login
Least k with 1^(k*m) + 2^(k*m) + ... + (k*m)^(k*m) == k (mod k*m) for m in A230311.
10

%I #11 Dec 02 2013 10:33:36

%S 1,1,1,1,1,5,5,39607528021345872635

%N Least k with 1^(k*m) + 2^(k*m) + ... + (k*m)^(k*m) == k (mod k*m) for m in A230311.

%C Least k with A031971(k*m) == k (mod k*m) for m in A230311.

%C See A031971 and A230311 for more comments and crossrefs.

%H J. M. Grau, A. M. Oller-Marcen, and J. Sondow, <a href="http://arxiv.org/abs/1309.7941">On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m</a>, arXiv:1309.7941 [math.NT].

%F a(2) = A229303(1), a(3) = A229302(1), a(4) = A229301(1), a(5) = A229300, a(6) = A229312(1).

%e 1^m + 2^m + ... + m^m == 1 (mod m) for the first 5 terms m = 1, 2, 6, 42, 1806 of A230311, so a(n) = 1 for n <= 5.

%Y Cf. A031971, A229300, A229301, A229302, A229303, A230311.

%K nonn,more,hard

%O 1,6

%A _Jonathan Sondow_, Nov 30 2013