A321518 - OEIS

%I #21 Jan 02 2023 12:30:54

%S 3,2,0,24

%N Smallest k > 1 such that n^k + k^n is prime, i.e., a Leyland prime, or 0 if no such k exists.

%C a(4) = 0. Proof: For k == 1 (mod 4), 4^k + k^4 = 4*x^4 + k^4 = (2*x^2 - 2*k*x + k^2)(2*x^2 + 2*k*x + k^2), where x = 4^((k-1)/4). For k == 3 (mod 4), 4^k + k^4 = 64*x^4 + k^4 = (8*x^2 - 4*k*x + k^2)(8*x^2 + 4*k*x + k^2), where x = 4^((k-3)/4) (cf. Israel, 2015).

%C Conjecture: a(6) = 0.

%C From _Jon E. Schoenfield_, Nov 13 2018: (Start)

%C Let t = 6^k + k^6.

%C If k is even, then 2|t.

%C If k is odd but not divisible by 7, then 7|t.

%C If k is divisible by 3, then 3|t.

%C If k == 7 or 63 (mod 70), then 5|t.

%C Thus, a(6) == 35, 49, 91, 119, 161, or 175 (mod 210) if a(6) > 0. (End)

%H Robert Israel, <a href="http://list.seqfan.eu/oldermail/seqfan/2015-December/015820.html">Re: Smallest k > 1 such that n^k+k^n is prime, or 0 if no such k exists</a>, Seqfan (Dec 11 2015).

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Leyland_number">Leyland number</a>

%e For n = 5: 5^24 + 24^5 = 59604644783353249 is prime, and 24 is the smallest k > 1 such that 5^k + k^5 is prime, so a(5) = 24.

%Y Cf. A076980, A094133.

%K nonn,hard,more

%O 2,1

%A _Felix Fröhlich_, Nov 12 2018