OFFSET
1,2
COMMENTS
From Jon E. Schoenfield, Nov 05 2014: (Start)
Several terms are close to 2^n / phi, where phi = (1 + sqrt(5))/2 = 1.6180339... (see A001622); e.g., 2^22/a(22) = 4194304/2592211 = 1.6180411... .
When a ratio r of two integers is expressed as a continued fraction, it cannot have a relatively large number of elements (i.e., relative to other fractions whose numerators and denominators are very roughly the same size as the numerator and denominator of r, respectively) if any of the elements of the continued fraction are large, so the elements of the continued fractions for 2^n / a(n) tend to consist only of small numbers, mostly ones; e.g., 131072/80983 = cf[1; 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 3] (20 elements, consisting of 17 ones, 2 threes, and 1 five).
For n = 1..34, the maximum number of elements is 1, 2, 4, 4, 5, 7, 9, 9, 10, 12, 12, 14, 15, 17, 19, 19, 20, 22, 22, 24, 26, 27, 27, 29, 30, 32, 33, 34, 35, 37, 39, 39, 41, 41.
If a(n) is even, then a(n) = 2*a(n-1), so 2^n/a(n) reduces to 2^(n-1)/a(n-1), and the maximum number of elements is the same at n as it is at n-1. Up to n=34, a(n) is even only at n = 8, 16, and 23. (End)
FORMULA
a(n) = A084242(2^n).
EXAMPLE
From Jon E. Schoenfield, Nov 05 2014: (Start)
The continued fractions for 2^3/k for k = 1..2^3 are
8/1 = 8 (1 element)
8/2 = 4 (1 element)
8/3 = 2 + 1/(1 + 1/2) = cf[2;1,2] (3 elements)
8/4 = 2 (1 element)
8/5 = 1 + 1/(1 + 1/(1 + 1/2)) = cf[1;1,1,2] (4 elements)
8/6 = 4/3 = 1 + 1/3 = cf[1;3] (2 elements)
8/7 = 1 + 1/7 = cf[1;7] (2 elements)
8/8 = 1 (1 element)
so the first (and, in this case, only) value of k at which the maximum number of elements (i.e., 4) occurs is k=5; thus, a(3)=5. (End)
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Jan 01 2004
EXTENSIONS
More terms from Jon E. Schoenfield, Nov 05 2014
STATUS
approved