A292493 p-INVERT of the odd positive integers, where p(S) = 1 + S - 3 S^2.

-1, 1, 12, 25, 61, 266, 963, 3053, 10220, 35413, 120345, 405682, 1376119, 4676201, 15859212, 53768225, 182400581, 618792826, 2098887003, 7119249973, 24149097580, 81915342653, 277858469505, 942504046562, 3197013067439, 10844389616401, 36784545696012

Offset: 0

Author

Clark Kimberling, Oct 04 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, -2, 11, 1)

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]  (* A292493 *)

PARI

x='x+O('x^99); Vec(((1+x)*(1-5*x-2*x^2))/(-1+3*x-2*x^2+11*x^3+x^4)) \\ Altug Alkan, Oct 05 2017

Formula

G.f.: -(((1 + x) (-1 + 5 x + 2 x^2))/(-1 + 3 x - 2 x^2 + 11 x^3 + x^4)).

a(n) = 3*a(n-1) - 2*a(n-2) + 11*a(n-3) + a(n-4) for n >= 5.