A292328 p-INVERT of the Fibonacci sequence (A000045), where p(S) = (1 - S)^3.

3, 9, 28, 84, 246, 707, 2001, 5592, 15461, 42357, 115122, 310716, 833472, 2223471, 5902415, 15598896, 41058423, 107673601, 281416248, 733229412, 1904957434, 4936026747, 12758472189, 32901998472, 84667043769, 217437602349, 557361593902, 1426167813324

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).

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (6, -9, -4, 9, 6, 1)

Formula

G.f.: (3 - 9*x + x^2 + 9*x^3 + 3*x^4)/(1 - 2*x - x^2)^3.

a(n) = 6*a(n-1) - 9*a(n-2) - 4*a(n-3) + 9*a(n-4) + 6*a(n-5) + a(n-6) for n >= 7.

Mathematica

z = 60; s = x/(1 - x - x^2); p = (1 - s)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000045 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292328 *)
LinearRecurrence[{6, -9, -4, 9, 6, 1}, {3, 9, 28, 84, 246, 707}, 30] (* Harvey P. Dale, Jul 01 2022 *)

Prog

(PARI) x='x+O('x^99); Vec((3-9*x+x^2+9*x^3+3*x^4)/(1-2*x-x^2)^3) \\ Altug Alkan, Oct 03 2017

Crossrefs

Cf. A000045, A292327.

Sequence in context: A047134 A184698 A027099 * A027090 A033139 A291731

Adjacent sequences: A292325 A292326 A292327 * A292329 A292330 A292331

Keyword

nonn,easy

Author

Clark Kimberling, Sep 15 2017

Status

approved