A291018 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6 S + S^2.

6, 41, 280, 1912, 13056, 89152, 608768, 4156928, 28385280, 193826816, 1323532288, 9037643776, 61712891904, 421401985024, 2877512744960, 19648886079488, 134170986676224, 916176804773888, 6256046544781312, 42718957920059392, 291703291002224640

Offset: 0

Author

Clark Kimberling, Aug 23 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 A291000 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 (8, -8)

Crossrefs

Cf. A000012, A289780, A291000.

Programs

Mathematica

z = 60; s = x/(1 - x); p = 1 - 6 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291018 *)

Formula

G.f.: (6 - 7 x)/(1 - 8 x + 8 x^2).

a(n) = 8*a(n-1) - 8*a(n-2) n >= 3.