OFFSET
1,2
COMMENTS
Alternative description: For each n, a(n) gives the first k such that n^k-1 has (n+1)^(n+1) as a factor.
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)).
Conjecture: a(n) <= 2*(n+1)^n.
EXAMPLE
For n=2: 2^18-1 has the factor 27=3^3.
For n=3: 3^64-1 has the factor 256=2^8=4^4.
MAPLE
a:= n-> (t-> numtheory[order](n, t^t))(n+1):
seq(a(n), n=1..20); # Alois P. Heinz, Sep 27 2019
PROG
(PARI) a(n) = znorder(Mod(n, (n+1)^(n+1))); \\ Daniel Suteu, Sep 27 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Matthias Baur, Sep 27 2019
EXTENSIONS
More terms from Daniel Suteu, Sep 27 2019
STATUS
approved