%I
%S 1,1,2,2,4,5,6,8,7,12,11,11,15,13,13,16,19,17,18,19,23,25,25,27,29,32,
%T 32,27,40,40,46,35,44,38,41,43,40,46,45,55,54,57,62,53,57,52,59,67,61,
%U 67,66,69,74,80,79,85,77,78,76,83,85,88,96,78,101,93,89,101,88,106,95
%N Number of 1's among the elements of the simple continued fraction for e(n)=sum(k=1,n,1/k!).
%C It seems that lim n >infinity a(n)/A069880(n) = C = 0.5... which is different from (log(4)log(3))/log(2)=0.415... the expected density of 1's (cf. measure theory of continued fractions).
%e e(10) has for continued fraction [1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 11, 1, 1, 29, 1, 1, 2] which contains 12 "1's" hence a(10)=12.
%o (PARI) for(n=1,150,if(prod(i=1,length(contfrac((1+1/n)^n)),ncomponent(contfrac((1+1/n)^n),i)) == 0,print1(n,",")))
%K easy,nonn
%O 1,3
%A _Benoit Cloitre_, Jun 02 2002
