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p-INVERT of the Fibonacci numbers (A000045, including 0), where p(S) = 1 - S - S^2.
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%I #6 Aug 21 2017 22:07:44

%S 0,1,1,4,7,18,37,85,183,407,888,1956,4284,9409,20630,45270,99289,

%T 217819,477776,1048053,2298912,5042783,11061455,24263687,53223023,

%U 116746272,256086074,561731936,1232174181,2702807740,5928681960,13004724921,28526216361

%N p-INVERT of the Fibonacci numbers (A000045, including 0), where p(S) = 1 - S - 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).

%C See A290890 for a guide to related sequences.

%H Clark Kimberling, <a href="/A289975/b289975.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 (2, 2, -3, -1)

%F G.f.: (x - x^2)/(1 - 2 x - 2 x^2 + 3 x^3 + x^4).

%F a(n) = 2*a(n-1) + 2*a(n-2) - 3*a(n-3) - a(n-4).

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289975 *)

%Y Cf. A000045, A289781, A289976, A289780.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Aug 21 2017