

A327836


Least k > 0 such that n^k == 1 (mod (n+1)^(n+1)).


0



1, 18, 64, 1250, 3888, 235298, 2097152, 86093442, 250000000, 51874849202, 743008370688, 46596170244962, 396857386627072, 58385852050781250, 1152921504606846976, 97322383751333736962, 273238944967337066496, 208254700595822483065682, 5242880000000000000000000
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OFFSET

1,2


COMMENTS

Alternative description: For each n, a(n) gives the first k such that n^k1 has (n+1)^(n+1) as a factor.
As n^(m*k)1 = (n^k)^m1 is divisible by n^k1 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.


LINKS

Table of n, a(n) for n=1..19.


EXAMPLE

For n=2: 2^181 has the factor 27=3^3.
For n=3: 3^641 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

Sequence in context: A232385 A275155 A259634 * A165029 A264652 A010006
Adjacent sequences: A327833 A327834 A327835 * A327837 A327838 A327839


KEYWORD

nonn


AUTHOR

Matthias Baur, Sep 27 2019


EXTENSIONS

More terms from Daniel Suteu, Sep 27 2019


STATUS

approved



