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A289781 p-INVERT of the positive Fibonacci numbers (A000045), where p(S) = 1 - S - S^2.

%I A289781 #24 Aug 18 2017 04:42:26
%S A289781 1,3,9,27,80,237,701,2073,6129,18120,53569,158367,468181,1384083,
%T A289781 4091760,12096453,35760689,105719157,312537041,923951760,2731474161,
%U A289781 8075043963,23872213729,70573310907,208635540400,616788246957,1823408134821,5390532719313
%N A289781 p-INVERT of the positive Fibonacci numbers (A000045), where p(S) = 1 - S - S^2.
%C A289781 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 A289781 See A289780 for a guide to related sequences.
%H A289781 Clark Kimberling, <a href="/A289781/b289781.txt">Table of n, a(n) for n = 0..1000</a>
%H A289781 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3, 1, -3, -1)
%F A289781 G.f.: (1 - x^2)/(1 - 3 x - x^2 + 3 x^3 + x^4).
%F A289781 a(n) = 3*a(n-1) + a(n-2) - 3*a(n-3) - a(n-4).
%t A289781 z = 60; s = x/(1 - x - x^2); p = 1 - s - s^2;
%t A289781 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
%t A289781 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289781 *)
%Y A289781 Cf. A000045, A289780, A289975, A289976, A289977.
%K A289781 nonn,easy
%O A289781 0,2
%A A289781 _Clark Kimberling_, Aug 10 2017