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A291247 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.

%I A291247 #6 Aug 31 2017 04:46:47
%S A291247 1,2,5,10,24,49,112,238,526,1142,2491,5442,11842,25873,56344,122975,
%T A291247 268042,584633,1274820,2779895,6062306,13219186,28827703,62861754,
%U A291247 137082358,298927682,651861824,1421488867,3099781932,6759580078,14740333285,32143687954
%N A291247 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.
%C A291247 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 A291247 See A291219 for a guide to related sequences.
%H A291247 Clark Kimberling, <a href="/A291247/b291247.txt">Table of n, a(n) for n = 0..1000</a>
%H A291247 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1, 5, -2, -9, 2, 5, -1, -1)
%F A291247 G.f.: (1 + x - 2 x^2 - 3 x^3 + 2 x^4 + x^5 - x^6)/(1 - x - 5 x^2 + 2 x^3 + 9 x^4 - 2 x^5 - 5 x^6 + x^7 + x^8).
%F A291247 a(n) = a(n-1) + 5*a(n-2) - 2*a(n-3) - 9*a(n-4) + 2*a(n-5) + 5*a(n-6) - a(n-7) - a(n-8) for n >= 9.
%t A291247 z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 + s^4;
%t A291247 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291247 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291247 *)
%Y A291247 Cf. A000035, A291219.
%K A291247 nonn,easy
%O A291247 0,2
%A A291247 _Clark Kimberling_, Aug 29 2017