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%I #30 Nov 11 2016 21:07:28
%S 3,4,11,76,3571,7881196,38388099893011,910763447271179530132922476,
%T 512653048485188394162163283930413917147479973138989971
%N a(n) = Lucas(2^n + 1).
%C Product_{n>0} (1 + 1/a(n)) = 3 - phi = A094874, where phi = (1+sqrt(5))/2 is the golden mean.
%C From _Peter Bala_, Oct 28 2013: (Start)
%C Compare with A230600(n) = Lucas(2^n - 1).
%C 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 has the value 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 Let Phi := 1/2*(sqrt(5) - 1) denote the reciprocal of the golden ratio. This sequence, apart from the initial term, provides a Pierce expansion of Phi^4 to the base Phi. That is we have the identity Phi^4 = Phi/4 - Phi^2/(4*11) + Phi^3/(4*11*76) - Phi^4/(4*11*76*3571) + ....
%C 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 Phi^(4*k) to the base Phi^k. Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] gives a Pierce expansion of Phi^(2^n + 2) to the base Phi. See below for some examples. (End)
%H J. Sondow, <a href="http://arxiv.org/abs/1106.4246">Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers</a>, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, vol. 1385, pp. 97-100.
%H Y. Tachiya, <a href="http://dx.doi.org/10.1016/j.jnt.2006.11.006">Transcendence of certain infinite products</a>, J. Number Theory 125 (2007), 182-200.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PierceExpansion.html">Pierce Expansion</a>
%F a(n) = A000032(2^n + 1).
%F From _Peter Bala_, Oct 28 2013: (Start)
%F a(n) = phi^(2^n + 1) - (1/phi)^(2^n + 1), where phi = 1/2*(1 + sqrt(5)) denotes the golden ratio A001622.
%F Recurrence equation: a(0) = 3, a(1) = 4 and a(n) = floor(1/phi*a(n-1)^2) + 2 for n >= 2. (End)
%e Pierce series expansion of Phi^(2^n + 2) to the base Phi for n = 1 to 4:
%e n = 1:
%e Phi^4 = Phi/4 - Phi^2/(4*11) + Phi^3/(4*11*76) - Phi^4/(4*11*76*3571) + ...
%e n = 2:
%e Phi^6 = Phi/11 - Phi^2/(11*76) + Phi^3/(11*76*3571) - ...
%e n = 3:
%e Phi^10 = Phi/76 - Phi^2/(76*3571) + Phi^3/(76*3571*7881196) - ...
%e n = 4:
%e Phi^18 = Phi/3571 - Phi^2/(3571*7881196) + ...
%t Table[LucasL[2^n + 1], {n, 0, 10}] (* _T. D. Noe_, Jan 11 2012 *)
%Y Cf. A000032 (Lucas numbers L(n)), A094874 (decimal expansion of 3 - phi), A192222 (Fibonacci(2^n + 1)). A001622, A058635, A230600, A230601, A230602.
%K nonn,easy
%O 0,1
%A _Jonathan Sondow_, Jun 26 2011
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