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A291025 p-INVERT of the positive integers, where p(S) = 1 - 3*S + S^2.

%I A291025 #9 Jun 22 2022 16:46:50
%S A291025 3,14,62,273,1200,5271,23146,101626,446181,1958880,8600043,37756502,
%T A291025 165760934,727733433,3194937360,14026596927,61580365906,270353629378,
%U A291025 1186921889997,5210892012480,22877154557139,100436585338334,440942410322894,1935850452749409
%N A291025 p-INVERT of the positive integers, where p(S) = 1 - 3*S + S^2.
%C A291025 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 A291025 See A290890 for a guide to related sequences.
%H A291025 Clark Kimberling, <a href="/A291025/b291025.txt">Table of n, a(n) for n = 0..1000</a>
%H A291025 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7, -13, 7, -1)
%F A291025 G.f.: (3 - 7 x + 3 x^2)/(1 - 7 x + 13 x^2 - 7 x^3 + x^4).
%F A291025 a(n) = 7*a(n-1) - 13*a(n-2) + 7*a(n-3) - a(n-4).
%t A291025 z = 60; s = x/(1 - x)^2; p = 1 - 3 s + s^2;
%t A291025 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A291025 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291025 *)
%t A291025 LinearRecurrence[{7,-13,7,-1},{3,14,62,273},30] (* _Harvey P. Dale_, Jun 22 2022 *)
%Y A291025 Cf. A000027, A290890.
%K A291025 nonn,easy
%O A291025 0,1
%A A291025 _Clark Kimberling_, Aug 19 2017