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A291729 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

%I A291729 #4 Sep 11 2017 20:04:29
%S A291729 2,5,14,39,106,290,794,2173,5946,16272,44530,121860,333480,912597,
%T A291729 2497400,6834349,18702782,51181767,140063294,383295214,1048920220,
%U A291729 2870460125,7855260268,21496593296,58827270844,160985870984,440551640160,1205607339709,3299247863502
%N A291729 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.
%C A291729 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 A291729 See A291728 for a guide to related sequences.
%H A291729 Clark Kimberling, <a href="/A291729/b291729.txt">Table of n, a(n) for n = 0..1000</a>
%H A291729 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2, 1, 2, 2, 0, 1)
%F A291729 G.f.: (-2 - x - 2 x^2 - 2 x^3 - x^5)/(-1 + 2 x + x^2 + 2 x^3 + 2 x^4 + x^6).
%F A291729 a(n) = 2*a(n-1) + a(n-2) + 2*a(n-3) + 2*a(n-4) + a(n-6) for n >= 7.
%t A291729 z = 60; s = x + x^3; p = 1 - 2 s - s^2;
%t A291729 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291729 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291729 *)
%Y A291729 Cf. A154272, A291728.
%K A291729 nonn,easy
%O A291729 0,1
%A A291729 _Clark Kimberling_, Sep 11 2017