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A230602 a(n) = Lucas(2^n - 2).

%I A230602 #9 Jan 03 2014 00:52:33
%S A230602 2,3,18,843,1860498,9062201101803,215002084978043708894524818,
%T A230602 121020968315000050139390193037122554865361969834971243,
%U A230602 38343921554607207587938114587587818441864732465057252794474861753545122655196096751375348482086938743684498
%N A230602 a(n) = Lucas(2^n - 2).
%C A230602 Let phi := 1/2*(1 + sqrt(5)) denote the golden ratio. This sequence, apart from the initial term, gives an Engel expansion of 1 to the base phi^2 (see A230601 for a definition of this term). The associated Engel series expansion of 1 to the base phi^2 begins 1 = phi^2/3 + phi^4/(3*18) + phi^6/(3*18*843) + phi^8/(3*18*843*1860498) + .... This result can be extended in two ways. Firstly, the sequence Lucas(2*k*(2^n - 1)) for k = 1,2,3,... is an Engel expansion of 1 to the base phi^(2*k). Secondly, for n = 1,2,3,... the sequence [a(n),a(n+1),a(n+2),...] is an Engel expansion of phi^(4 - 2^n) to the base phi^2. Some example are given below.
%H A230602 Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>
%F A230602 a(n) = A000032(2^n-2) = phi^(2^n-2) + (1/phi)^(2^n-2), where phi := 1/2*(1 + sqrt(5)).
%F A230602 Recurrence equation: a(1) = 2, a(2) = 3 and a(n) = floor(phi^2*a(n-1)^2) - 5 for n >= 3.
%e A230602 Engel series expansion of phi^(4 - 2^n) to the base phi^2 for n = 1 to 5.
%e A230602 n = 1:
%e A230602 phi^2 = phi^2/2 + phi^4/(2*3) + phi^6/(2*3*18) + phi^8/(2*3*18*843) + ...
%e A230602 n = 2:
%e A230602 1 = phi^2/3 + phi^4/(3*18) + phi^6/(3*18*843) + phi^8/(3*18*843*1860498) + ...
%e A230602 n = 3:
%e A230602 1/phi^4 = phi^2/18 + phi^4/(18*843) + phi^6/(18*843*1860498) + ...
%e A230602 n = 4:
%e A230602 1/phi^12 = phi^2/843 + phi^4/(843*1860498) + phi^6/(843*1860498*9062201101803) + ...
%e A230602 n = 5:
%e A230602 1/phi^28 = phi^2/1860498 + phi^4/(1860498*9062201101803) + ...
%t A230602 Table[LucasL[2^n - 2], {n, 1, 10}]
%Y A230602 Cf. A000032, A001622, A192223, A230600, A230601.
%K A230602 nonn,easy
%O A230602 1,1
%A A230602 _Peter Bala_, Oct 28 2013