A069203 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.

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, 53, 91, 48, 139, 187, 326, 171, 497, 668, 1165, 611, 592, 401, 331, 244, 575, 273, 848, 1121, 1969, 1030, 2999, 1343, 4342, 1895, 2079, 3974, 6053, 10027, 5360, 5129

Offset: 1

Author

Benoit Cloitre, Apr 11 2002

Keywords

• easy
• nonn

Comments

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)?

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Programs

Mathematica

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