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a(1)=1, a(2)=8; for n >= 1, a(n+2)=(a(n+1)+a(n))/3 if (a(n+1)+a(n)==0 (mod 3)); a(n+2)=(a(n+1)+a(n))/2 if (a(n+1)+a(n)==0 (mod 2)); a(n+2)=a(n+1)+a(n) otherwise.
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%I #6 Mar 30 2012 18:38:57

%S 1,8,3,11,7,6,13,19,16,35,17,26,43,23,22,15,37,26,21,47,34,27,61,44,

%T 35,79,38,39,77,58,45,103,74,59,133,64,197,87,142,229,371,200,571,257,

%U 276,533,809,671,740,1411,717,1064,1781,2845,1542,4387,5929,5158,11087

%N a(1)=1, a(2)=8; for n >= 1, a(n+2)=(a(n+1)+a(n))/3 if (a(n+1)+a(n)==0 (mod 3)); a(n+2)=(a(n+1)+a(n))/2 if (a(n+1)+a(n)==0 (mod 2)); a(n+2)=a(n+1)+a(n) otherwise.

%C A Collatz-Fibonacci mixture. Does this sequence diverge to infinity? Conjecture: if a(1)=1 and a(2)=2 sequence is constant = 1, if a(2)=5 sequence is cyclic = (5,2,7,3) but if a(2)=m, different from 2 or 5, sequence diverges.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Apr 11 2002