A292486 p-INVERT of the odd positive integers, where p(S) = 1 - S - 3 S^2.

1, 7, 36, 165, 747, 3420, 15705, 72063, 330516, 1515933, 6953283, 31893516, 146289393, 671000247, 3077745156, 14117009877, 64751939163, 297004363452, 1362300384969, 6248602953135, 28661108314356, 131462846314317, 602994126047283, 2765815028667756

Offset: 0

Author

Clark Kimberling, Oct 03 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 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 (5, -4, 9, 3)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 3 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292486 *)
LinearRecurrence[{5,-4,9,3},{1,7,36,165},30] (* Harvey P. Dale, Sep 29 2024 *)

PARI

x='x+O('x^99); Vec(((1+x)*(1+x+4*x^2))/(1-5*x+4*x^2-9*x^3-3*x^4)) \\ Altug Alkan, Oct 03 2017

Formula

G.f.: -(((1 + x) (1 + x + 4 x^2))/(-1 + 5 x - 4 x^2 + 9 x^3 + 3 x^4)).

a(n) = 5*a(n-1) - 4*a(n-2) + 9*a(n-3) + 3*a(n-4) for n >= 5.