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%I #19 Sep 28 2019 22:56:05
%S 1,18,64,1250,3888,235298,2097152,86093442,250000000,51874849202,
%T 743008370688,46596170244962,396857386627072,58385852050781250,
%U 1152921504606846976,97322383751333736962,273238944967337066496,208254700595822483065682,5242880000000000000000000
%N Least k > 0 such that n^k == 1 (mod (n+1)^(n+1)).
%C Alternative description: For each n, a(n) gives the first k such that n^k-1 has (n+1)^(n+1) as a factor.
%C As n^(m*k)-1 = (n^k)^m-1 is divisible by n^k-1 for all m >= 1, all integer multiples k = m*a(n), m >= 1, also give n^k == 1 (mod (n+1)^(n+1)).
%C Conjecture: a(n) <= 2*(n+1)^n.
%e For n=2: 2^18-1 has the factor 27=3^3.
%e For n=3: 3^64-1 has the factor 256=2^8=4^4.
%p a:= n-> (t-> numtheory[order](n, t^t))(n+1):
%p seq(a(n), n=1..20); # _Alois P. Heinz_, Sep 27 2019
%o (PARI) a(n) = znorder(Mod(n, (n+1)^(n+1))); \\ _Daniel Suteu_, Sep 27 2019
%K nonn
%O 1,2
%A _Matthias Baur_, Sep 27 2019
%E More terms from _Daniel Suteu_, Sep 27 2019
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