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A153441
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Numbers k such that k^27*(k^27+1)+1 is prime.
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6
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1, 21, 50, 77, 153, 191, 194, 311, 405, 440, 462, 557, 638, 659, 690, 746, 852, 887, 944, 945, 1140, 1326, 1344, 1452, 1463, 1607, 1632, 1652, 1659, 1683, 1710, 1788, 1812, 1851, 1925, 1943, 1992, 2157, 2294, 2309, 2352, 2402, 2621, 2687, 2700, 2733, 2756
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OFFSET
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1,2
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COMMENTS
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It seems numbers of the form k^n*(k^n+1)+1 with n > 0, k > 1 may be primes only if n has the form 3^j. When n is even, k^(4*n)+k^(2*n)+1=(k^(2*n)+1)^2-(k^n)^2=(k^(2*n)+k^n+1)*(k^(2*n)-k^n+1) so composite. But why if n odd > 3 and not a power of 3, k^n*(k^n+1)+1 is always composite ??
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LINKS
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PROG
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(PARI) isok(k) = isprime(k^27*(k^27+1)+1); \\ Michel Marcus, Sep 20 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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