mersenneforum.org A useful function.
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 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

2×761 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

2×1,949 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

25×3 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 6016 Posts M( only for odd integer number) M() Only for odd numbers.
 2018-02-26, 17:08 #6 JM Montolio A   Feb 2018 9610 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

203008 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 6016 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

173416 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 25×3 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.

 Similar Threads Thread Thread Starter Forum Replies Last Post Batalov And now for something completely different 24 2018-02-27 17:03 rula Homework Help 3 2017-01-18 01:41 Calvin Culus Analysis & Analytic Number Theory 6 2010-12-23 22:18 flouran Miscellaneous Math 23 2009-01-04 20:03 dsouza123 Math 16 2004-03-04 13:57

All times are UTC. The time now is 11:41.

Sat Dec 5 11:41:58 UTC 2020 up 2 days, 7:53, 0 users, load averages: 1.79, 1.84, 1.65