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%I #6 Jul 08 2022 11:09:51
%S 2,5,14,38,102,271,714,1868,4858,12569,32374,83058,212350,541219,
%T 1375570,3487384,8821170,22266413,56098206,141087934,354268502,
%U 888238903,2223968666,5561234916,13889778218,34652529473,86361653126,215021205770,534861620718
%N p-INVERT of the Fibonacci sequence (A000045), where p(S) = (1 - S)^2.
%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="/A292327/b292327.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, -2, -4, -1)
%F G.f.: -(2 + x) *(-1 + 2 *x))/(-1 + 2 x + x^2)^2.
%F a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.
%F a(n) = A006645(n+1) +2*A000129(n+1). - _R. J. Mathar_, Jul 08 2022
%t z = 60; s = x/(1 - x - x^2); p = (1 - s)^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292327 *)
%Y Cf. A000045, A292328.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Sep 15 2017
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