

A134834


Let {b_n(m)} be a sequence defined by b_n(0)=1, b_n(m) is 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.


1



2, 3, 2, 4, 3, 8, 2, 3, 4, 6, 2, 6, 3, 3, 2, 5, 4, 10, 3, 4, 3, 2, 2, 3, 3, 3, 8, 6, 8, 22, 2, 5, 5, 8, 2, 8, 4, 9, 2, 4, 3, 12, 3, 3, 6, 6, 2, 7, 4, 6, 3, 6, 4, 3, 3, 5, 12, 14, 3, 14, 4, 12, 3, 8, 6, 21, 5, 5, 7, 7, 2, 9, 5, 10, 11, 10, 7, 8, 3, 3, 14, 4, 7
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OFFSET

1,1


LINKS



EXAMPLE

Sequence {b_9(m)} is 1, 5, 7, 2, 11, 5, 7, 2, 11, ... (5, 7, 2, 11) repeats. So a(9) = 4, the length of the cycle in {b_9(m)}.


PROG

(PARI) a(n) = my(b=1, k, v=List([1])); until(k<#v, k=1; listput(v, b=vecmax(factor(b+n)[, 1])); until(v[k]==bk==#v, k++)); #vk; \\ Jinyuan Wang, Aug 22 2021


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



