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A291227 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - 2*S^2.

%I A291227 #15 Jan 02 2021 04:54:07
%S A291227 1,3,6,17,37,96,221,551,1302,3189,7625,18528,44537,107835,259830,
%T A291227 628105,1515053,3659808,8832085,21328159,51481638,124302381,300068689,
%U A291227 724468416,1748959153,4222461747,10193761254,24610180673,59413804789,143438304480,346289581709
%N A291227 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - 2*S^2.
%C A291227 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 A291227 See A291219 for a guide to related sequences.
%H A291227 Clark Kimberling, <a href="/A291227/b291227.txt">Table of n, a(n) for n = 0..1000</a>
%H A291227 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,4,-1,-1).
%F A291227 G.f.: (1 + 2 x - x^2)/(1 - x - 4 x^2 + x^3 + x^4).
%F A291227 a(n) = a(n-1) + 4*a(n-2) - a(n-3) - a(n-4) for n >= 5.
%F A291227 a(n-1) = (4*A000129(n) + (-1)^n*A000045(n))/3 for n >= 1. - _Greg Dresden_, Jan 01 2021
%t A291227 z = 60; s = x/(1 - x^2); p = 1 - s - 2 s^2;
%t A291227 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291227 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291227 *)
%Y A291227 Cf. A000035, A291219.
%K A291227 nonn,easy
%O A291227 0,2
%A A291227 _Clark Kimberling_, Aug 25 2017