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A291015 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^2.

%I A291015 #8 Jun 06 2023 08:18:16
%S A291015 2,7,23,75,244,793,2576,8366,27167,88215,286439,930072,3019941,
%T A291015 9805712,31838986,103380599,335674791,1089929347,3538978588,
%U A291015 11490991649,37311016064,121148109014,393365440335,1277249563655,4147203285279,13465884484800,43723452275981
%N A291015 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^2.
%C A291015 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 A291015 See A291000 for a guide to related sequences.
%H A291015 Clark Kimberling, <a href="/A291015/b291015.txt">Table of n, a(n) for n = 0..1000</a>
%H A291015 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-6,1).
%F A291015 G.f.: (2 - 3*x)/(1 - 5*x + 6*x^2 - x^3).
%F A291015 a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3) n >= 4.
%t A291015 z = 60; s = x/(1-x); p = (1 - s^3)^2;
%t A291015 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291015 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291015 *)
%t A291015 LinearRecurrence[{5,-6,1}, {2,7,23}, 50] (* _G. C. Greubel_, Jun 06 2023 *)
%o A291015 (Magma) I:=[2,7,23]; [n le 3 select I[n] else 5*Self(n-1) -6*Self(n-2) +Self(n-3): n in [1..50]]; // _G. C. Greubel_, Jun 06 2023
%o A291015 (SageMath)
%o A291015 @CachedFunction
%o A291015 def a(n): # a = A291015
%o A291015     if (n<3): return (2,7,23)[n]
%o A291015     else: return 5*a(n-1) - 6*a(n-2) + a(n-3)
%o A291015 [a(n) for n in range(51)] # _G. C. Greubel_, Jun 06 2023
%Y A291015 Cf. A000012, A033453, A289780, A291000.
%K A291015 nonn,easy
%O A291015 0,1
%A A291015 _Clark Kimberling_, Aug 23 2017