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Least number k such that (n^k-k^n)/(k-n) is prime, or 0 if no such number exists.
0

%I #6 Jul 29 2014 21:04:17

%S 0,6,4,3,3,2,3,0,5,7,3,13,11,0,17,0,15,0,7,0,0,15,5,0,79,0,0,0,15,0,0,

%T 0,65,0,47,0,39,0,37,0,9,0,0,45,44,0,11,0,103,0,71,0,11,0,119,0,5,0,0,

%U 0,0,0,0,0,33,0,75,0,77,0,51,143,0,0,67,0,69,0,25,0,131,0,0,0,57,0,8887,0,221,0,291,0,0,0,0,0,101,0,0

%N Least number k such that (n^k-k^n)/(k-n) is prime, or 0 if no such number exists.

%C a(1) = 0 is the only confirmed 0 in this sequence.

%C a(n) = 0 for n > 1 is confirmed for k < 10000.

%C If a(n) = m, then a(m) <= n for m > 0 and n > 0.

%e (2^1-1^2)/(1-2) = -1 is not prime.

%e (2^3-3^2)/(3-2) = -1 is not prime.

%e (2^4-4^2)/(4-2) = 0 is not prime.

%e (2^5-5^2)/(5-2) = 7/3 is not prime.

%e (2^6-6^2)/(6-2) = 7 is prime. Thus a(2) = 6.

%o (PARI)

%o a(n)=for(k=1,10^4,if(k!=n,s=(n^k-k^n)/(k-n);if(floor(s)==s,if(ispseudoprime(s),return(k)))))

%o n=1;while(n<100,print1(a(n),", ");n++)

%Y Cf. A242922.

%K nonn

%O 1,2

%A _Derek Orr_, Jul 28 2014