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%I #13 Sep 20 2019 18:07:47
%S 1,21,50,77,153,191,194,311,405,440,462,557,638,659,690,746,852,887,
%T 944,945,1140,1326,1344,1452,1463,1607,1632,1652,1659,1683,1710,1788,
%U 1812,1851,1925,1943,1992,2157,2294,2309,2352,2402,2621,2687,2700,2733,2756
%N Numbers k such that k^27*(k^27+1)+1 is prime.
%C 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 ??
%H Pierre CAMI, <a href="/A153441/b153441.txt">Table of n, a(n) for n=1,...,37719</a>
%o (PARI) isok(k) = isprime(k^27*(k^27+1)+1); \\ _Michel Marcus_, Sep 20 2019
%Y Cf. A153438.
%K nonn
%O 1,2
%A _Pierre CAMI_, Dec 26 2008
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