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A321518
Smallest k > 1 such that n^k + k^n is prime, i.e., a Leyland prime, or 0 if no such k exists.
0
OFFSET
2,1
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)
EXAMPLE
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.
CROSSREFS
Sequence in context: A244134 A105629 A085075 * A267883 A333437 A058257
KEYWORD
nonn,hard,more
AUTHOR
Felix Fröhlich, Nov 12 2018
STATUS
approved