A226961 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 3 (mod n).

1, 2, 3, 18, 126, 5418

Offset

1,2

Comments

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

Links

Table of n, a(n) for n=1..6.

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:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT]

Mathematica

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 3 &]

Prog

(PARI) is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-3 \\ Charles R Greathouse IV, Nov 13 2013

Crossrefs

Cf. A031971.

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).

Sequence in context: A288492 A184719 A076016 * A374865 A107095 A102939

Adjacent sequences: A226958 A226959 A226960 * A226962 A226963 A226964

Keyword

nonn,fini,full

Author

José María Grau Ribas, Jun 24 2013

Extensions

1, 2, 3 prepended by Max Alekseyev, Aug 25 2013

Status

approved