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A290999 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6*S^2.

%I A290999 #15 Apr 26 2023 06:27:32
%S A290999 0,6,12,54,168,606,2052,7134,24528,84726,292092,1007814,3476088,
%T A290999 11991246,41362932,142682094,492178848,1697768166,5856430572,
%U A290999 20201701974,69685556808,240379623486,829187031012,2860272179454,9866479513968,34034319925206,117401037420252
%N A290999 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6*S^2.
%C A290999 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 A290999 See A291000 for a guide to related sequences.
%H A290999 Clark Kimberling, <a href="/A290999/b290999.txt">Table of n, a(n) for n = 0..1000</a>
%H A290999 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,5).
%F A290999 G.f.: 6*x/(1 - 2*x - 5*x^2).
%F A290999 a(n) = 2*a(n-1) + 5*a(n-2) for n >= 3.
%F A290999 a(n) = 6*A002532(n).
%F A290999 a(n) = sqrt(3/2)*((1+sqrt(6))^n - (1-sqrt(6))^n). - _Colin Barker_, Aug 23 2017
%t A290999 z = 60; s = x/(1 - x); p = 1 - s^6;
%t A290999 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A290999 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290999 *)
%t A290999 LinearRecurrence[{2,5},{0,6},30] (* _Harvey P. Dale_, Mar 25 2018 *)
%o A290999 (Magma) [n le 2 select 6*(n-1) else 2*Self(n-1) +5*Self(n-2): n in [1..41]]; // _G. C. Greubel_, Apr 25 2023
%o A290999 (SageMath)
%o A290999 A290999=BinaryRecurrenceSequence(2,5,0,6)
%o A290999 [A290999(n) for n in range(41)] # _G. C. Greubel_, Apr 25 2023
%Y A290999 Cf. A000012, A002532, A033453, A289780, A290997, A290998, A291000.
%K A290999 nonn,easy
%O A290999 0,2
%A A290999 _Clark Kimberling_, Aug 22 2017