A056826 Primes p such that (p^p + 1)/(p + 1) is a prime.

3, 5, 17, 157

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(1) = 1, b(2) = 3; b(n+2) = (b(n+1)^2 + b(n)^2)/2. Conjecture: if b(n) = p is prime then (p^p+1)/(p+1) is prime. Note that b(2) = 3, b(3) = 5, b(4) = 17, b(5) = 157 and b(10) is 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

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

Table of n, a(n) for n=1..4.

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 *)

Prog

(PARI) forprime(p=3, 1000, if(isprime((p^p+1)/(p+1)), print1(p", "))) \\ Seiichi Manyama, Mar 24 2018

Crossrefs

Cf. A088790 ((n^n-1)/(n-1) is prime), A088817 (cyclotomic(2n, n) is prime).

Sequence in context: A107312 A083213 A171271 * A370879 A278138 A273870

Adjacent sequences: A056823 A056824 A056825 * A056827 A056828 A056829

Keyword

hard,nonn,more

Author

Robert G. Wilson v, Aug 29 2000

Extensions

Definition corrected by Alexander Adamchuk, Nov 12 2006

Status

approved