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A292324 p-INVERT of (1,0,0,1,0,0,0,0,0,...), where p(S) = (1 - S)^2.

%I A292324 #6 Mar 19 2024 08:29:31
%S A292324 2,3,4,7,12,19,28,42,64,97,144,212,312,459,672,979,1422,2062,2984,
%T A292324 4308,6206,8925,12816,18376,26310,37620,53728,76648,109230,155507,
%U A292324 221184,314325,446320,633249,897804,1271993,1800942,2548242,3603468,5092747,7193604
%N A292324 p-INVERT of (1,0,0,1,0,0,0,0,0,...), where p(S) = (1 - S)^2.
%C A292324 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).
%H A292324 Clark Kimberling, <a href="/A292324/b292324.txt">Table of n, a(n) for n = 0..1000</a>
%H A292324 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, 0, 2, -2, 0, 0, -1)
%F A292324 G.f.: -(((-1 + x) (1 + x) (1 - x + x^2) (2 + x + x^2 + x^3))/(-1 + x + x^4)^2).
%F A292324 a(n) = 2*a(n-1) - a(n-2) + 2*a(n-4) - 2*a(n-5) - a(n-8) for n >= 9.
%F A292324 a(n) = a(n-1)+a(n-4)+A003269(n+2). - _R. J. Mathar_, Mar 19 2024
%t A292324 z = 60; s = x + x^4; p = (1 - s)^2;
%t A292324 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
%t A292324 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292324 *)
%Y A292324 Cf. A079978, A292322.
%K A292324 nonn,easy
%O A292324 0,1
%A A292324 _Clark Kimberling_, Sep 15 2017