20180226, 11:26  #1 
Feb 2018
60_{16} Posts 
A useful function.
Hi,
Define M(n) as: for (p^e), M( p^e ) = M(p)*(p ^ (e1) ) 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 
20180226, 14:22  #2 
Dec 2012
The Netherlands
1,621 Posts 

20180226, 14:53  #3  
Feb 2017
Nowhere
2^{4}·271 Posts 
Quote:
(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)? 

20180226, 17:00  #4 
Feb 2018
2^{5}·3 Posts 
yes , gcd . En Español "El Máximo".
Yes, gcd(). En español "El máximo". JM M Last fiddled with by JM Montolio A on 20180226 at 17:01 
20180226, 17:05  #5 
Feb 2018
2^{5}×3 Posts 
M( only for odd integer number)
M() Only for odd numbers.

20180226, 17:08  #6 
Feb 2018
2^{5}×3 Posts 
and more: properties
d  n, then M(d)  M(n).

20180226, 17:15  #7 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
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 20180226 at 17:20 
20180226, 17:36  #8 
Feb 2018
140_{8} Posts 
well, for p prime, M(p) must be the correct value.
 N*D = 2^M(n) 1  for p prime , M(p)(p1) JM M Last fiddled with by JM Montolio A on 20180226 at 17:37 
20180226, 17:55  #9 
Aug 2006
2·29·103 Posts 

20180226, 18:03  #10 
Feb 2018
140_{8} Posts 
well, is only one axiomatic definition.

20180226, 18:06  #11 
Feb 2018
2^{5}×3 Posts 
other propertie M( 2^e  1 ) = e.
other property, M( 2^e  1 ) = e.

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