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A057856
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Least k such that (n+1)^k + n^k is a prime.
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4
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1, 1, 1, 2, 1, 1, 2, 1, 1, 32, 1, 2, 4, 1, 1, 4, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 4
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OFFSET
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1,4
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COMMENTS
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Note: k must be of the form 2^m (see A058064 for the m values).
Conjecture: For every pair of relatively prime numbers (x, y) there exist at least one number n=2^m and one prime number p such that p = x^n + y^n. This sequence shows one case of this conjecture where y = x + 1. - Tomas Xordan, Jun 02 2007
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LINKS
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EXAMPLE
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a(101)=16 because 101^16 + 102^16 = 254536435001431070450581794495937.
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MATHEMATICA
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Do[ k = 0; While[ !PrimeQ[ (n + 1)^(2^k) + n^(2^k) ], k++ ]; Print[ 2^k ], {n, 1, 60} ].
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PROG
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(PARI) a(n) = my(k=1); while (!isprime(p=(n+1)^k + n^k), k++); k; \\ Michel Marcus, Sep 16 2018
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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STATUS
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approved
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