A290904 p-INVERT of the positive integers, where p(S) = 1 - 2*S^2.

0, 2, 8, 24, 72, 222, 688, 2128, 6576, 20322, 62808, 194120, 599960, 1854270, 5730912, 17712288, 54742624, 169190722, 522910632, 1616137848, 4994929128, 15437616926, 47712391952, 147462678768, 455756685840, 1408587979170, 4353463496440, 13455066133672

Offset: 0

Author

Clark Kimberling, Aug 17 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, -4, 4, -1)

Crossrefs

Cf. A000027, A290890, A290905.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290903 *)
u/2 (* A290905 *)

Formula

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

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

a(n) = 2*A290905(n) for n >= 0.