A291004 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 3*S)^2.

6, 33, 168, 816, 3840, 17664, 79872, 356352, 1572864, 6881280, 29884416, 128974848, 553648128, 2365587456, 10066329600, 42681237504, 180388626432, 760209211392, 3195455668224, 13400297963520, 56075093016576, 234195976716288, 976366325465088, 4063794976260096

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,-16).

Crossrefs

Cf. A000012, A289780, A291000.

Programs

Magma

[3*(4^(n-1)*(3*n+8)): n in [0..30]]; // Vincenzo Librandi, Aug 27 2017

Mathematica

z = 60; s = x/(1-x); p = (1 - 3 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291004 *)
LinearRecurrence[{8, -16}, {6, 33}, 25] (* Vincenzo Librandi, Aug 27 2017 *)

SageMath

[3*4^n*(3*n+8)//4 for n in range(41)] # G. C. Greubel, Jun 01 2023

Formula

G.f.: 3*(2 - 5*x)/(1 - 4*x)^2.

a(n) = 8*a(n-1) - 16*a(n-2) for n >= 3.

a(n) = 3*A006234(n+3) for n >= 0.

a(n) = 3 * 4^(n-1) * (3*n+8). - Colin Barker, Aug 23 2017