cp's OEIS Frontend

A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2*k*a(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

%I A174161 #13 Oct 16 2024 16:40:36
%S A174161 2,3,5,19,37,73,29,23,7,13,17,11,43,257,79,157,313,139,277,41,163,31,
%T A174161 61,487,59,47,281,1123,449,359,239,53,211,421,101,67,89,71,283,113,
%U A174161 677,2707,5413,433,173,691,1381,251,167,1669,5563,7417,44501,431,1723,2297
%N A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2*k*a(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.
%C A174161 Conjectures: 1) The sequence is a permutation of prime numbers; 2) k = k(n) runs all positive integers.
%H A174161 Ivan Neretin, <a href="/A174161/b174161.txt">Table of n, a(n) for n = 1..10000</a>
%t A174161 a = {2}; Do[k = 1; While[(d = Complement[FactorInteger[2 k a[[-1]] - 1][[All, 1]], a]) == {}, k++]; AppendTo[a, Min@d], {n, 50}]; a (* _Ivan Neretin_, Dec 04 2018 *)
%Y A174161 Cf. A005132, A063733, A078943, A174162.
%K A174161 nonn
%O A174161 1,1
%A A174161 _Vladimir Shevelev_, Mar 10 2010
%E A174161 Terms from a(22) onwards corrected by _Ivan Neretin_, Dec 04 2018