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%I #35 May 08 2021 22:57:58
%S 3,5,17,157
%N Primes p such that (p^p + 1)/(p + 1) is a prime.
%C 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
%C (3^2 + 5^2)/2 = 17, (5^2 + 17^2)/2 = 157. - _Thomas Ordowski_, Jul 28 2013
%C 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
%C Next term > 3000. - _Seiichi Manyama_, Mar 24 2018
%C No more terms through 6000. - _Jon E. Schoenfield_, Mar 25 2018
%D J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 157, p. 51, Ellipses, Paris 2008.
%D R. K. Guy, Unsolved Problems in Theory of Numbers, 1994 A3.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CyclotomicPolynomial.html">Cyclotomic Polynomial</a>
%t Do[ If[ PrimeQ[ (Prime[ n ]^Prime[ n ] + 1)/(Prime[ n ] + 1) ], Print[ Prime[ n ] ] ], {n, 1, 213} ]
%t Do[p=Prime[n]; If[PrimeQ[(p^p+1)/(p+1)], Print[p]], {n, 100}] (* _T. D. Noe_ *)
%o (PARI) forprime(p=3, 1000, if(isprime((p^p+1)/(p+1)), print1(p", "))) \\ _Seiichi Manyama_, Mar 24 2018
%Y Cf. A088790 ((n^n-1)/(n-1) is prime), A088817 (cyclotomic(2n, n) is prime).
%K hard,nonn,more
%O 1,1
%A _Robert G. Wilson v_, Aug 29 2000
%E Definition corrected by _Alexander Adamchuk_, Nov 12 2006
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