OFFSET
1,2
COMMENTS
Also, integers n such that B(n)*n == 7 (mod n), where B(n) is the n-th Bernoulli number, or SUM[prime p, (p-1) divides n] n/p == -7 (mod n). It is easy to see that for n>1, every prime divisor p of n, except p=7, must appear in first power, while p=7 may appear in first or second power. Moreover, the multiset P of prime divisors of all such n satisfies the property: if p is in P, then p-1 is the product of distinct elements of P. This multiset is P = {2, 3, 7, 7, 43}, implying that the sequence is finite and complete. - Max Alekseyev, Aug 25 2013
LINKS
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, #}], #] == 7&]
PROG
(PARI) is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-7 \\ Charles R Greathouse IV, Nov 13 2013
CROSSREFS
KEYWORD
nonn,full,fini
AUTHOR
José María Grau Ribas, Jun 24 2013
EXTENSIONS
Corrected and keywords full,fini added by Max Alekseyev, Aug 25 2013
STATUS
approved