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Solutions to the congruence 1^n + 2^n + ... + n^n == 79 (mod n).
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%I #14 Sep 06 2018 09:56:51

%S 1,2,6,79,158,474,3318,142674

%N Solutions to the congruence 1^n + 2^n + ... + n^n == 79 (mod n).

%C Also, integers n such that B(n)*n == 79 (mod n), where B(n) is the n-th Bernoulli number.

%C Also, integers n such that Sum_{prime p, (p-1) divides n} n/p == -79 (mod n).

%C Although this sequence is finite, the prime 79 does not belong to A302345.

%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]

%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), this sequence (m=79), A302344 (m=193).

%Y Cf. A302345.

%K nonn,full,fini

%O 1,2

%A _Max Alekseyev_, Apr 05 2018