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%I #16 Feb 05 2021 14:09:56
%S 67,97,127,163,307,317,337,349,409,521,523,547,643,709,757,811,839,
%T 857,919,967,997,1021,1069,1087,1093,1153,1277,1291,1297,1301,1399,
%U 1429,1459,1483,1619,1627,1637,1697,1709,1721,1741,1789,1877,1933,1949,1999,2017,2029,2083,2131,2179,2239,2269,2311,2383,2389,2437,2503,2539,2557,2591,2659,2671,2707,2731
%N Primes p such that the set { 1+2p, 1+6p, 1+14p, 1+42p, 1+86p, 1+258p, 1+602p, 1+1806p } does not contain any primes.
%C For each term p, the solutions n to the congruence 1^n + 2^n + ... + n^n == p (mod n) form a subset of A014117 U p*A014117. In particular, there are at most 10 solutions for each such p.
%C The coefficients { 2, 6, 14, 42, 86, 258, 602, 1806 } are the even divisors of 1806 = 2 * 3 * 7 * 43.
%H Amiram Eldar, <a href="/A302345/b302345.txt">Table of n, a(n) for n = 1..10000</a>
%H M. A. Alekseyev, J. M. Grau, A. M. Oller-Marcén, <a href="https://doi.org/10.1016/j.dam.2018.05.022">Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n)</a>, Discrete Applied Mathematics 286 (2020), 3-9. Preprint: <a href="http://arxiv.org/abs/1602.02407">arXiv:1602.02407 [math.NT]</a>, 2016.
%t Select[Range[3000], PrimeQ[#] && AllTrue[{2, 6, 14, 42, 86, 258, 602, 1806}*# + 1, ! PrimeQ[#1] &] &] (* _Amiram Eldar_, Aug 09 2020 *)
%o (PARI) { is_A302345(p) = !vecmax( apply( x->ispseudoprime(1+x*p), 2*divisors(3*7*43) ) ); }
%Y Solutions to 1^n+2^n+...+n^n == m (mod n): A005408 (m=0), A014117 (m=1), A226960 (m=2), A226961 (m=3), A226962 (m=4), A226963 (m=5), A226964 (m=6), A226965 (m=7), A226966 (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).
%K nonn
%O 1,1
%A _Max Alekseyev_, Apr 05 2018