A153441 Numbers k such that k^27*(k^27+1)+1 is prime.

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

Offset: 1

Author

Pierre CAMI, Dec 26 2008

Keywords

• nonn

Comments

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 ??

Links

• Pierre CAMI, Table of n, a(n) for n=1,...,37719

Crossrefs

Cf. A153438.

Programs

PARI

isok(k)  = isprime(k^27*(k^27+1)+1); \\ Michel Marcus, Sep 20 2019