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A291409 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S^2)(1 - S)^2.

%I A291409 #10 May 12 2024 02:01:27
%S A291409 2,6,14,31,66,136,272,534,1030,1958,3678,6837,12594,23016,41768,75325,
%T A291409 135084,241032,428112,757236,1334292,2342892,4100676,7155937,12453170,
%U A291409 21616242,37432010,64675099,111512574,191893120,329605760,565166682,967491754,1653659282
%N A291409 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S^2)(1 - S)^2.
%C A291409 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). See A291382 for a guide to related sequences.
%H A291409 Clark Kimberling, <a href="/A291409/b291409.txt">Table of n, a(n) for n = 0..1000</a>
%H A291409 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2, 2, -2, -5, -2, 4, 4, 1)
%F A291409 G.f.: -(((1 + x) (2 - 2 x^2 - 3 x^3 + x^4 + 3 x^5 + x^6))/((-1 + x + x^2)^3 (1 + x + x^2))).
%F A291409 a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - 5*a(n-4) - 2*a(n-5) + 4*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.
%t A291409 z = 60; s = x + x^2; p = (1 - s^2)(1 - s)^2;
%t A291409 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291409 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291409 *)
%t A291409 LinearRecurrence[{2,2,-2,-5,-2,4,4,1},{2,6,14,31,66,136,272,534},40] (* _Harvey P. Dale_, May 12 2024 *)
%Y A291409 Cf. A019590, A291382.
%K A291409 nonn,easy
%O A291409 0,1
%A A291409 _Clark Kimberling_, Sep 07 2017