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A239279
Smallest k such that n^k - k^n is prime, or 0 if no such number exists.
2
5, 1, 1, 14, 1, 20, 1, 10, 273, 14, 1, 38, 1, 68, 0
OFFSET
2,1
COMMENTS
It is believed that for all n > 4 and not in A097764, a(n) > 0.
a(n+1) = 1 if and only if n is prime.
If a(n) > 0 then a(n) and n are coprime.
If n is in the sequence A097764, then a(n) = 0 or 1 since n^k-k^n is factorable.
33^2570 - 2570^33 is a probable prime, so a(33) is probably 2570. - Jon E. Schoenfield, Mar 20 2014
Unknown a(n) values checked for k <= 10000 using PFGW. a(97) = 6006 found by Donovan Johnson in 2005. The Lifchitz link shows some large candidates for larger n but a smaller k exists in many of those cases. - Jens Kruse Andersen, Aug 13 2014
Unknown a(n) values checked for k <= 15000 using PFGW.
EXAMPLE
2^1-1^2 = 1 is not prime. 2^2-2^2 = 0 is not prime. 2^3-3^2 = -1 is not prime. 2^4-4^2 = 0 is not prime. 2^5-5^2 = 7 is prime. So a(2) = 5.
PROG
(Python)
import sympy
from sympy import isprime
from sympy import gcd
def Min(x):
..k = 1
..while k < 5000:
....if gcd(k, x) == 1:
......if isprime(x**k-k**x):
........return k
......else:
........k += 1
....else:
......k += 1
x = 1
while x < 100:
..print(Min(x))
..x += 1
(PARI)
a(n)=k=1; if(n>4, forprime(p=1, 100, if(ispower(n)&&ispower(n)%p==0&&n%p==0, return(0)); if(n%p==n, break))); k=1; while(!ispseudoprime(n^k-k^n), k++); return(k)
vector(15, n, a(n+1))
KEYWORD
nonn,hard,more
AUTHOR
Derek Orr, Mar 14 2014
STATUS
approved