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A290915 p-INVERT of the positive integers, where p(S) = 1 - 8*S^2.

%I A290915 #8 Aug 19 2017 13:29:36
%S A290915 0,8,32,144,672,3096,14272,65824,303552,1399848,6455520,29770160,
%T A290915 137287520,633112632,2919650688,13464207936,62091296128,286339090504,
%U A290915 1320476135328,6089483698896,28082152132128,129503141377112,597214328432960,2754102721315680
%N A290915 p-INVERT of the positive integers, where p(S) = 1 - 8*S^2.
%C A290915 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 A290915 See A290890 for a guide to related sequences.
%H A290915 Clark Kimberling, <a href="/A290915/b290915.txt">Table of n, a(n) for n = 0..1000</a>
%H A290915 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, 2, 4, -1)
%F A290915 G.f.: (8 x)/(1 - 4 x - 2 x^2 - 4 x^3 + x^4).
%F A290915 a(n) = 4*a(n-1) + 2*a(n-2) + 4*a(n-3) - a(n-4).
%F A290915 a(n) = 8*A290916(n) for n >= 0.
%t A290915 z = 60; s = x/(1 - x)^2; p = 1 - 8 s^2;
%t A290915 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290915 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290915 *)
%t A290915 u/8 (* A290916 *)
%Y A290915 Cf. A000027, A290890, A290916.
%K A290915 nonn,easy
%O A290915 0,2
%A A290915 _Clark Kimberling_, Aug 18 2017