# No subject

Wed Sep 3 00:41:26 CEST 2003

```Many thanks for this very instructive answer. Furthermore, I built an
array defined as follows :  T(p,k) is the least integer s such that p
divides k^s-1,  p is an odd prime and k is ranging from 1 to p-1 :

[1, 2]
[1, 4, 4, 2]
[1, 3, 6, 3, 6, 2]
[1, 10, 5, 5, 5, 10, 10, 10, 5, 2]
[1, 12, 3, 6, 4, 12, 12, 4, 3, 6, 12, 2]
[1, 8, 16, 4, 16, 16, 16, 8, 8, 16, 16, 16, 4, 16, 8, 2]
[1, 18, 18, 9, 9, 9, 3, 6, 9, 18, 3, 6, 18, 18, 18, 9, 9, 2]
[1, 11, 11, 11, 22, 11, 22, 11, 11, 22, 22, 11, 11, 22, 22, 11, 22, 11,
22, 22, 22, 2]
[1, 28, 28, 14, 14, 14, 7, 28, 14, 28, 28, 4, 14, 28, 28, 7, 4, 28, 28,
7, 28, 14, 7, 7, 7, 28, 28, 2]
[1, 5, 30, 5, 3, 6, 15, 5, 15, 15, 30, 30, 30, 15, 10, 5, 30, 15, 15,
15, 30, 30, 10, 30, 3, 6, 10, 15, 10, 2]

I just noticed 2 simple rules :

(i) T(p,1)=1 and  T(p,p-1)=2

(ii) T(p,k) 1<k<p-1  take values among divisors >2  of (p-1) (does each
divisor appear at least once?)

Considering the case : p and (p-1)/2  primes
(http://www.research.att.com/projects/OEIS?Anum=A005385) :

T(p,2)=(p-1)/2  for p=7,23,47,167,263,359,383,479,503,... not in EIS
T(p,2)=(p-1)     for p=5,11,59,83,107,179,227,347,467,563,587,... not
in EIS

I tried also to find the sequence of primes p such that there is at
least one k<=p-1,  with p^2 dividing k^T(p,k)-1 :

11,29,37,43,59,71,79,97,103,....not in EIS

Benoit Cloitre

>     The situation has been understood for a long time.  Let's ask
> for the exact power of some prime  p  that divides  a^K - 1.  Then
> the assertion is that if  k  is the smallest positive number for which
> p  itself divides  a^k - 1,  and  a^k - 1  is exactly divisible by
> p^i,  then  a^K - 1  will be divisible by  p  precisely when  K  is
> a multiple of  k,  and then the exact power of  p  that divides it
> will be  p^(i+j),  where  p^j  is the exact power of  p  that divides
> K/k.
>
>     In other words, the first time you get a multiple of  p  you
> can "accidentally" get a higher power than the first,  but from
> then on you can only get more p's by putting them into the exponent.
>
>    Examples:  the first time  3^K - 1  is divisible by  11  is at
> 3^5 - 1, which is divisible precisely by  11^2.  So  3^K - 1  will
> be divisible by  11^(2+j)  only when  KI  is divisible by  5 times
> 11^j.
>
>     Similarly,  2^1092 - 1  happens to be divisible by just 1093^2,
> so  2^(1092.1093^j) - 1  will be divisible by just  1093^(2+j).
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