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%I #11 Aug 24 2021 14:46:53
%S 1,1,4,1,1,1,3,1,1,1,6,1,1,1,5,1,10,1,1,1,1,1,1,1,1,8,1,1,22,1,5,1,1,
%T 1,8,1,1,1,1,1,1,1,1,1,1,1,9,1,6,1,1,1,6,1,1,1,1,1,10,1,1,1,8,1,1,1,1,
%U 1,7,1,9,1,1,1,1,1,1,1,5,14,1,1,6,1,1,1
%N Let {b_n(m)} be a sequence defined by b_n(0)=0, b_n(m) is the largest prime dividing (b_n(m-1) + 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)}.
%o (PARI) a(n) = my(b, k, v=List([0])); until(k<#v, k=1; listput(v, b=vecmax(factor(b+n)[, 1])); until(v[k]==b||k==#v, k++)); #v-k; \\ _Jinyuan Wang_, Aug 22 2021
%Y Cf. A006530, A134834.
%K nonn
%O 2,3
%A _Leroy Quet_, Nov 12 2007
%E More terms from _Jinyuan Wang_, Aug 22 2021
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