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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.
1
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
1,1
COMMENTS
Conjectures: 1) The sequence is a permutation of prime numbers; 2) k = k(n) runs all positive integers.
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Mar 10 2010
EXTENSIONS
More terms from Jinyuan Wang, Feb 26 2020
STATUS
approved