A292487 p-INVERT of the odd positive integers, where p(S) = 1 - S - 4 S^2.

1, 8, 44, 212, 1020, 4980, 24348, 118868, 580156, 2831924, 13824092, 67481876, 329408892, 1607991540, 7849328028, 38316090836, 187038012604, 913016364980, 4456842098396, 21755843899028, 106200025265148, 518409923170932, 2530591191342108, 12352949840710484

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, -3, 11, 4)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 4 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292487 *)

PARI

x='x+O('x^99); Vec(((1+x)*(1+2*x+5*x^2))/(1-5*x+3*x^2-11*x^3-4*x^4)) \\ Altug Alkan, Oct 03 2017

Formula

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

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