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A063904
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a(1) = 2, a(2) = 3 and a(k+1) is the least prime not already chosen that divides some a(i)*a(j)+1, where 1<=i<j<=k.
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1
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2, 3, 7, 5, 11, 13, 17, 23, 29, 19, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271
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OFFSET
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1,1
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COMMENTS
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"Does the sequence [above] contain every prime? Is the sequence infinite?" ... "The sequence of problem [above] is not even known to be infinite, though it almost surely contains every prime. We do not know whether anyone has attacked the problem computationally; perhaps you, the reader, would like to give it a try. The problem is due to M. Newman at the Australian Nation University." - Crandall and Pomerance
Indices of primes in this sequence: 1, 2, 4, 3, 5, 6, 7, 9, 10, 8, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, ..., . It appears that only the primes 7 and 19 are out of order in the first 1000. - Robert G. Wilson v, Apr 12 2006 [corrected by Sergey Pavlov, Apr 26 2017]
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REFERENCES
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R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, NY, 2001; see p. 6.
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LINKS
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EXAMPLE
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a(3) is equal to 7 because a(1)*a(2)+1 = 2*3+1 = 7.
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MATHEMATICA
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a = {2, 3}; Do[len = Length@a; AppendTo[a, Complement[a, Union@ Flatten@ Table[ First /@ FactorInteger[a[[i]]*a[[j]] + 1], {i, len}, {j, i}]] [[1]]], {n, 3, 60}]; a
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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