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A289804 p-INVERT of the even bisection (A001519) of the Fibonacci numbers, where p(S) = 1 - S - S^2.

%I A289804 #9 Aug 13 2017 17:24:16
%S A289804 1,3,9,29,96,321,1077,3617,12149,40802,137009,459991,1544169,5183201,
%T A289804 17396800,58387097,195950657,657602545,2206838633,7405775266,
%U A289804 24852220929,83398067755,279861976377,939138581941,3151475258656,10575403936625,35487807890381
%N A289804 p-INVERT of the even bisection (A001519) of the Fibonacci numbers, where p(S) = 1 - S - S^2.
%C A289804 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 A289804 See A289780 for a guide to related sequences.
%H A289804 Clark Kimberling, <a href="/A289804/b289804.txt">Table of n, a(n) for n = 0..1000</a>
%H A289804 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7, -15, 9, 1)
%F A289804 G.f.: (-1 + 4 x - 3 x^2 - 2 x^3)/(-1 + 7 x - 15 x^2 + 9 x^3 + x^4).
%F A289804 a(n) = 7*a(n-1) - 15*a(n-2) + 9*a(n-3) + a(n-4).
%t A289804 z = 60; s = x (1 - 2*x)/(1 - 3*x + x^2); p = 1 - s - s^2;
%t A289804 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A001519 shifted *)
%t A289804 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289804 *)
%Y A289804 Cf. A001519, A289780, A289803.
%K A289804 nonn,easy
%O A289804 0,2
%A A289804 _Clark Kimberling_, Aug 12 2017