

A226965


Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 7 (mod n).


10




OFFSET

1,2


COMMENTS

Also, integers n such that B(n)*n == 7 (mod n), where B(n) is the nth Bernoulli number, or SUM[prime p, (p1) 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 p1 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

Table of n, a(n) for n=1..7.
M. A. Alekseyev, J. M. Grau, A. M. OllerMarcen. 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

Cf. A031971.
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), this sequence (m=7), A226966 (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).
Sequence in context: A281167 A210660 A344343 * A128918 A078113 A286054
Adjacent sequences: A226962 A226963 A226964 * A226966 A226967 A226968


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



