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%I #4 Sep 23 2017 19:22:10
%S 0,0,1,3,9,23,57,138,332,798,1920,4626,11157,26925,64997,156921,
%T 378861,914692,2208324,5331444,12871324,31074180,75019701,181113471,
%U 437246349,1055605659,2548456957,6152518758,14853493752,35859505946,86572506132,209004519918
%N p-INVERT of the Fibonacci sequence (A000045), where p(S) = 1 - S^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).
%H Clark Kimberling, <a href="/A292329/b292329.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3, 0, -4, 0, 3, 1)
%F G.f.: -(x^2/((-1 + 2 x + x^2) (1 - x - x^2 + x^3 + x^4))).
%F a(n) = 3*a(n-1) - 4*a(n-3) + 3*a(n-5) + a(n-6) for n >= 7.
%t z = 60; s = x/(1 - x - x^2); p = 1 - s^3;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292329 *)
%Y Cf. A000045, A292328.
%K nonn,easy
%O 0,4
%A _Clark Kimberling_, Sep 15 2017
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