A136263 Integral form based on A122188: Coefficients of recursive polynomial from: p(x,n)=x*IP(x,n)=-(1+x)*B(x,n)-B'(x,n).

1, 0, -1, 0, -2, 1, 0, 0, 3, -1, 0, 0, 1, -4, 1, 0, 0, -1, -2, 5, -1, 0, 0, 1, 2, 3, -6, 1, 0, 0, -1, -2, -3, -4, 7, -1, 0, 0, 1, 2, 3, 4, 5, -8, 1, 0, 0, -1, -2, -3, -4, -5, -6, 9, -1, 0, 0, 1, 2, 3, 4, 5, 6, 7, -10, 1, 0, 0, -1, -2, -3, -4, -5, -6, -7, -8, 11, -1

Offset

1,5

Comments

Row sum:

{1, -1, -1, 2, -2, 1, 1, -4, 8, -13, 19, -26}

Derivation:

The Bonacci polynomial recursion:

B(x, n) = -(1 + x)*B(x, n - 1) - x*B(x, n - 2);

Implies the differential equation:

B''(x, n) = -(1 + x)*B'(x, n) - x*B(x, n);

Integration to a constant of that differential equation gives:

B'(x,n)=-(1+x)*B(x,n)-x*IB(x,n)

which rearranged gives the output form given here.

Links

Table of n, a(n) for n=1..78.

Formula

Bonacci polynomial recursion: B(x, n) = -(1 + x)*B(x, n - 1) - x*B(x, n - 2); Differentiation Polynomials: B'(x, n) := D[B[x, n + 1], x]=dB[x,n+1)/dx; Output form: p(x,n)=x*IP(x,n)=-(1+x)*B(x,n)-B'(x,n)

Example

{1},
{0, -1},
{0, -2, 1},
{0, 0, 3, -1},
{0, 0, 1, -4, 1},
{0, 0, -1, -2, 5, -1},
{0, 0, 1, 2, 3, -6, 1},
{0, 0, -1, -2, -3, -4, 7, -1},
{0, 0, 1, 2, 3,4, 5, -8, 1},
{0, 0, -1, -2, -3, -4, -5, -6, 9, -1},
{0, 0, 1, 2, 3, 4,5, 6, 7, -10, 1},
{0, 0, -1, -2, -3, -4, -5, -6, -7, -8, 11, -1}

Mathematica

Clear[B, x, n] B[x, 0] = 1; B[x, 1] = -1*x + 1; B[x_, n_] := B[x, n] = -(1 + x)*B[x, n - 1] - x*B[x, n - 2]; P[x_, n_] := D[B[x, n + 1], x]; Table[ExpandAll[ -(1 + x)*B[x, n] - P[x, n]], {n, 0, 10}]; a = Join[{{1}}, Table[CoefficientList[ -(1 + x)* B[x, n] - P[x, n], x], {n, 0, 10}]]; Flatten[a] Table[Apply[Plus, CoefficientList[ -(1 + x)*B[x, n] - P[x, n], x]], {n, 0, 10}];

Crossrefs

Cf. A122188.

Sequence in context: A128229 A132013 A105820 * A105593 A029371 A114374

Adjacent sequences: A136260 A136261 A136262 * A136264 A136265 A136266

Keyword

uned,tabl,sign

Author

Roger L. Bagula, Mar 18 2008

Status

approved