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A291734 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S).

%I A291734 #4 Sep 11 2017 20:05:03
%S A291734 3,7,18,45,108,258,615,1459,3453,8164,19287,45540,107496,253695,
%T A291734 598659,1412587,3332970,7863853,18553752,43774722,103279023,243668295,
%U A291734 574890057,1356344056,3200033343,7549859464,17812425600,42024945087,99149648967,233924207559
%N A291734 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S).
%C A291734 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 A291734 See A291728 for a guide to related sequences.
%H A291734 Clark Kimberling, <a href="/A291734/b291734.txt">Table of n, a(n) for n = 0..1000</a>
%H A291734 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3, -2, 3, -4, 0, -2)
%F A291734 G.f.: -(((1 + x^2) (-3 + 2 x + 2 x^3))/((-1 + x + x^3) (-1 + 2 x + 2 x^3))).
%F A291734 a(n) = 3*a(n-1) - 2*a(n-2) + 3*a(n-3) - 4*a(n-4) - 2*a(n-6) for n >= 7.
%t A291734 z = 60; s = x + x^3; p = (1 - s) (1 - 2 s);
%t A291734 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291734 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291734 *)
%Y A291734 Cf. A154272, A291728.
%K A291734 nonn,easy
%O A291734 0,1
%A A291734 _Clark Kimberling_, Sep 11 2017