A136093 A generalized ( Shabat) type tree transformed Chebyshev recursion where: P(x,n)->C*P(c*x+d,n)+D: C=-1;c=-1;D=1;d=1: with substitution: x-1->x; as a triangular sequence of coefficients.
1, 0, 1, -1, -1, 1, 0, -3, -1, 1, 1, 0, -4, -1, 1, 0, 3, 1, -5, -1, 1, -1, -1, 7, 2, -6, -1, 1, 0, -5, -2, 12, 3, -7, -1, 1, 1, 0, -12, -4, 18, 4, -8, -1, 1, 0, 5, 2, -24, -7, 25, 5, -9, -1, 1, -1, -1, 17, 6, -42, -11, 33, 6, -10, -1, 1
Offset: 1
Examples
{1}, {0, 1}, {-1, -1, 1}, {0, -3, -1, 1}, {1, 0, -4, -1, 1}, {0, 3, 1, -5, -1, 1}, {-1, -1, 7, 2, -6, -1, 1}, {0, -5, -2, 12, 3, -7, -1, 1}, {1, 0, -12, -4, 18, 4, -8, -1, 1}, {0, 5, 2, -24, -7,25, 5, -9, -1, 1}, {-1, -1, 17, 6, -42, -11, 33, 6, -10, -1, 1}
References
- http : // logic.pdmi.ras.ru/~yumat/personaljournal/chebyshev/chebysh.htm Quote:"It is easy to see that if P is a generalized Chebyshev Polynomial, then so is polynomial CP(cz + d) + D, moreover, it represents the same tree (of course, provided that both C and c are different from zero)."
Programs
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Mathematica
Clear[B, a] B[x, 0] = 1; B[x, 1] = x; B[x_, n_] := B[x, n] = -x + 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]
Formula
p(x,0)=1;p(x,1)=x; p(x,n)=-x+x*p(x,n-1)-p(x,n-2)
Comments