A292485 p-INVERT of the odd positive integers, where p(S) = 1 - S - 2 S^2.

1, 6, 28, 120, 504, 2128, 9016, 38208, 161864, 685648, 2904408, 12303264, 52117544, 220773552, 935211704, 3961620096, 16781691912, 71088388112, 301135245080, 1275629368416, 5403652717288, 22890240236144, 96964613663352, 410748694893888, 1739959393240264

Offset: 0

Author

Clark Kimberling, Oct 02 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, -5, 7, 2)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292485 *)
LinearRecurrence[{5,-5,7,2},{1,6,28,120},30] (* Harvey P. Dale, Oct 14 2023 *)

Formula

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

a(n) = 5*a(n-1) - 5*a(n-2) + 7*a(n-3) + 2*a(n-4) for n >= 5.