A292328 - OEIS

%I #10 Jul 01 2022 13:12:26

%S 3,9,28,84,246,707,2001,5592,15461,42357,115122,310716,833472,2223471,

%T 5902415,15598896,41058423,107673601,281416248,733229412,1904957434,

%U 4936026747,12758472189,32901998472,84667043769,217437602349,557361593902,1426167813324

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

%C 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).

%H Clark Kimberling, <a href="/A292328/b292328.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6, -9, -4, 9, 6, 1)

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

%F 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.

%t z = 60; s = x/(1 - x - x^2); p = (1 - s)^3;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000045 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292328 *)

%t LinearRecurrence[{6,-9,-4,9,6,1},{3,9,28,84,246,707},30] (* _Harvey P. Dale_, Jul 01 2022 *)

%o (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

%Y Cf. A000045, A292327.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Sep 15 2017