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A291026 p-INVERT of the positive integers, where p(S) = 1 - 4*S + S^2.

%I A291026 #11 May 18 2024 15:50:14
%S A291026 4,23,128,711,3948,21920,121700,675673,3751296,20826953,115629868,
%T A291026 641969344,3564171060,19788040311,109861881472,609945846247,
%U A291026 3386378699324,18800948912352,104381615697460,579519775642745,3217455182279552,17863096800262569
%N A291026 p-INVERT of the positive integers, where p(S) = 1 - 4*S + S^2.
%C A291026 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 A291026 See A290890 for a guide to related sequences.
%H A291026 Clark Kimberling, <a href="/A291026/b291026.txt">Table of n, a(n) for n = 0..1000</a>
%H A291026 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (8,-15,8,-1).
%F A291026 G.f.: (4 - 9 x + 4 x^2)/(1 - 8 x + 15 x^2 - 8 x^3 + x^4).
%F A291026 a(n) = 8*a(n-1) - 15*a(n-2) + 8*a(n-3) - a(n-4).
%t A291026 z = 60; s = x/(1 - x)^2; p = 1 - 4 s + s^2;
%t A291026 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A291026 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291026 *)
%t A291026 LinearRecurrence[{8,-15,8,-1},{4,23,128,711},30] (* _Harvey P. Dale_, May 18 2024 *)
%Y A291026 Cf. A000027, A033453, A290890.
%K A291026 nonn,easy
%O A291026 0,1
%A A291026 _Clark Kimberling_, Aug 19 2017