login
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).
2

%I #7 Mar 30 2012 18:39:04

%S 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,

%T 20,21,21,22,22,23,23,24,25,26,26,27,28,29,29,30,31,32,32,33,33,34,34,

%U 35,36,37,37,38,38,39,39,40,41,42,42,43,43,44,44,45,46,47,47,48,49,50,50

%N 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).

%C 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 ?

%e 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.

%Y First differences are in A098725.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Jul 29 2002