A088334 Expansion of 1/phi (phi being the golden ratio) as an infinite product: 1/phi = Product_{k=0..n} (1-1/a(k)).

3, 14, 611, 1346270, 6557470319843, 155576970220531065681649694, 87571595343018854458033386304178158174356588264390371

Offset

0,1

Comments

The next term is too large to include.

Links

Table of n, a(n) for n=0..6.

J. Shallit, Problem B-423, The Fibonacci Quarterly 18,1 Feb.(1980)85. Solution 19,1 Feb. (1981) 92. [Wolfdieter Lang, Nov 04 2010]

Formula

a(0) = 3, a(n+1) = (a(n)-1)*A001566(n+1)

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

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

Prog

(PARI) a(n)=if(n<0, 0, fibonacci(2^(n+2)-1)+1)

Crossrefs

Cf. A001566, A001622, A094214, A101342.

Sequence in context: A081397 A264748 A092987 * A304059 A305453 A305016

Adjacent sequences: A088331 A088332 A088333 * A088335 A088336 A088337

Keyword

nonn

Author

Thomas Baruchel, Nov 07 2003

Status

approved