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p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)*(1 - 2*S).
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%I #12 Jun 05 2023 07:21:53

%S 2,7,22,68,208,632,1912,5768,17368,52232,156952,471368,1415128,

%T 4247432,12746392,38247368,114758488,344308232,1032990232,3099101768,

%U 9297567448,27893226632,83680728472,251044282568,753137042008,2259419514632,6778275321112

%N p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)*(1 - 2*S).

%C 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 See A291000 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291012/b291012.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-6).

%F G.f.: (2 - 3 x - x^2)/(1 - 5*x + 6*x^2).

%F a(n) = 5*a(n-1) - 6*a(n-2) for n >= 4.

%F a(n) = (16*3^n - 3*2^n) / 6 for n > 0. - _Colin Barker_, Aug 23 2017

%F E.g.f.: (1/6)*(-1 - 3*exp(2*x) + 16*exp(3*x)). - _G. C. Greubel_, Jun 04 2023

%t z = 60; s = x/(1-x); p = (1-s)^2(1-2s);

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* this sequence *)

%t LinearRecurrence[{5,-6}, {2,7,22}, 40] (* _G. C. Greubel_, Jun 04 2023 *)

%o (PARI) Vec((2 -3*x -x^2)/((1-2*x)*(1-3*x)) + O(x^30)) \\ _Colin Barker_, Aug 23 2017

%o (Magma) [2] cat [8*3^(n-1) - 2^(n-1): n in [1..40]]; // _G. C. Greubel_, Jun 04 2023

%o (SageMath) [8*3^(n-1) - 2^(n-1) - int(n==0)/6 for n in range(41)] # _G. C. Greubel_, Jun 04 2023

%Y Cf. A000012, A033453, A289780, A291000.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 23 2017