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%I #15 Feb 08 2024 09:48:54
%S 1,1,2,1,1,2,1,3,2,1,1,1,3,2,1,1,2,1,1,2,3,1,1,1,2,2,2,1,3,1,1,1,3,2,
%T 1,1,2,2,2,1,2,1,2,4,1,3,1,1,1,2,2,1,4,1,1,2,3,1,1,1,3,1,1,2,3,1,2,2,
%U 5,1,1,1,4,1,2,2,2,2,1,4,3,1,1,1,1,2,1
%N Define the sequence {b_n(m)} by b_n(0)=1; b_n(m) = A000005(b_n(m-1) + n), for all m >= 1. Then a(n) is the smallest positive integer such that b_n(m) = b_n(m + a(n)) for all m > some positive integer.
%e {b_8(m)} is 1, 3, 2, 4, 6, 4, 6, ..., with (4, 6) repeating thereafter. So a(8) = 2, the length of the repeating subsequence (4, 6).
%t nmax = 100; mmax = 10; (* sequence shows zeroes if mmax insufficient *)
%t b[n_, m_] := b[n, m] = If[m == 0, 1, DivisorSigma[0, b[n, m-1] + n]];
%t perLen[lst_] := FindTransientRepeat[lst, 2] // Last // Length;
%t a[n_] := Table[b[n, m], {m, 0, mmax}] // perLen;
%t Table[a[n], {n, 0, nmax}] (* _Jean-François Alcover_, Feb 08 2024 *)
%o (PARI) a(n) = my(b=1, k, v=List([1])); until(k<#v, k=1; listput(v, b=numdiv(b+n)); until(v[k]==b||k==#v, k++)); #v-k; \\ _Jinyuan Wang_, Aug 22 2021
%Y Cf. A000005, A135063.
%K nonn
%O 0,3
%A _Leroy Quet_, Nov 15 2007
%E More terms from _Jinyuan Wang_, Aug 22 2021
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