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A289799 p-INVERT of (n^3), where p(S) = 1 - S - S^2.

%I A289799 #11 Jan 07 2024 15:34:14
%S A289799 1,10,62,377,2232,13015,75898,444014,2601503,15244128,89303905,
%T A289799 523084546,3063814838,17945741321,105115487400,615706236199,
%U A289799 3606449444722,21124456768934,123734572586495,724763983514112,4245239506761217,24866107799273146,145650985218990062
%N A289799 p-INVERT of (n^3), where p(S) = 1 - S - S^2.
%C A289799 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 A289799 See A289780 for a guide to related sequences.
%H A289799 Clark Kimberling, <a href="/A289799/b289799.txt">Table of n, a(n) for n = 0..1000</a>
%H A289799 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (9, -27, 55, -36, 55, -27, 9, -1)
%F A289799 G.f.: (1 + x - x^2 + 34 x^3 - x^4 + x^5 + x^6)/(1 - 9 x + 27 x^2 - 55 x^3 + 36 x^4 - 55 x^5 + 27 x^6 - 9 x^7 + x^8).
%F A289799 a(n) = 9*a(n-1) - 27*a(n-2) + 55*a(n-3) - 36*a(n-4) + 55*a(n-5) - 27*a(n-6) + 9*a(n-7) - a(n-8).
%t A289799 z = 60; s = x*(1 + 4*x + x^2)/(1 - x)^4; p = 1 - s - s^2;
%t A289799 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000578 *)
%t A289799 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289799 *)
%t A289799 LinearRecurrence[{9,-27,55,-36,55,-27,9,-1},{1,10,62,377,2232,13015,75898,444014},30] (* _Harvey P. Dale_, Jan 07 2024 *)
%Y A289799 Cf. A000578, A289780.
%K A289799 nonn,easy
%O A289799 0,2
%A A289799 _Clark Kimberling_, Aug 12 2017