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.

2, 3, 5, 19, 37, 73, 29, 23, 7, 13, 17, 11, 43, 257, 79, 157, 313, 139, 277, 41, 163, 31, 61, 487, 59, 47, 281, 1123, 449, 359, 239, 53, 211, 421, 101, 67, 89, 71, 283, 113, 677, 2707, 5413, 433, 173, 691, 1381, 251, 167, 1669, 5563, 7417, 44501, 431, 1723, 2297

Offset: 1

Author

Vladimir Shevelev, Mar 10 2010

Keywords

• nonn

Comments

Conjectures: 1) The sequence is a permutation of prime numbers; 2) k = k(n) runs all positive integers.

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005132, A063733, A078943, A174162.

Programs

Mathematica

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 *)

Extensions

Terms from a(22) onwards corrected by Ivan Neretin, Dec 04 2018