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A289783 p-INVERT of the (3^n), where p(S) = 1 - S - S^2.

%I A289783 #28 Aug 05 2025 16:19:49
%S A289783 1,5,24,113,527,2446,11325,52369,242008,1117997,5163891,23849270,
%T A289783 110142089,508652653,2349005592,10847859961,50095958215,231345247934,
%U A289783 1068361195173,4933730638937,22784141325656,105217952251285,485900111176779,2243903303473318
%N A289783 p-INVERT of the (3^n), where p(S) = 1 - S - S^2.
%C A289783 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 A289783 Clark Kimberling, <a href="/A289783/b289783.txt">Table of n, a(n) for n = 0..1000</a>
%H A289783 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,-11).
%F A289783 G.f.: (1 - 2 x)/(1 - 7 x + 11 x^2).
%F A289783 a(n) = 7*a(n-1) - 11*a(n-2).
%F A289783 a(n) = (2^(-n-1)*((7-sqrt(5))^(n+1)*(-4+sqrt(5)) + (4+sqrt(5))*(7+sqrt(5))^(n+1))) / (11*sqrt(5)). - _Colin Barker_, Aug 11 2017
%F A289783 a(n) = A099453(n) - 2*A099453(n-1). - _R. J. Mathar_, Jul 08 2022
%F A289783 E.g.f.: exp(7*x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5. - _Stefano Spezia_, Aug 05 2025
%t A289783 z = 60; s = x/(1 - 3*x); p = 1 - s - s^2;
%t A289783 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000244 *)
%t A289783 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289783 *)
%o A289783 (PARI) Vec(x*(1 - 2*x) / (1 - 7*x + 11*x^2) + O(x^30)) \\ _Colin Barker_, Aug 11 2017
%Y A289783 Cf. A000244, A289780.
%K A289783 nonn,easy
%O A289783 0,2
%A A289783 _Clark Kimberling_, Aug 10 2017