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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
1, 2, 3, 6, 8, 11, 13, 18, 21, 23, 28, 31, 33, 43, 46, 51, 53, 56, 58, 66, 71, 73, 78, 83, 86, 88, 91, 93, 96, 98, 101, 106, 111, 113, 121, 123, 128, 131, 133, 136, 138, 141, 146, 153, 158, 161, 171, 173, 176, 178, 181, 183, 188, 193, 201, 203, 206, 211, 216, 218
OFFSET
1,2
COMMENTS
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}.
EXAMPLE
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.
CROSSREFS
Sequence in context: A099798 A230108 A097383 * A378162 A349662 A371002
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Jul 29 2002
STATUS
approved