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A230600 a(n) = Lucas(2^n - 1).

%I A230600 #30 Feb 25 2025 01:45:40
%S A230600 2,1,4,29,1364,3010349,14662949395604,347880681146567910619198829,
%T A230600 195816040085094172011386545446645681141059001652009364,
%U A230600 62041768337314169100816125405238438263014895218124648720624536859920496229610874552659773465281966850403949
%N A230600 a(n) = Lucas(2^n - 1).
%C A230600 Compare with A192223(n) = Lucas(2^n + 1).
%C A230600 Let x and b be positive real numbers. We define a Pierce expansion of x to the base b to be a (possibly infinite) increasing sequence of positive integers [a(1), a(2), a(3), ...] such that we have the alternating series representation x = b/a(1) - b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) - .... This definition generalizes the ordinary Pierce expansion of a real number 0 < x < 1, where the base b is taken equal to 1. Depending on the values of x and b such a generalized Pierce expansion to the base b may not exist, and if it does exist it may not be unique.
%C A230600 Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence, apart from the initial term, provides a Pierce expansion of 1 to the base phi. That is, we have the identity 1 = phi/1 - phi^2/(1*4) + phi^3/(1*4*29) - phi^4/(1*4*29*1364) + ....
%C A230600 This result can be extended in two ways. Firstly, for k odd, the sequence {Lucas(k*(2^n - 1))} n>=1 gives a Pierce expansion of 1 to the base phi^k. Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] provides a Pierce expansion of the quadratic irrational 1/phi^(2^n - 2) = Fibonacci(2^n - 1) - Fibonacci(2^n - 2)*phi to the base phi. Some examples are given below.
%H A230600 Iain Fox, <a href="/A230600/b230600.txt">Table of n, a(n) for n = 0..12</a>
%H A230600 Iain Fox, <a href="/A230600/a230600.txt">Table of n, a(n) for n = 0..24</a> (terms too large for b-file)
%H A230600 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PierceExpansion.html">Pierce Expansion</a>
%F A230600 a(n) = Lucas(2^n - 1) = A000032(2^n-1) = phi^(2^n-1) + (-1/phi)^(2^n-1).
%F A230600 Recurrence equation: a(0) = 2, a(1) = 1 and a(n) = floor(phi*a(n-1)^2) + 4 for n >= 2.
%F A230600 Product {n = 2..k} (1 - 1/a(n)) = 1/2 + Fibonacci(2^k - 1)/(2*Lucas(2^k - 1)).
%F A230600 Product {n >= 2} (1 - 1/a(n)) = (5 + sqrt(5))/10.
%F A230600 a(n) = A000032(A000225(n)). - _Omar E. Pol_, Oct 28 2013
%e A230600 A Pierce expansion of 1/phi^(2^n - 2) = Fibonacci(2^n - 1) - Fibonacci(2^n - 2)*phi to the base phi for n = 1 to 4.
%e A230600 n = 1: 1        = phi/1 - phi^2/(1*4) + phi^3/(1*4*29) - phi^4/(1*4*29*1364) + ...
%e A230600 n = 2: 1/phi^2  = phi/4 - phi^2/(4*29) + phi^3/(4*29*1364) - phi^4/(4*29*1364*3010349) + ...
%e A230600 n = 3: 1/phi^6  = phi/29 - phi^2/(29*1364) + phi^3/(29*1364*3010349) - ...
%e A230600 n = 4: 1/phi^14 = phi/1364 - phi^2/(1364*3010349) + ...
%t A230600 Table[LucasL[2^n - 1], {n, 0, 10}]
%o A230600 (PARI) for(n=0,10, print1(fibonacci(2^n) + fibonacci(2^n -2), ", ")) \\ _G. C. Greubel_, Dec 22 2017
%o A230600 (Magma) [Lucas(2^n -1): n in [0..10]]; // _G. C. Greubel_, Dec 22 2017
%Y A230600 Cf. A000032, A001622, A058635, A192223, A230601, A230602.
%K A230600 nonn,easy
%O A230600 0,1
%A A230600 _Peter Bala_, Oct 28 2013