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A289784 p-INVERT of the (4^n), where p(S) = 1 - S - S^2.

%I A289784 #24 Jul 08 2022 12:56:36
%S A289784 1,6,35,201,1144,6477,36557,205950,1158967,6517653,36638504,205911129,
%T A289784 1157068585,6501305814,36527449211,205222232433,1152978556888,
%U A289784 6477584595765,36391668781013,204450911709582,1148616498546991,6452981164440861,36253117007574920
%N A289784 p-INVERT of the (4^n), where p(S) = 1 - S - S^2.
%C A289784 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 A289784 Clark Kimberling, <a href="/A289784/b289784.txt">Table of n, a(n) for n = 0..1000</a>
%H A289784 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (9, -19)
%F A289784 G.f.: (1 - 3 x)/(1 - 9 x + 19 x^2).
%F A289784 a(n) = 9*a(n-1) - 19*a(n-2).
%F A289784 a(n) = (2^(-2-n)*((9-sqrt(5))^(n+1)*(-11+3*sqrt(5)) + (9+sqrt(5))^(n+1)*(11+3*sqrt(5)))) / (19*sqrt(5)). - _Colin Barker_, Aug 11 2017
%F A289784 a(n) = A081574(n+1)-3*A081574(n). - _R. J. Mathar_, Jul 08 2022
%t A289784 z = 60; s = x/(1 - 4*x); p = 1 - s - s^2;
%t A289784 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000302 *)
%t A289784 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289784 *)
%o A289784 (PARI) Vec(x*(1 - 3*x) / (1 - 9*x + 19*x^2) + O(x^30)) \\ _Colin Barker_, Aug 11 2017
%Y A289784 Cf. A000302, A289780.
%K A289784 nonn,easy
%O A289784 0,2
%A A289784 _Clark Kimberling_, Aug 10 2017