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 2018-02-26, 11:26 #1 JM Montolio A   Feb 2018 25×3 Posts A useful function. Hi, 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) ¿ useful function ? JM M Spain
2018-02-26, 14:22   #2
Nick

Dec 2012
The Netherlands

150110 Posts

Quote:
 Originally Posted by JM Montolio A for (m,n ) coprimes, M(n*m)= (M(n)*M(m))/(mcd(M(n),M(m))
What is the mcd function that you are using?

2018-02-26, 14:53   #3
Dr Sardonicus

Feb 2017
Nowhere

2·3·5·127 Posts

Quote:
 Originally Posted by JM Montolio A Hi, 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)
I object, on the following grounds:

(1) The requirement p | (2^M(p)-1) is not a definition. Assuming M(p) takes positive integer values, M(2) is problematic. The only possible integer value of M(2) is zero. For odd p, M(p) merely has to be divisible by the multiplicative order of 2 (mod p).

(2) The expression (mcd(M(n),M(m)) has an extra left parenthesis.

(3) The function mcd() is undefined. Do you perhaps mean gcd() (greatest common divisor)?

2018-02-26, 17:00   #4
JM Montolio A

Feb 2018

1408 Posts
yes , gcd . En Español "El Máximo".

Quote:
 Originally Posted by Nick What is the mcd function that you are using?

Yes, gcd(). En español "El máximo".
JM M

Last fiddled with by JM Montolio A on 2018-02-26 at 17:01

 2018-02-26, 17:05 #5 JM Montolio A   Feb 2018 25×3 Posts M( only for odd integer number) M() Only for odd numbers.
 2018-02-26, 17:08 #6 JM Montolio A   Feb 2018 25·3 Posts and more: properties d | n, then M(d) | M(n).
2018-02-26, 17:15   #7
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by JM Montolio A Yes, gcd(). En español "El máximo". JM M
Think the reason m isn't used in english is it could be maximal or minimal. Also without a definition at the primes I'm not sure the definition is complete M(p)=M(p)*1 is not all that helpful.

Last fiddled with by science_man_88 on 2018-02-26 at 17:20

 2018-02-26, 17:36 #8 JM Montolio A   Feb 2018 11000002 Posts well, for p prime, M(p) must be the correct value. - N*D = 2^M(n) -1 - for p prime , M(p)|(p-1) JM M Last fiddled with by JM Montolio A on 2018-02-26 at 17:37
2018-02-26, 17:55   #9
CRGreathouse

Aug 2006

2·2,969 Posts

Quote:
 Originally Posted by JM Montolio A ¿ useful function ?
How do you compute it?

 2018-02-26, 18:03 #10 JM Montolio A   Feb 2018 9610 Posts well, is only one axiomatic definition.
 2018-02-26, 18:06 #11 JM Montolio A   Feb 2018 25·3 Posts other propertie M( 2^e - 1 ) = e. other property, M( 2^e - 1 ) = e.

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