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Search: id:A137277
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| A137277 |
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Triangle of the coefficients [x^k] P_n(x) of the polynomials P_n(x) = sum_{j=0..[n/2]} (-1)^j binomial(n,j) (n-4j) x^(n-2j)/n. |
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| 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, -6, 0, 0, 0, 1, 0, -6, 0, -1, 0, 1, 20, 0, -5, 0, -2, 0, 1, 0, 25, 0, -3, 0, -3, 0, 1, -70, 0, 28, 0, 0, 0, -4, 0, 1, 0, -98, 0, 28, 0, 4, 0, -5, 0, 1, 252, 0, -126, 0, 24, 0, 9, 0, -6, 0, 1, 0, 378, 0, -150, 0, 15, 0, 15, 0, -7, 0, 1, -924, 0, 528, 0, -165, 0, 0, 0, 22, 0, -8, 0, 1, 0, -1452
(list; table; graph; listen)
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OFFSET
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0,4
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COMMENT
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The first four P_n(x) are the same as the first four Boubaker Polynomials A137276.
Row sums are 1, 1, 3, 2, -5, -6, 14, 20, -45, -70, 154, a signed variant of A047074.
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FORMULA
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P(0,n)=1. P_n(x) = sum_{j=0..floor(n/2)} (-1)^j binomial(n,j) (n-4j) x^(n-2j)/n.
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EXAMPLE
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{1}, = 1
{0, 1}, = x
{2, 0, 1}, = 2+x^2
{0, 1, 0, 1}, = x+x^3
{-6, 0, 0, 0, 1}, = -6+x^4
{0, -6, 0, -1, 0, 1},
{20, 0, -5, 0, -2, 0, 1},
{0, 25, 0, -3,0, -3, 0, 1},
{-70, 0, 28, 0, 0, 0, -4, 0, 1},
{0, -98, 0, 28, 0,4, 0, -5, 0, 1},
{252, 0, -126, 0, 24, 0, 9, 0, -6, 0, 1}
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MAPLE
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A137277 := proc(n, k) if n = 0 then 1; else add( (-1)^j*binomial(n, j)*(n-4*j)*x^(n-2*j), j=0..n/2)/n ; coeftayl(%, x=0, k) ; fi; end:
seq( seq(A137277(n, k), k=0..n), n=0..15) ;
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MATHEMATICA
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B[x_, n_] = If[n > 0, Sum[(-1)^p*Binomial[n, p]*(n - 4*p)*x^(n - 2*p)/ n, {p, 0, Floor[n/2]}], 1]; a = Table[CoefficientList[B[x, n], x], {n, 0, 10}]; Flatten[a]
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CROSSREFS
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Cf. A138034.
Sequence in context: A116927 A137276 A140581 this_sequence A039975 A016253 A117188
Adjacent sequences: A137274 A137275 A137276 this_sequence A137278 A137279 A137280
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KEYWORD
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sign,easy,tabl
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AUTHOR
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Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 13 2008
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EXTENSIONS
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Edited by the Associate Editors of the OEIS, Aug 27 2009
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