OFFSET
1,1
COMMENTS
Note that (k^k+1)/(k+1) is prime only if k is prime, in which case it equals cyclotomic(2k,k), the 2k-th cyclotomic polynomial evaluated at x=k. This sequence is a subsequence of A088817. Are there only a finite number of these primes? - T. D. Noe, Oct 20 2003
(3^2 + 5^2)/2 = 17, (5^2 + 17^2)/2 = 157. - Thomas Ordowski, Jul 28 2013
Let b(0) = 1, b(1) = 3; b(n+1) = (b(n)^2 + b(n-1)^2)/2. Conjecture: if b(n) = p is prime, then (p^p+1)/(p+1) is prime. Note that b(1) = 3, b(2) = 5, b(3) = 17, b(4) = 157 and b(9) is also prime. - Thomas Ordowski, Jul 29 2013
Next term > 3000. - Seiichi Manyama, Mar 24 2018
No more terms through 6000. - Jon E. Schoenfield, Mar 25 2018
No more terms through 20000. - Michael S. Branicky, Jul 30 2024
From Harsh R. Aggarwal, Jan 23 2026: (Start)
Other than the first term 3, all prime terms must be p==1 (mod 4) as all other prime terms have an Aurifeuillean factorization.
No more terms through 230000. (End)
REFERENCES
J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 157, p. 51, Ellipses, Paris 2008.
R. K. Guy, Unsolved Problems in Theory of Numbers, 1994 A3.
LINKS
Eric Weisstein's World of Mathematics, Cyclotomic Polynomial.
MATHEMATICA
Do[ If[ PrimeQ[ (Prime[ n ]^Prime[ n ] + 1)/(Prime[ n ] + 1) ], Print[ Prime[ n ] ] ], {n, 1, 213} ]
Do[p=Prime[n]; If[PrimeQ[(p^p+1)/(p+1)], Print[p]], {n, 100}] (* T. D. Noe, Oct 20 2003 *)
PROG
(PARI) forprime(p=3, 1000, if(isprime((p^p+1)/(p+1)), print1(p", "))) \\ Seiichi Manyama, Mar 24 2018
CROSSREFS
KEYWORD
hard,nonn,more
AUTHOR
Robert G. Wilson v, Aug 29 2000
EXTENSIONS
Definition corrected by Alexander Adamchuk, Nov 12 2006
STATUS
approved