A useful function.

[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)


¿ useful function ?


JM M
Spain

[Nick]

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?

[Dr Sardonicus]

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)?

[JM Montolio A]

Originally Posted by Nick

What is the mcd function that you are using?

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

[JM Montolio A]

M() Only for odd numbers.

[JM Montolio A]

d | n, then M(d) | M(n).

[science_man_88]

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.

[JM Montolio A]

- N*D = 2^M(n) -1


- for p prime , M(p)|(p-1)

JM M

[CRGreathouse]

Originally Posted by JM Montolio A

¿ useful function ?

How do you compute it?

[JM Montolio A]

well, is only one axiomatic definition.

[JM Montolio A]

other property, M( 2^e - 1 ) = e.