A071528 Number of 1's among the elements of the simple continued fraction for e(n)=sum(k=1,n,1/k!).

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, 32, 27, 40, 40, 46, 35, 44, 38, 41, 43, 40, 46, 45, 55, 54, 57, 62, 53, 57, 52, 59, 67, 61, 67, 66, 69, 74, 80, 79, 85, 77, 78, 76, 83, 85, 88, 96, 78, 101, 93, 89, 101, 88, 106, 95

Offset: 1

Author

Benoit Cloitre, Jun 02 2002

Keywords

• easy
• nonn

Comments

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).

Examples

			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.
		

Programs

PARI

for(n=1,150,if(prod(i=1,length(contfrac((1+1/n)^n)),n-component(contfrac((1+1/n)^n),i)) == 0,print1(n,",")))