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A072893 Let c(k) be defined as follows: c(1)=1, c(2)=x, 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) - c(k) otherwise. Sequence gives values of x such that c(k)=1 for any k large enough. 0

%I

%S 1,2,3,6,8,11,13,18,21,23,28,31,33,43,46,51,53,56,58,66,71,73,78,83,

%T 86,88,91,93,96,98,101,106,111,113,121,123,128,131,133,136,138,141,

%U 146,153,158,161,171,173,176,178,181,183,188,193,201,203,206,211,216,218

%N Let c(k) be defined as follows: c(1)=1, c(2)=x, 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) - c(k) otherwise. Sequence gives values of x such that c(k)=1 for any k large enough.

%C Conjectures: (1) Lim n ->infinity a(n)/n = C = 3.5.... (2) For any n > 2, a(n+1)-a(n) = 2, 3, 5, 7, 8 or 10 only. First differences a(n+1)-a(n) for n>2 are 3, 2, 3, 2, 5, 3, 2, 5, 3, 2, 10, 3, 5, 2, 3, 2, 8... (3) If x is not in this sequence, for k large enough, c(k)= -1 or c(k) reaches one of the two cycles {3, 1, 2, -1} or {0, -1, 1, 0, 1, -1}.

%e If x = 6, c(3)=5, c(4)=-1, c(5)=2, c(6)=3, c(7)=1, c(8)=2, c(9)=1, c(10)=-1, c(11)=0, c(12)=1, c(13)=1, c(14)=1, c(15)=1, ... Hence 6 is in the sequence.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Jul 29 2002

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Last modified December 2 17:30 EST 2021. Contains 349445 sequences. (Running on oeis4.)