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a(1)=0 a(2)=3 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) otherwise.
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%I #8 Nov 21 2013 12:47:46

%S 0,3,1,4,5,3,8,11,19,10,29,13,14,9,23,32,55,29,28,19,47,22,23,15,38,

%T 53,91,48,139,187,326,171,497,668,1165,611,592,401,331,244,575,273,

%U 848,1121,1969,1030,2999,1343,4342,1895,2079,3974,6053,10027,5360,5129

%N a(1)=0 a(2)=3 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) otherwise.

%C A Collatz-Fibonacci mixture. If a(1)=0 and a(2)=m not congruent to 3, a(n) is cyclic (m,m,2m). Does this sequence diverge to infinity if a(1)=0 and a(2)=m == 0 (mod 3)?

%H Harvey P. Dale, <a href="/A069203/b069203.txt">Table of n, a(n) for n = 1..1000</a>

%t Transpose[NestList[{Last[#],If[Divisible[Total[#],3],Total[#]/3, Total[ #]]}&,{0,3},60]][[1]] (* _Harvey P. Dale_, Dec 08 2011 *)

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Apr 11 2002