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A292326 p-INVERT of (1,1,1,0,0,0,0,0,0,0,0,...), where p(S) = (1 - S)^3.

%I A292326 #9 Nov 01 2019 13:40:41
%S A292326 3,9,25,63,153,359,819,1830,4018,8694,18582,39298,82350,171186,353338,
%T A292326 724719,1478061,2999175,6057687,12183945,24411935,48740193,96998325,
%U A292326 192459996,380812692,751557756,1479686972,2906717460,5698014924,11147786740,21769549380
%N A292326 p-INVERT of (1,1,1,0,0,0,0,0,0,0,0,...), where p(S) = (1 - S)^3.
%C A292326 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 A292326 Clark Kimberling, <a href="/A292326/b292326.txt">Table of n, a(n) for n = 0..1000</a>
%H A292326 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (3, 0, -2, -6, 0, 4, 6, 3, 1)
%F A292326 G.f.: -(((1 + x + x^2) (3 - 3 x - 2 x^2 - x^3 + 3 x^4 + 2 x^5 + x^6))/(-1 + x + x^2 + x^3)^3).
%F A292326 a(n) = 3*a(n-1) - 2*a(n-3) - 6*a(n-4) + 4*a(n-6) + 6*a(n-7) + 3*a(n-8) + a(n-9) for n >= 10.
%t A292326 z = 60; s = x + x^2 + x^3; p = (1 - s)^3;
%t A292326 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
%t A292326 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292326 *)
%t A292326 LinearRecurrence[{3,0,-2,-6,0,4,6,3,1},{3,9,25,63,153,359,819,1830,4018},40] (* _Harvey P. Dale_, Nov 01 2019 *)
%Y A292326 Cf. A292324.
%K A292326 nonn,easy
%O A292326 0,1
%A A292326 _Clark Kimberling_, Sep 15 2017