A291011 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)^2 * (1 - 2*S).

4, 15, 52, 172, 552, 1736, 5384, 16536, 50440, 153112, 463176, 1397720, 4210568, 12668568, 38083528, 114414424, 343587336, 1031482904, 3095956040, 9291013848, 27879595144, 83652416920, 250985562312, 753015407192, 2259167856392, 6777755227416, 20333785775944

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..999

Index entries for linear recurrences with constant coefficients, signature (7,-16,12).

Formula

G.f.: (4 - 13*x + 11*x^2)/((1-2*x)^2 * (1-3*x)).

a(n) = 7*a(n-1) - 16*a(n-2) + 12*a(n-3) for n >= 4.

a(n) = 8*3^n - 2^(n-1)*(8+n). - Colin Barker, Aug 23 2017

E.g.f.: 8*exp(3*x) - (4 + x)*exp(2*x). - G. C. Greubel, Jun 04 2023

Mathematica

z = 60; s = x/(1-x); p = (1-s)^2*(1-2*s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* this sequence *)
LinearRecurrence[{7, -16, 12}, {4, 15, 52}, 30] (* Harvey P. Dale, Sep 23 2017 *)

Prog

(PARI) Vec((4 -13*x +11*x^2)/((1-2*x)^2*(1-3*x)) + O(x^30)) \\ Colin Barker, Aug 23 2017
(Magma) [8*3^n - 2^(n-1)*(8+n): n in [0..40]]; // G. C. Greubel, Jun 04 2023
(SageMath) [8*3^n - 2^(n-1)*(8+n) for n in range(41)] # G. C. Greubel, Jun 04 2023

Crossrefs

Cf. A000012, A033453, A289780, A291000.

Sequence in context: A005492 A003013 A117202 * A137213 A027853 A132894

Adjacent sequences: A291008 A291009 A291010 * A291012 A291013 A291014

Keyword

nonn,easy

Author

Clark Kimberling, Aug 23 2017

Status

approved