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A291399 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - S^3 - S^4.

%I A291399 #6 Oct 06 2018 14:07:54
%S A291399 1,3,8,22,59,156,412,1093,2903,7707,20453,54275,144035,382255,1014469,
%T A291399 2692284,7144989,18961928,50322686,133550412,354426839,940606403,
%U A291399 2496256771,6624766692,17581338025,46658767166,123826784175,328621466028,872122042693
%N A291399 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - S^3 - S^4.
%C A291399 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 A291399 See A291382 for a guide to related sequences.
%H A291399 Clark Kimberling, <a href="/A291399/b291399.txt">Table of n, a(n) for n = 0..1000</a>
%H A291399 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1, 2, 3, 5, 7, 7, 4, 1)
%F A291399 G.f.: -(((1 + x) (1 + x + x^2) (1 + x^2 + 2 x^3 + x^4))/(-1 + x + 2 x^2 + 3 x^3 + 5 x^4 + 7 x^5 + 7 x^6 + 4 x^7 + x^8)).
%F A291399 a(n) = a(n-1) + 2*a(n-2) + 3*a(n-3) + 5*a(n-4) + 7*a(n-5) + 7*a(n-6) +4*a(n-7) + a(n-8) for n >= 9.
%t A291399 z = 60; s = x + x^2; p = 1 - s - s^2 - s^3 - s^4;
%t A291399 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291399 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291399 *)
%t A291399 LinearRecurrence[{1,2,3,5,7,7,4,1},{1,3,8,22,59,156,412,1093},40] (* _Harvey P. Dale_, Oct 06 2018 *)
%Y A291399 Cf. A019590, A291382.
%K A291399 nonn,easy
%O A291399 0,2
%A A291399 _Clark Kimberling_, Sep 06 2017