|
Search: id:A137276
|
|
|
| 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
|
|
|
Search completed in 0.002 seconds
|