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A137276 Triangle of coefficients of Recursive Polynomials for Boubaker polynomials: you need to go to b(x,4) as input so that it doesn't give A049310: b(x,b)=x*b(x,n-1)-b(x,n-2). +0
3
1, 0, 1, 2, 0, 1, -2, 0, 0, 0, 1, 0, -3, 0, -1, 0, 1, 2, 0, -3, 0, -2, 0, 1, 0, 5, 0, -2, 0, -3, 0, 1, -2, 0, 8, 0, 0, 0, -4, 0, 1, 0, -7, 0, 10, 0, 3, 0, -5, 0, 1, 2, 0, -15, 0, 10, 0, 7, 0, -6, 0, 1, 0, 9, 0, -25, 0, 7, 0, 12, 0, -7, 0, 1 (list; graph; listen)
OFFSET

1,4

COMMENT

It took me several tries even with the equations on the web for the Boubaker polynomials to get a working recursion. They are intitial four polynomial dependent.

They are also closely related to the Steinbach type polynomial Gary Adamson and I explored in the past as well as Chebyshev polynomials.

Row sums are repeating

{1, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1}

Again Gary Adamson started me on this work.

REFERENCES

P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.

http://planetmath.org/encyclopedia/BoubakerPolynomials.html

FORMULA

b(x,b)=x*b(x,n-1)-b(x,n-2)

EXAMPLE

{1},

{0, 1},

{2, 0, 1},

{-2, 0, 0, 0, 1},

{0, -3, 0, -1, 0, 1},

{2, 0, -3, 0, -2, 0, 1},

{0, 5, 0, -2, 0, -3, 0, 1},

{-2, 0, 8, 0, 0, 0, -4, 0, 1},

{0, -7, 0, 10, 0, 3, 0, -5, 0, 1},

{2, 0, -15, 0, 10, 0, 7, 0, -6, 0, 1},

{0, 9, 0, -25, 0, 7, 0, 12, 0, -7, 0, 1}

MATHEMATICA

Clear[B, x, n]; B[x, 0] = 1; B[x, 1] = x; B[x, 2] = 2 + x^2; B[x, 3] = -2 + x^4; B[ x, 4] = -3 *x - x^3 + x^5; B[x_, n_] := B[x, n] = x*B[x, n - 1] - B[x, n - 2]; Table[ExpandAll[B[x, n]], {n, 0, 10}]; a = Table[CoefficientList[B[x, n], x], {n, 0, 10}]; Flatten[a]

CROSSREFS

Cf. A123956.

Adjacent sequences: A137273 A137274 A137275 this_sequence A137277 A137278 A137279

Sequence in context: A102564 A077762 A085496 this_sequence A101661 A079644 A072705

KEYWORD

uned,sign

AUTHOR

Roger L. Bagula and Gary Adamson (rlbagulatftn(AT)yahoo.com), Mar 13 2008

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Last modified October 11 13:47 EDT 2008. Contains 144830 sequences.


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