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%I #15 Oct 25 2019 16:57:17
%S 4,5,31,106,53,582,318,9528,11201,19174,142177,315394,648675
%N Let f(i) = prime( f(i - 1) (modulo 10^n) ) with f(0) = 1; a(n) is the length of the period of the sequence f(i).
%e For a(1), the sequence is 1, 2, 3, 5, 11, 2, 3, 5, 11, 2, 3,
%e 5, 11, ... The sequence has period {2, 3, 5, 11} so a(1) = 4.
%e For a(2) see the A112279: 1, 2, 3, 5, 11, 31, 127, 103, 5,
%e 11, 31, 127, 103, 5, 11, ..., . This sequence has a cyclic length of 5.
%t f[n_] := Block[{k = 1, a}, a[0] = 1; a[i_] := a[i] = Prime[Mod[a[i - 1], 10^n]]; While[t = Table[a[i], {i, 0, k - 1}]; MemberQ[t, a[k]] == False, k++ ]; k + 1 - Flatten[ Position[ t, a[k]]][[1]]]; Array[ f, 10]
%K nonn,more
%O 1,1
%A _Robert G. Wilson v_, Apr 05 2007
%E a(11) from _Chai Wah Wu_, Sep 26 2019
%E a(12)-a(13) from _Chai Wah Wu_, Oct 02 2019
%E a(11) and a(12) verified by _Robert G. Wilson v_, Oct 22 2019