|
%I #19 Sep 06 2018 10:11:06
%S 1,16,48,336,14448
%N Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 8 (mod n).
%C Also, numbers n such that B(n)*n == 8 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -8 (mod n). There are no other terms below 10^15. - _Max Alekseyev_, Aug 26 2013
%t Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 8 &]
%o (PARI) is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-8 \\ _Charles R Greathouse IV_, Nov 13 2013
%Y Cf. A031971.
%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), this sequence (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).
%K nonn,more
%O 1,2
%A _José María Grau Ribas_, Jun 24 2013
%E a(1)=1 prepended by _Max Alekseyev_, Aug 26 2013
|
