A292484 p-INVERT of the odd positive integers, where p(S) = 1 + S - S^2.

-1, -1, 4, 9, 5, 8, 63, 183, 348, 745, 2061, 5456, 12991, 30831, 76660, 192137, 472597, 1155032, 2843007, 7024935, 17315404, 42592489, 104847389, 258355104, 636507775, 1567442143, 3859933668, 9507231753, 23417547813, 57675809960, 142047927231, 349856144791

Offset: 0

Author

Clark Kimberling, Oct 02 2017

Keywords

• easy
• sign

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 A292480 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 (3, -4, 7, -1)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 + s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292484 *)
LinearRecurrence[{3,-4,7,-1},{-1,-1,4,9},40] (* Harvey P. Dale, Sep 22 2024 *)

Formula

G.f.: ((1 + x) (-1 + 3 x))/(1 - 3 x + 4 x^2 - 7 x^3 + x^4).

a(n) = 3*a(n-1) - 4*a(n-2) + 7*a(n-3) - a(n-4) for n >= 5.