A292327 -id:A292327 - cp's OEIS frontend

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

Clark Kimberling, Sep 15 2017

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).

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)

Cf. A000045, A292327.

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 *)

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

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.

cp's OEIS Frontend

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

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