p | f(m,n)
 

There is this function f(m,n), where:

p | m & p | n => p^2 | f(m,n)

p !| m & p | n => p !| f(m,n)

p | m & p !| n => p | f(m,n)

p !| m & p !| n => p !| f(m,n)

For integers m, n and p

I have an intuitive feeling that there is a theoretical factoring significance to this function. But I can't formulate anything useful out of it.

Thank you in advance for any comments/inputs.:smile:

[QUOTE=a1call;462855]There is this function f(m,n), where:

p | m & p | n => p^2 | f(m,n)

p !| m & p | n => p !| f(m,n)

p | m & p !| n => p | f(m,n)

p !| m & p !| n => p !| f(m,n)

For integers m, n and p[/QUOTE]

p can be any integer, not just a prime?

[QUOTE=a1call;462855]I have an intuitive feeling that there is a theoretical factoring significance to this function. But I can't formulate anything useful out of it.[/QUOTE]

Me either.

[QUOTE=a1call;462855]There is this function f(m,n), where:

p | m & p | n => p^2 | f(m,n)

p !| m & p | n => p !| f(m,n)

p | m & p !| n => p | f(m,n)

p !| m & p !| n => p !| f(m,n)

For integers m, n and p

I have an intuitive feeling that there is a theoretical factoring significance to this function. But I can't formulate anything useful out of it.

Thank you in advance for any comments/inputs.:smile:[/QUOTE]

The second one, doesn't hold for a straight product mn, being the output of the function ( assuming !| is does not divide).

Thank you for the reply SM.
The function is not a product. And that's the significant of the function that it only is divisible by factor of n if n is also divisible by that factor. And m and n are not interchangeable.

[QUOTE=CRGreathouse;462875]p can be any integer, not just a prime?



Me either.[/QUOTE]

Yes p can be any integer. The function does not have a practical value as is, because it is exponential and yields very large integers. But I think the concept of a function having non interchangeable inputs has theoretical use in factoring size notwithstanding.

[QUOTE=a1call;462855]There is this function f(m,n), where:

p | m & p | n => p^2 | f(m,n)

p !| m & p | n => p !| f(m,n)

p | m & p !| n => p | f(m,n)

p !| m & p !| n => p !| f(m,n)

For integers m, n and p[/QUOTE]

Let m = 2 and n = 4. Then 4 | f(m,n) since 2 | m and 2 | n. But also 4 !| f(m,n) since 4 !| m and 4 | n. So there is no such function.

If you intended for p to be restricted to primes rather than integers as you wrote, then you can just use f(m, n) = m^2.

[QUOTE=a1call;462855]There is this function f(m,n), where:

p | m & p | n => p^2 | f(m,n)

p !| m & p | n => p !| f(m,n)

p | m & p !| n => p | f(m,n)

p !| m & p !| n => p !| f(m,n)

[/QUOTE]

The problems with this, as CRG's example points out, is that some of these combined are true only over certain subsets of integers.

the first is true as long as [TEX]\left|\frac{m}{p}\right| \& \left|\frac{n}{p}\right|[/TEX] are both integer. ( in this case m=p=n is possible, as well as them being any positive powers of each other etc. ordering doesn't really matter for m and n in this one to be true, as long long as p divides into the minimum of them.)

the second and third ones only care if one ( though a specific one) of those is integer (m can't be a positive power of n in the second, n can't be a positive power of m in the third, p could be a positive power of m in the second, or a positive power of n in the third) .

the fourth is there for when neither of that set are integer. ( both m,n can't be positive powers of p, p can be any integer whose absolute value isn't a divisor of m or n).

[QUOTE=CRGreathouse;462897]Let m = 2 and n = 4. Then 4 | f(m,n) since 2 | m and 2 | n. But also 4 !| f(m,n) since 4 !| m and 4 | n. So there is no such function.

If you intended for p to be restricted to primes rather than integers as you wrote, then you can just use f(m, n) = m^2.[/QUOTE]




All my reply was wrong so I deleted it pending further figuring out.

[QUOTE=a1call;462925]All my reply was wrong so I deleted it pending further figuring out.[/QUOTE]

The fourth, also doesn't work for a general difference or a sum of m and n, because 6-1=5 and 6+1=7 which are divisible by 5 and 7 but 6 doesn't divide by either and neither does 1. Also, if m and n are opposite class mod p, then their sum is a multiple of p. If they are the same class, then their difference is a multiple of p.

The truth table I gave in OP is false. Please disregard it.

Here are some actual results:

------------------------
m=7;n=11;
valuation(f(m,n),5)=0
------------------------
m=5;n=11;
valuation(f(m,n),5)=1
------------------------
m=7;n=5;
valuation(f(m,n),5)=0
------------------------
m=5;n=5;
valuation(f(m,n),5)=2
------------------------
m=25;n=5;
valuation(f(m,n),5)=3
------------------------
m=5;n=25;
valuation(f(m,n),5)=2
------------------------
m=7;n=25;
valuation(f(m,n),5)=0

So m is similar to an enable signal in digital electronics, and if enabled then n can contribute to valuation increase. But only as much as the valuation of m.

[QUOTE=a1call;462965]The truth table I gave in OP is false. Please disregard it.

Here are some actual results:

------------------------
[COLOR="Green"]m=7;n=11;
valuation(f(m,n),5)=0[/COLOR]
------------------------
[COLOR="Green"]m=5;n=11;
valuation(f(m,n),5)=1[/COLOR]
------------------------
[COLOR="yellow"]m=7;n=5;
valuation(f(m,n),5)=0[/COLOR]
------------------------
[COLOR="Green"]m=5;n=5;
valuation(f(m,n),5)=2[/COLOR]
------------------------
[COLOR="Green"]m=25;n=5;
valuation(f(m,n),5)=3[/COLOR]
------------------------
[COLOR="Red"]m=5;n=25;
valuation(f(m,n),5)=2[/COLOR]
------------------------
[COLOR="Yellow"]m=7;n=25;
valuation(f(m,n),5)=0[/COLOR]

So m is similar to an enable signal in digital electronics, and if enabled then n can contribute to valuation increase. But only as much as the valuation of m.[/QUOTE]
Is this, some stupid guess the computer science function game ? BTW, Green = valuation as though f(m,n) is their product ; Red=f(m,n) has the same valuation as n ; Yellow= same valuation as m. What's important here ?