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A242884
Least number k such that (k^k-n^n)/(k-n) is prime or 0 if no such number exists.
0
2, 1, 1, 3, 3, 7, 3, 0, 0, 11, 5, 0, 4, 0, 0, 7, 16, 0, 1, 0, 281, 0, 19, 0, 0, 0, 7, 0, 35, 0, 1, 0, 113, 0, 29, 91, 19, 0, 19, 0, 23, 0, 0, 0, 0, 37, 65, 0, 0, 0, 0, 0, 153, 0, 199, 0, 0, 115, 0, 0, 0, 0, 319, 0, 47, 0, 0, 0, 13, 0, 47, 0, 0, 0, 539, 0, 0, 0, 0, 0, 13, 0, 147
OFFSET
1,1
COMMENTS
a(n) = 0 is confirmed for k <= 5000. (Which means that the zeros are at present only conjectural. - N. J. A. Sloane, May 26 2014)
If a(i) = j, then a(j) <= i for all i and j.
EXAMPLE
(2^2-1^1)/(2-1) = 3 is prime. Thus a(1) = 2.
PROG
(PARI) a(n)=for(k=1, 5000, if(k!=n, s=(k^k-n^n)/(k-n); if(floor(s)==s, if(ispseudoprime(s), return(k)))))
n=1; while(n<100, print(a(n)); n+=1)
CROSSREFS
Sequence in context: A276777 A219876 A216655 * A197219 A300508 A120013
KEYWORD
nonn,hard,more
AUTHOR
Derek Orr, May 25 2014
EXTENSIONS
We don't normally allow conjectural terms, except in special circumstances. This is one of those exceptions, for if we included only terms that are known for certain, not much of this sequence would remain. - N. J. A. Sloane, May 31 2014
STATUS
approved