cp's OEIS Frontend

A087037 Smallest integer x > 0 such that x^x + n is prime, or 0 if no such x exists.

%I A087037 #37 Apr 01 2018 20:27:40
%S A087037 1,1,2,1,444,1,2
%N A087037 Smallest integer x > 0 such that x^x + n is prime, or 0 if no such x exists.
%C A087037 The sequence with the unknown terms a(8), a(11), a(17), a(24), a(41), a(53), a(59), a(65) indicated by ? (each of which exceeds 6000) begins: 1, 1, 2, 1, 444, 1, 2, ?, 2, 1, ?, 1, 2, 3, 2, 1, ?, 1, 2, 3, 4, 1, 6, ?, 2, 3, 2, 1, 30, 1, 6, 3, 2, 3, 6, 1, 2, 5, 2, 1, ?, 1, 2, 3, 58, 1, 6, 7, 2, 3017, 4, 1, ?, 35, 2, 3, 2, 1, ?, 1, 4, 3, 2, 19, ?, 1, 2, 27, 2, 1, 6, 1, 8, 3, 2, ..., where the value a(50)=3017 corresponds to a probable prime. [extended by _Jon E. Schoenfield_, Mar 17 2018, Mar 19 2018]
%C A087037 It is conjectured that such x always exists. - _Dean Hickerson_
%C A087037 From _Farideh Firoozbakht_ and _M. F. Hasler_, Nov 27 2009: (Start)
%C A087037 We can show that for all n=(6k-1)^3, k > 0, there is no such x, which disproves the conjecture:
%C A087037 Since n=(6k-1)^3 is odd, x must be even, else x^x+n is even and composite.
%C A087037 If x == +-1 (mod 3), then x^x + n == (+-1)^2 + (-1)^3 == 0 (mod 3), i.e., divisible by 3 and therefore composite.
%C A087037 Finally, if x == 0 (mod 3), then x^x + n = (x^(x/3) + 6k-1)*(x^(2x/3) - x^(x/3)*(6k-1) + (6k-1)^2) is again composite. (End)
%C A087037 a(8) >= 36869. - _Max Alekseyev_, Sep 16 2013
%H A087037 OpenPFGW Project, <a href="http://www.primeform.net/openpfgw/">Primality Tester</a>
%e A087037 a(7)=2 because 2^2 + 7 = 11 is prime.
%o A087037 (PARI) a(n) = {my(x = 1); while (!isprime(x^x+n), x++); x;} \\ _Michel Marcus_, Mar 20 2018
%Y A087037 Cf. A000312 (n^n), A087038 (x^x+n is prime, x>1).
%Y A087037 Cf. A166853 (x^x-n is prime). - _Farideh Firoozbakht_ and _M. F. Hasler_, Nov 27 2009
%K A087037 nonn,more,hard
%O A087037 1,3
%A A087037 _Hugo Pfoertner_, Jul 31 2003
%E A087037 Name edited by _Altug Alkan_, Apr 01 2018