

A071529


Number of 1's among the elements of the simple continued fraction for (1+1/n)^n.


2



0, 0, 1, 1, 1, 4, 7, 7, 12, 5, 8, 10, 23, 18, 25, 14, 18, 17, 14, 24, 22, 22, 35, 15, 21, 30, 29, 33, 37, 30, 27, 47, 49, 44, 54, 55, 53, 51, 46, 46, 43, 60, 64, 65, 79, 64, 64, 67, 73, 66, 79, 68, 60, 76, 86, 85, 85, 83, 86, 74, 90, 84, 93, 106, 90, 85, 98, 107, 88, 104, 86
(list;
graph;
refs;
listen;
history;
text;
internal format)



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



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(1component(contfrac(s), i), 0, 1)), ", "))


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



STATUS

approved



