A292490 p-INVERT of the odd positive integers, where p(S) = 1 - S - 7 S^2.

1, 11, 68, 365, 2019, 11328, 63321, 353483, 1974124, 11026373, 61584323, 343956104, 1921047729, 10729356747, 59925127764, 334691142941, 1869302113507, 10440343236752, 58310941508105, 325675681470731, 1818949357172988, 10159115194159989, 56740239146359107

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, 0, 17, 7)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 7 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292490 *)

PARI

x='x+O('x^99); Vec(((1+x)*(1+5*x+8*x^2))/(1-5*x-17*x^3-7*x^4)) \\ Altug Alkan, Oct 03 2017

Formula

G.f.: -(((1 + x) (1 + 5 x + 8 x^2))/(-1 + 5 x + 17 x^3 + 7 x^4)).

a(n) = 5*a(n-1) + 17*a(n-3) + 7*a(n-4) for n >= 5.