%I #30 Sep 06 2018 10:03:57
%S 1,2,3,18,126,5418
%N Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 3 (mod n).
%C Equivalently, numbers n such that B(n)*n == 3 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -3 (mod n). - _Max Alekseyev_, Aug 25 2013
%H M. A. Alekseyev, J. M. Grau, A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:<a href="http://doi.org/10.1016/j.dam.2018.05.022">10.1016/j.dam.2018.05.022</a> arXiv:<a href="http://arxiv.org/abs/1602.02407">1602.02407</a> [math.NT]
%t Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 3 &]
%o (PARI) is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-3 \\ _Charles R Greathouse IV_, Nov 13 2013
%Y Cf. A031971.
%Y Solutions to 1^n+2^n+...+n^n == m (mod n): A005408 (m=0), A014117 (m=1), A226960 (m=2), this sequence (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,fini,full
%O 1,2
%A _José María Grau Ribas_, Jun 24 2013
%E 1, 2, 3 prepended by _Max Alekseyev_, Aug 25 2013