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%I #19 Jan 22 2019 14:22:33
%S 3,14,611,1346270,6557470319843,155576970220531065681649694,
%T 87571595343018854458033386304178158174356588264390371
%N Expansion of 1/phi (phi being the golden ratio) as an infinite product: 1/phi = Product_{k=0..n} (1-1/a(k)).
%C The next term is too large to include.
%H J. Shallit, <a href="https://www.fq.math.ca/Scanned/18-1/elementary18-1.pdf">Problem B-423</a>, The Fibonacci Quarterly 18,1 Feb.(1980)85. <a href="https://www.fq.math.ca/Scanned/19-1/elementary19-1.pdf">Solution</a> 19,1 Feb. (1981) 92. [_Wolfdieter Lang_, Nov 04 2010]
%F a(0) = 3, a(n+1) = (a(n)-1)*A001566(n+1)
%F a(n) = 1+ceiling(1/2*(1-1/sqrt(5))*phi^(2^(n+2))) where phi=(1+sqrt(5))/2. a(n)==2 (mod 3) for n>0. - _Benoit Cloitre_, Nov 09 2003
%F a(n) = b(n+2)+1, n>=0, with b(n):= A101342(n) = F(2^n-1). See the reciprocal of the infinite product of this entry. For a proof see the J. Shallit reference. - _Wolfdieter Lang_, Nov 04 2010
%o (PARI) a(n)=if(n<0,0,fibonacci(2^(n+2)-1)+1)
%Y Cf. A001566, A001622, A094214, A101342.
%K nonn
%O 0,1
%A _Thomas Baruchel_, Nov 07 2003
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