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A291032 p-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3 + S^4.

%I A291032 #6 Aug 21 2017 13:02:39
%S A291032 1,4,15,54,188,645,2208,7570,25982,89190,306095,1050268,3603276,
%T A291032 12361763,42409154,145491117,499126660,1712311759,5874263702,
%U A291032 20152234481,69134134820,237171010852,813636681973,2791253840066,9575645985794,32850107071454,112695214040224
%N A291032 p-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3 + S^4.
%C A291032 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 A291032 See A290890 for a guide to related sequences.
%H A291032 Clark Kimberling, <a href="/A291032/b291032.txt">Table of n, a(n) for n = 0..1000</a>
%H A291032 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (9, -33, 68, -87, 68, -33, 9, -1)
%F A291032 G.f.: (1 - 5 x + 12 x^2 - 17 x^3 + 12 x^4 - 5 x^5 + x^6)/(1 - 9 x + 33 x^2 - 68 x^3 + 87 x^4 - 68 x^5 + 33 x^6 - 9 x^7 + x^8).
%F A291032 a(n) = 9*a(n-1) - 33*a(n-2) + 68*a(n-3) - 87*a(n-4) + 68*a(n-5) - 33*a(n-6) + 9*a(n-7) - a(n-8).
%t A291032 z = 60; s = x/(1 - x)^2; p =1 - s - s^2 - s^3 + s^4;
%t A291032 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A291032 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291032 *)
%Y A291032 Cf. A000027, A290890.
%K A291032 nonn,easy
%O A291032 0,2
%A A291032 _Clark Kimberling_, Aug 19 2017