A290915 p-INVERT of the positive integers, where p(S) = 1 - 8*S^2.

0, 8, 32, 144, 672, 3096, 14272, 65824, 303552, 1399848, 6455520, 29770160, 137287520, 633112632, 2919650688, 13464207936, 62091296128, 286339090504, 1320476135328, 6089483698896, 28082152132128, 129503141377112, 597214328432960, 2754102721315680

Offset: 0

Author

Clark Kimberling, Aug 18 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 A290890 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 (4, 2, 4, -1)

Crossrefs

Cf. A000027, A290890, A290916.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 8 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290915 *)
u/8 (* A290916 *)

Formula

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

a(n) = 4*a(n-1) + 2*a(n-2) + 4*a(n-3) - a(n-4).

a(n) = 8*A290916(n) for n >= 0.