%I #15 Jun 06 2023 08:18:01
%S 0,3,6,15,36,85,198,456,1040,2352,5280,11776,26112,57600,126464,
%T 276480,602112,1306624,2826240,6094848,13107200,28114944,60162048,
%U 128450560,273678336,581959680,1235222528,2617245696,5536481280,11693719552,24662507520,51942260736
%N p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)^3.
%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="/A291013/b291013.txt">Table of n, a(n) for n = 0..186</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-12,8).
%F G.f.: x*(3 - 12*x + 15*x^2 - 6*x^3 + x^4)/(1 - 2*x)^3.
%F a(n) = 6*a(n-1) - 12*a(n-2) + 8*a(n-3) for n >= 7.
%F a(n) = 2^(n-6) * (60 + 17*n + n^2) for n>2. - _Colin Barker_, Aug 23 2017
%F E.g.f.: -(15/16) + (9/16)*x - x^2/16 + (1/16)*(15 +9*x +x^2)*exp(2*x). - _G. C. Greubel_, Jun 05 2023
%t z = 60; s = x/(1-x); p = (1 - s^2)^3;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291013 *)
%t LinearRecurrence[{6,-12,8}, {0,3,6,15,36,85}, 41] (* _G. C. Greubel_, Jun 05 2023 *)
%o (PARI) concat(0, Vec(x*(3 -12*x +15*x^2 -6*x^3 +x^4)/(1-2*x)^3 + O(x^50))) \\ _Colin Barker_, Aug 23 2017
%o (Magma) [0,3,6] cat [2^(n-6)*(60 +17*n +n^2): n in [3..40]]; // _G. C. Greubel_, Jun 05 2023
%o (SageMath) [(2^(n-2)*(60 +17*n +n^2) -15*int(n==0) + 9*int(n==1))//16 for n in range(41)] # _G. C. Greubel_, Jun 05 2023
%Y Cf. A000012, A033453, A289780, A291000.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 23 2017
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