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Let c(1)=x, c(n+1) = c(n)/2 + n if c(n) is even, c(n+1)= 2c(n) - n otherwise; then a(n)=c(n) for c(1)=3.
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%I #11 Sep 02 2024 22:46:18

%S 3,5,8,7,10,10,11,15,22,20,20,21,30,28,28,29,42,38,37,55,90,66,55,87,

%T 150,100,76,65,102,80,70,66,65,97,160,115,194,134,105,171,302,192,138,

%U 112,100,95,144,119,190,144,122,112,108,107,160,135,214,164,140

%N Let c(1)=x, c(n+1) = c(n)/2 + n if c(n) is even, c(n+1)= 2c(n) - n otherwise; then a(n)=c(n) for c(1)=3.

%C It seems that for any n, 2n <= a(n) <16n. If x=0,1,2,4 or 6 we have c(k+1)-c(k)=2 for k large enough and then lim k -> oo c(k)/k=2. For x=3,5 and for any x >6 there is a conjectured constant 4 < C < 5 such that lim N -> oo (1/N)*sum(k=1,N,c(k)/k) = C. Hence lim N -> oo (1/N)*sum(k=1,N,a(k)/k) should be C=4.6...

%e a(1)=3 is odd hence a(2) = 2*a(1)-1 = 2*3-1 = 5.

%K easy,nonn

%O 1,1

%A _Benoit Cloitre_, Jul 30 2002

%E a(34) onward corrected by _Sean A. Irvine_, Sep 02 2024