A321518 Smallest k > 1 such that n^k + k^n is prime, i.e., a Leyland prime, or 0 if no such k exists.

3, 2, 0, 24

Offset: 2

Author

Felix Fröhlich, Nov 12 2018

Keywords

• nonn
• hard
• more

Comments

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).

Conjecture: a(6) = 0.

From Jon E. Schoenfield, Nov 13 2018: (Start)

Let t = 6^k + k^6.

If k is even, then 2|t.

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

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

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

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

Examples

			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.
		

Links

• Robert Israel, Re: Smallest k > 1 such that n^k+k^n is prime, or 0 if no such k exists, Seqfan (Dec 11 2015).
• Wikipedia, Leyland number

Crossrefs

Cf. A076980, A094133.