A174162 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, 5, 11, 23, 47, 19, 3, 7, 29, 59, 17, 103, 619, 2477, 991, 661, 3967, 2267, 907, 191, 383, 13, 53, 107, 43, 173, 347, 139, 31, 83, 167, 67, 269, 359, 719, 1439, 2879, 443, 887, 71, 61, 41, 137, 823, 37, 149, 199, 797, 1063, 709, 2837, 227, 101, 607, 3643, 21859

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.

Crossrefs

Cf. A005132, A063733, A078943, A174161.

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 (* Jinyuan Wang, Feb 26 2020 *)

Extensions

More terms from Jinyuan Wang, Feb 26 2020