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A292532 p-INVERT of the squares (A000290), where p(S) = 1 - S^3.

%I A292532 #7 Sep 27 2023 12:58:27
%S A292532 0,0,1,12,75,329,1158,3606,10971,35601,126168,467541,1722714,6173070,
%T A292532 21563906,74452230,257613930,899546303,3166966692,11185908147,
%U A292532 39459021883,138761604786,486746839758,1705955898935,5982257083623,20999661326520,73772324787965
%N A292532 p-INVERT of the squares (A000290), where p(S) = 1 - S^3.
%C A292532 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 A292532 See A292479 for a guide to related sequences.
%H A292532 Clark Kimberling, <a href="/A292532/b292532.txt">Table of n, a(n) for n = 0..1000</a>
%H A292532 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9, -36, 85, -123, 129, -83, 36, -9, 1)
%F A292532 G.f.: -((x^2 (1 + x)^3)/((-1 + 4 x - 2 x^2 + x^3) (1 - 5 x + 14 x^2 - 18 x^3 + 18 x^4 - 7 x^5 + x^6))).
%F A292532 a(n) = 9*a(n-1) - 36*a(n-2) + 85*a(n-3) - 123*a(n-4) + 129*a(n-5) - 83*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n >= 10.
%t A292532 z = 60; s = x (x + 1)/(1 - x)^3; p = 1 - s^3;
%t A292532 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292532 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292532 *)
%t A292532 LinearRecurrence[{9,-36,85,-123,129,-83,36,-9,1},{0,0,1,12,75,329,1158,3606,10971},30] (* _Harvey P. Dale_, Sep 27 2023 *)
%Y A292532 Cf. A000290, A292479.
%K A292532 nonn,easy
%O A292532 0,4
%A A292532 _Clark Kimberling_, Oct 04 2017