A292488 p-INVERT of the odd positive integers, where p(S) = 1 - S - 5 S^2.

1, 9, 52, 261, 1323, 6814, 35077, 180261, 926348, 4761289, 24472527, 125783886, 646502873, 3322895889, 17079026852, 87782799261, 451186103523, 2319006747614, 11919233055677, 61262485125261, 314876977751548, 1618404981969089, 8318279426249127

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, -2, 13, 5)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 5 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292488 *)
LinearRecurrence[{5,-2,13,5},{1,9,52,261},30] (* Harvey P. Dale, Jul 31 2025 *)

PARI

x='x+O('x^99); Vec(((1+x)*(1+3*x+6*x^2))/(1-5*x+2*x^2-13*x^3-5*x^4)) \\ Altug Alkan, Oct 03 2017

Formula

G.f.: -(((1 + x) (1 + 3 x + 6 x^2))/(-1 + 5 x - 2 x^2 + 13 x^3 + 5 x^4)).

a(n) = 5*a(n-1) - 2*a(n-2) + 13*a(n-3) + 5*a(n-4) for n >= 5.