A291379 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^4.

0, 0, 0, 1, 4, 6, 4, 2, 8, 28, 56, 71, 68, 94, 228, 497, 808, 1044, 1352, 2316, 4608, 8264, 12592, 17717, 26968, 47044, 83912, 138417, 211052, 319850, 517356, 881918, 1483336, 2377252, 3700648, 5853067, 9605596, 15991378, 26154700, 41734433, 66214096

Offset: 0

Author

Clark Kimberling, Sep 04 2017

Keywords

• nonn
• easy

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291382 for a guide to related sequences.

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 1, 4, 6, 4, 1)

Crossrefs

Cf. A019590, A291382.

Programs

Mathematica

z = 60; s = x + x^2; p = (1 - s)^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291379 *)

Formula

G.f.: -((x^3 (1 + x)^4)/((-1 + x + x^2) (1 + x + x^2) (1 + x^2 + 2 x^3 + x^4))).

a(n) = a(n-4) + 4*a(n-5) + 6*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.