%I #17 Jan 01 2024 02:01:22
%S 1,209,210,842,1176,1358,1370,1608,1707,1845,1850,2594,2880,2882,3123,
%T 3384,4085,4457,4469,4808,5090,5186,5516,5529,5867,5991,6123,6144,
%U 6606,6906,7001,7019,7119,7430,7541,7719,8031,8463,8471,8486,8595,8609,8627
%N Numbers k such that k^81*(k^81+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="/A153442/b153442.txt">Table of n, a(n) for n=1,...,9439</a>
%t k81Q[k_]:=Module[{k81=k^81},PrimeQ[k81(k81+1)+1]]; Select[Range[9000], k81Q] (* _Harvey P. Dale_, Aug 28 2011 *)
%t Select[Range[9000], PrimeQ[(#^81 (#^81 + 1)) + 1] &] (* _Vincenzo Librandi_, Jan 17 2015 *)
%Y Cf. A153438.
%K nonn
%O 1,2
%A _Pierre CAMI_, Dec 26 2008