A291012 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)*(1 - 2*S).

2, 7, 22, 68, 208, 632, 1912, 5768, 17368, 52232, 156952, 471368, 1415128, 4247432, 12746392, 38247368, 114758488, 344308232, 1032990232, 3099101768, 9297567448, 27893226632, 83680728472, 251044282568, 753137042008, 2259419514632, 6778275321112

Offset

0,1

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 (5,-6).

Formula

G.f.: (2 - 3 x - x^2)/(1 - 5*x + 6*x^2).

a(n) = 5*a(n-1) - 6*a(n-2) for n >= 4.

a(n) = (16*3^n - 3*2^n) / 6 for n > 0. - Colin Barker, Aug 23 2017

E.g.f.: (1/6)*(-1 - 3*exp(2*x) + 16*exp(3*x)). - G. C. Greubel, Jun 04 2023

Mathematica

z = 60; s = x/(1-x); p = (1-s)^2(1-2s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* this sequence *)
LinearRecurrence[{5, -6}, {2, 7, 22}, 40] (* G. C. Greubel, Jun 04 2023 *)

Prog

(PARI) Vec((2 -3*x -x^2)/((1-2*x)*(1-3*x)) + O(x^30)) \\ Colin Barker, Aug 23 2017
(Magma) [2] cat [8*3^(n-1) - 2^(n-1): n in [1..40]]; // G. C. Greubel, Jun 04 2023
(SageMath) [8*3^(n-1) - 2^(n-1) - int(n==0)/6 for n in range(41)] # G. C. Greubel, Jun 04 2023

Crossrefs

Cf. A000012, A033453, A289780, A291000.

Sequence in context: A171847 A037552 A308113 * A369314 A294005 A333494

Adjacent sequences: A291009 A291010 A291011 * A291013 A291014 A291015

Keyword

nonn,easy

Author

Clark Kimberling, Aug 23 2017

Status

approved