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A289789 p-INVERT of A016777, where p(S) = 1 - S - S^2.

%I A289789 #12 Mar 14 2025 02:35:28
%S A289789 1,6,26,111,460,1905,7910,32880,136675,568050,2360825,9811650,
%T A289789 40777750,169474875,704348000,2927312625,12166086250,50562982500,
%U A289789 210142784375,873366003750,3629761440625,15085506018750,62696266831250,260569441284375,1082942209562500
%N A289789 p-INVERT of A016777, where p(S) = 1 - S - S^2.
%C A289789 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).
%C A289789 See A289780 for a guide to related sequences.
%H A289789 Clark Kimberling, <a href="/A289789/b289789.txt">Table of n, a(n) for n = 0..1000</a>
%H A289789 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -5, 5, 5)
%F A289789 G.f.: (-1 - x - x^2 - 6 x^3)/(-1 + 5 x - 5 x^2 + 5 x^3 + 5 x^4).
%F A289789 a(n) = 5*a(n-1) - 5*a(n-2) + 5*a(n-3) + 5*a(n-4).
%t A289789 z = 60; s = x (1 + 2*x)/(1 - x)^2; p = 1 - s - s^2;
%t A289789 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A016777 *)
%t A289789 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289789 *)
%Y A289789 Cf. A016777, A289780.
%K A289789 nonn,easy
%O A289789 0,2
%A A289789 _Clark Kimberling_, Aug 11 2017