A291398 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^3.

0, 1, 3, 5, 9, 19, 39, 76, 150, 301, 600, 1191, 2370, 4721, 9396, 18696, 37212, 74069, 147417, 293398, 583956, 1162257, 2313237, 4604037, 9163443, 18238042, 36299229, 72246487, 143792475, 286190708, 569606421, 1133689810, 2256387135, 4490895817, 8938246848

Offset: 0

Author

Clark Kimberling, Sep 06 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, 1, 3, 4, 3, 1)

Crossrefs

Cf. A019590, A291382.

Programs

Mathematica

z = 60; s = x + x^2; p = 1 - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291398 *)
LinearRecurrence[{0,1,3,4,3,1},{0,1,3,5,9,19},40] (* Harvey P. Dale, Dec 13 2017 *)

Formula

G.f.: -((x (1 + x)^2 (1 + x + x^2))/(-1 + x^2 + 3 x^3 + 4 x^4 + 3 x^5 + x^6)).

a(n) = a(n-2) + 3*a(n-3) + 4*a(n-4) + 3*a(n-5) + a(n-6) for n >= 7.