OFFSET
1,6
COMMENTS
It seems that lim_{n->infinity} a(n)/A069887(n) = C = 0.41..., which is close to (log(4)-log(3))/log(2)=0.415..., the expected density of 1's (cf. measure theory of continued fractions).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
EXAMPLE
(1+1/14)^14 has for continued fraction [2, 1, 1, 1, 2, 6, 1, 7, 1, 6, 2, 1, 4, 21, 1, 1, 7, 1, 1, 1, 3, 2, 7, 2, 7, 1, 2, 4, 1, 3, 2, 1, 1, 1, 5, 1, 2, 5, 1, 2] which contains 18 "1's" hence a(14)=18.
MATHEMATICA
Table[Count[ContinuedFraction[(1+1/n)^n], 1], {n, 80}] (* Harvey P. Dale, Mar 11 2013 *)
PROG
(PARI) for(n=1, 100, s=(1+1/n)^n; print1(sum(i=1, length(contfrac(s)), if(1-component(contfrac(s), i), 0, 1)), ", "))
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Jun 02 2002
STATUS
approved