, 120, -210, 252, -210, 120, -45, 10, -1}
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MATHEMATICA
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Clear[p] p[x, 0] = 1; p[x, 1] = x - 1; p[x_, n_] := x^n*(1/x^n - (1 - 1/x)^n); a = Table[ExpandAll[p[x, n]], {n, 0, 10}]; b = Table[CoefficientList[ExpandAll[p[x, n]], x], {n, 0, 10}]; Flatten[b]
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CROSSREFS
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Sequence in context: A054758 A077876 A095056 this_sequence A101933 A117127 A136624
Adjacent sequences: A140572 A140573 A140574 this_sequence A140576 A140577 A140578
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KEYWORD
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tabl,uned,sign
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AUTHOR
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Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Jul 05 2008
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