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a(1)=1 a(2)=6 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 1,6,7,13,10,23,11,17,14,31,15,23,19,14,11,25,12,37,49,43,46,89,45,67,

%T 56,41,97,46,143,63,103,83,62,145,69,107,88,65,51,58,109,167,92,259,

%U 117,188,305,493,266,253,173,142,105,247,176,141,317,229,182,137,319

%N a(1)=1 a(2)=6 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