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Old 2018-03-05, 16:46   #15
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by JM Montolio A View Post
I tell us a secret. Dont waste time looking for. They are all square free.
The Mersenne numbers, the Wagstaff numbers, and the Fermat numbers.
That follows of the properties of the M() function.
Unfortunately, you have yet to completely define M(p) for p prime.

In the Initial post to the thread A useful function. you wrote:

Quote:
Define M(n) as:


for (p^e), M( p^e ) = M(p)*(p ^ (e-1) )


for (m,n ) coprimes, M(n*m)= (M(n)*M(m))/(mcd(M(n),M(m))


for p prime, p | (2^M(p)-1)
In Post #8 you added
Quote:
for p prime, M(p)|(p-1)
and in Post #11 you added

Quote:
other property, M( 2^e - 1 ) = e.
As far as I can tell, for p prime, M(p) is only well defined generally when either (a) p is a Mersenne prime, or (b) 2 is a primitive root (mod p). In Post #15 you also individually defined the specific values

Quote:
M( 1093 ) = 364

M( 3511 ) = 1755
I also note that, due to your condition

Quote:
for (p^e), M( p^e ) = M(p)*(p ^ (e-1) )
quoted above, if p is a Wieferich prime, your M(p^e) for e > 1 is no longer the multiplicative order of 2 (mod p^e). For example, 346 is the multiplicative order of 2 (mod 1093) and also (mod 193^2).

So your M(1093^2) = 1093*346 is the multiplicative order of 2 (mod 1093^3).

Last fiddled with by wblipp on 2018-03-06 at 13:53 Reason: fixed link for Post #11
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