A072894 Let c(k) be defined as follows: c(1)=1, c(2)=n, c(k+2) = c(k+1)/2 + c(k)/2 if c(k+1) and c(k) have the same parity; c(k+2) = c(k+1)/2 + c(k)/2 + 1/2 otherwise; a(n) = limit_{ k -> infinity} c(k).

1, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 10, 10, 11, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 21, 21, 22, 22, 23, 23, 24, 25, 26, 26, 27, 28, 29, 29, 30, 31, 32, 32, 33, 33, 34, 34, 35, 36, 37, 37, 38, 38, 39, 39, 40, 41, 42, 42, 43, 43, 44, 44, 45, 46, 47, 47, 48, 49, 50, 50

Offset: 1

Author

Benoit Cloitre, Jul 29 2002

Keywords

• easy
• nonn

Comments

Conjectures : (1) a(n+1)-a(n) = 0 or 1; (2) lim n ->infinity a(n)/n = 2/3; (3) 1/2 < (3a(n)-2n)/Log(n) <3/2 for any n > 1000. Does lim n -> infinity (3a(n)-2n)/Log(n) = 1 ?

Examples

			If n=5, c(3)=(1+5)/2=3, c(4)=(3+5)/2=4, c(5)=(4+3+1)/2=4, ..., hence a(5)=4.
		

Crossrefs

First differences are in A098725.