OFFSET
0,1
COMMENTS
The next term is too large to include.
LINKS
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
KEYWORD
nonn
AUTHOR
Thomas Baruchel, Nov 07 2003
STATUS
approved