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A291017 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 5 S + S^2.

%I A291017 #4 Aug 23 2017 23:37:13
%S A291017 5,29,168,973,5635,32634,188993,1094513,6338640,36708889,212591743,
%T A291017 1231179978,7130117645,41292563669,239137122168,1384911909493,
%U A291017 8020423511275,46448581212474,268997103908393,1557839658871433,9021897884741280,52248407581088929
%N A291017 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 5 S + S^2.
%C A291017 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 A291017 See A291000 for a guide to related sequences.
%H A291017 Clark Kimberling, <a href="/A291017/b291017.txt">Table of n, a(n) for n = 0..1000</a>
%H A291017 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7, -7)
%F A291017 G.f.: (5 - 6 x)/(1 - 7 x + 7 x^2).
%F A291017 a(n) = 7*a(n-1) - 7*a(n-2) n >= 3.
%t A291017 z = 60; s = x/(1 - x); p = 1 - 5 s + s^2;
%t A291017 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291017 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291017 *)
%Y A291017 Cf. A000012, A289780, A291000.
%K A291017 nonn,easy
%O A291017 0,1
%A A291017 _Clark Kimberling_, Aug 23 2017