%I
%S 1,1,4,1,1,1,3,1,1,1,6,1
%N Let {b_n(m)} be a sequence defined by b_n(0)=0, b_n(m) = the largest prime dividing (b_n(m1) +n). Then a(n) is the smallest positive integer such that b_n(m+a(n)) = b_n(m), for all integers m that are greater than some positive integer M.
%e Sequence {b_8(m)} is 0,2,5,13,7,5,13,7,...(5,13,7) repeats. So a(8) = 3, the length of the cycle in {b_8(m)}.
%Y Cf. A134834.
%K more,nonn
%O 2,3
%A _Leroy Quet_, Nov 12 2007
