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%I #20 Sep 08 2022 08:46:23
%S 3,9,261,
%T 1852673427797059126777135760139006525652319754650249024631321344126610074238981
%N a(1) = 3; for n > 1, a(n) = 2^(a(n-1) - 1) + 5.
%C a(n) divides a(n+1) for n <= 4, but it is unknown if this divisibility holds for larger n. In other words, it is unknown if this sequence is a subsequence of A245594.
%C Modulo any m > 1, the sequence stabilizes within the first A227944(m) <= log_2(m) terms. That is, for any n >= A227944(m), we have a(n) == a(A227944(m)) == A318989(m) (mod m).
%C It follows that the prime divisors of the terms (cf. A318971) are very sparse: if prime p does not divide any of the first log_2(p) terms, then p does not divide any term.
%H Max Alekseyev, <a href="https://mathoverflow.net/q/251717">Iterations of 2^(n-1)+5: the strong law of small numbers, or something bigger?</a>, MathOverflow, 2016.
%t RecurrenceTable[{a[1]==3, a[n]==2^(a[n-1] - 1) + 5}, a, {n, 4}] (* _Vincenzo Librandi_, Sep 07 2018 *)
%o (Magma) [n le 1 select 3 else 2^(Self(n-1)-1)+5: n in [1..4]]; // _Vincenzo Librandi_, Sep 07 2018
%Y Cf. A245594, A318971, A318989.
%K nonn
%O 1,1
%A _Max Alekseyev_, Sep 06 2018
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