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A291250 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - 2 S^2 + 2 S^3.

%I A291250 #10 May 13 2019 13:51:48
%S A291250 1,3,4,13,17,52,69,203,272,781,1053,2976,4029,11267,15296,42469,57765,
%T A291250 159596,217361,598499,815860,2241165,3057025,8383872,11440897,
%U A291250 31340691,42781588,117100285,159881873,437378260,597260133,1633244795,2230504928,6097779229
%N A291250 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - 2 S^2 + 2 S^3.
%C A291250 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 A291250 See A291219 for a guide to related sequences.
%H A291250 Clark Kimberling, <a href="/A291250/b291250.txt">Table of n, a(n) for n = 0..1000</a>
%H A291250 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1, 5, -4, -5, 1, 1)
%F A291250 G.f.: (-1 - 2 x + 4 x^2 + 2 x^3 - x^4)/(-1 + x + 5 x^2 - 4 x^3 - 5 x^4 + x^5 + x^6).
%F A291250 a(n) = a(n-1) + 5*a(n-2) - 4*a(n-3) - 5*a(n-4) + a(n-5) + a(n-6) for n >= 7.
%t A291250 z = 60; s = x/(1 - x^2); p = 1 - s - 2 s^2 + 2 s^3;
%t A291250 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291250 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291250 *)
%t A291250 LinearRecurrence[{1,5,-4,-5,1,1},{1,3,4,13,17,52},40] (* _Harvey P. Dale_, May 13 2019 *)
%Y A291250 Cf. A000035, A291219.
%K A291250 nonn,easy
%O A291250 0,2
%A A291250 _Clark Kimberling_, Aug 29 2017