A290911 p-INVERT of the positive integers, where p(S) = 1 - 6*S^2.

0, 6, 24, 96, 408, 1722, 7248, 30528, 128592, 541638, 2281416, 9609504, 40475976, 170487930, 718108320, 3024727680, 12740386464, 53663491206, 226034767224, 952075887072, 4010217126648, 16891344084282, 71147645118192, 299679373092288, 1262272651579632

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, 0, 4, -1)

Crossrefs

Cf. A000027, A290890, A290912.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 6 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290911 *)
u/6 (* A290912 *)

Formula

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

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

a(n) = 6*A290912(n) for n >= 0.