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A280041 Solutions to the congruence 1^n+2^n+...+n^n == 19 (mod n). 11

%I #14 Sep 06 2018 09:56:06

%S 1,2,6,19,38,114,798,34314

%N Solutions to the congruence 1^n+2^n+...+n^n == 19 (mod n).

%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 f[n_] := Mod[Sum[PowerMod[k, n, n], {k, 1, n}] - 19, n];

%t For[n = 1, n < 40000, n++, If[f[n] == 0, Print[n]]] (* _Jean-François Alcover_, Sep 06 2018 *)

%K nonn,fini,full

%O 1,2

%A _N. J. A. Sloane_, Dec 29 2016

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Last modified March 28 20:05 EDT 2024. Contains 371254 sequences. (Running on oeis4.)