A290923 p-INVERT of the positive integers, where p(S) = 1 - 2*S - 2*S^2.

2, 10, 46, 208, 938, 4230, 19078, 86048, 388106, 1750490, 7895302, 35610480, 160615298, 724429270, 3267420814, 14737172032, 66469626002, 299800475370, 1352201455582, 6098885514512, 27508034668634, 124070532153830, 559600027205398, 2523985228499040

Offset: 0

Author

Clark Kimberling, Aug 19 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 (6, -8, 6, -1)

Crossrefs

Cf. A000027, A033453, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290923 *)
u/2   (* A290924 *)

Formula

G.f.: (2 (1 - x + x^2))/(1 - 6 x + 8 x^2 - 6 x^3 + x^4).

a(n) = 6*a(n-1) - 8*a(n-2) + 6*a(n-3) - a(n-4).