Divisors from a Prime factorization

[Savage]

Say you gave me the prime factorization 3^5 x 5^5
A system to list every possible divisor would be (of course 1 is included)
start with one (3) and do 3^1, 3^2, 3^3...3^5      do the next single digit 5^1, 5^2, 5^3...5^5
I know how to do the rest...
but what happens when we want to know say (2^a)(3^b)(5^c)? I don't know a system to help me list all possible divisors

I would apply the 2 digit system in doing all 2's and 3's as well as all 3's and 5's and doing all 2's and 5's

but is there a non confusing way to list doing it when all 3 are in it?

[CuCN]

So you want a way to list all the possible ordered triples (x,y,z) such that 0<=x<=a, 0<=y<=b, 0<=z<=c. You can set up a correspondence between those and the integers from 0 to (a+1)(b+1)(c+1)-1 by relating (x,y,z) to x(b+1)(c+1)+y(c+1)+z.
This order lists all the factors sorted first by the power of two, then the power of 3, then the power of 5.

[iDaire]

I love math, but even I have to ask if you both could slow down.

[Savage]

So you want a way to list all the possible ordered triples (x,y,z) such that 0<=x<=a, 0<=y<=b, 0<=z<=c. You can set up a correspondence between those and the integers from 0 to (a+1)(b+1)(c+1)-1 by relating (x,y,z) to x(b+1)(c+1)+y(c+1)+z.
This order lists all the factors sorted first by the power of two, then the power of 3, then the power of 5.

Can you please give an example with a=3, b=3, c=3 and show it your way please?

[CuCN]

0->(0,0,0)->(2^0)(3^0)(5^0)=1
1->(0,0,1)->(2^0)(3^0)(5^1)=5
2->(0,0,2)->(2^0)(3^0)(5^2)=25
3->(0,0,3)->(2^0)(3^0)(5^3)=125
4->(0,1,0)->(2^0)(3^1)(5^0)=3
5->(0,1,1)->(2^0)(3^1)(5^1)=15
6->(0,1,2)->(2^0)(3^1)(5^2)=75
7->(0,1,3)->(2^0)(3^1)(5^3)=375
8->(0,2,0)->(2^0)(3^2)(5^0)=9
...
15->(0,3,3)->(2^0)(3^3)(5^3)=3375
16->(1,0,0)->(2^1)(3^0)(5^0)=2
17->(1,0,1)->(2^1)(3^0)(5^1)=10
...
62->(3,3,2)->(2^3)(3^3)(5^2)=5400
63->(3,3,3)->(2^3)(3^3)(5^3)=27000

[andretimpa]

Also, since N (the set of natural numbers) and N^3 have the same cardinality you could find correspondences in the case where you didn't limited a, b and c (to get all possible products of powers of 2, 3 and 5).

On the other hand, the fundamental theorem of arithmetic implies that you could make a listing by simply ordering 2^a.3^b.5^c in a growing order (this is easier to define, but much more cumbersome to use though):

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 32, 36, ..., 4500, 5400, 6750, 9000, 13500, 27000

corresponding to (in CuCN's notation):

(0,0,0), (1,0,0), (0,1,0), (2,0,0), (0,0,1), (1,1,0), (3,0,0), (0,2,0), (1,0,1), (2,1,0), (0,1,1), (4,0,0), (1,2,0), (2,0,1), (3,1,0), (0,0,2), (1,1,1), (5,0,0), (2,2,0), ..., (4,4,5), (5,5,4), (3,5,5), (5,4,5), (4,5,5), (5,5,5)


To make it for an unlimited a, b and c it is easier to see it with just 2 and them generalize:

(0,0), (0,1), (1,0), (2,0), (1,1), (0,2), (0,3), (1,2), (2,1), (3,0), ... If you draw these points you'll see that we are covering triangles. To go to 3 we need to cover pyramids:

(0,0,0), (0,0,1), (0,1,0), (1,0,0), (2,0,0), (1,1,0), (0,2,0), (0,1,1), (1,0,1), (0,0,2), (0,0,3), (0,1,2), (1,0,2), (2,0,1), (1,1,1), (0,2,1), (0,3,0), (1,2,0), (2,1,0), (3,0,0), (4,0,0), (3,1,0), (2,2,0), (1,3,0), (0,4,0), (0,3,1), (1,2,1), (2,1,1), (3,0,1), (2,0,2), (1,1,2), (0,2,2), (0,1,3), (1,0,3), (0,0,4), (0,0,5), (0,1,4), (1,0,4), (2,0,3), (1,1,3), (0,2,3), (0,3,2), (1,2,2), (2,1,2), (3,0,2), (4,0,1), (3,1,1), (2,2,1), (1,3,1), (0,4,1), (0,5,0), (1,4,0), (2,3,0), (3,2,0), (4,1,0), (5,0,0), (6,0,0), (5,1,0), ....

[Savage]

0->(0,0,0)->(2^0)(3^0)(5^0)=1
1->(0,0,1)->(2^0)(3^0)(5^1)=5
2->(0,0,2)->(2^0)(3^0)(5^2)=25
3->(0,0,3)->(2^0)(3^0)(5^3)=125
4->(0,1,0)->(2^0)(3^1)(5^0)=3
5->(0,1,1)->(2^0)(3^1)(5^1)=15
6->(0,1,2)->(2^0)(3^1)(5^2)=75
7->(0,1,3)->(2^0)(3^1)(5^3)=375
8->(0,2,0)->(2^0)(3^2)(5^0)=9
...
15->(0,3,3)->(2^0)(3^3)(5^3)=3375
16->(1,0,0)->(2^1)(3^0)(5^0)=2
17->(1,0,1)->(2^1)(3^0)(5^1)=10
...
62->(3,3,2)->(2^3)(3^3)(5^2)=5400
63->(3,3,3)->(2^3)(3^3)(5^3)=27000

Thank you very much...that helps quite a bit.
Now, referring to andre's post regarding no restrictions on a,b,c I am still confused on your process in both your system and how you applied it to CuCN's notation.

You start off with (0,0,0), (0,0,1), (0,1,0), (1,0,0), then do (2,0,0)...but then you next say (1,1,0)...why do this step next as opposed to say...after (1,0,0) go to (0,1,1)?
Again, I am just trying to understand the system so I don't get confused if I were to list 100 different sets of (a,b,c) and lost count (that would suck)

[CuCN]

To make it for an unlimited a, b and c it is easier to see it with just 2 and them generalize:

(0,0), (0,1), (1,0), (2,0), (1,1), (0,2), (0,3), (1,2), (2,1), (3,0), ... If you draw these points you'll see that we are covering triangles. To go to 3 we need to cover pyramids:

(0,0,0), (0,0,1), (0,1,0), (1,0,0), (2,0,0), (1,1,0), (0,2,0), (0,1,1), (1,0,1), (0,0,2), (0,0,3), (0,1,2), (1,0,2), (2,0,1), (1,1,1), (0,2,1), (0,3,0), (1,2,0), (2,1,0), (3,0,0), (4,0,0), (3,1,0), (2,2,0), (1,3,0), (0,4,0), (0,3,1), (1,2,1), (2,1,1), (3,0,1), (2,0,2), (1,1,2), (0,2,2), (0,1,3), (1,0,3), (0,0,4), (0,0,5), (0,1,4), (1,0,4), (2,0,3), (1,1,3), (0,2,3), (0,3,2), (1,2,2), (2,1,2), (3,0,2), (4,0,1), (3,1,1), (2,2,1), (1,3,1), (0,4,1), (0,5,0), (1,4,0), (2,3,0), (3,2,0), (4,1,0), (5,0,0), (6,0,0), (5,1,0), ....

This isn't necessary at all, because a, b, and c will always be finite. The ordering I provided works for all finite a, b, and c.

[andretimpa]

0->(0,0,0)->(2^0)(3^0)(5^0)=1
1->(0,0,1)->(2^0)(3^0)(5^1)=5
2->(0,0,2)->(2^0)(3^0)(5^2)=25
3->(0,0,3)->(2^0)(3^0)(5^3)=125
4->(0,1,0)->(2^0)(3^1)(5^0)=3
5->(0,1,1)->(2^0)(3^1)(5^1)=15
6->(0,1,2)->(2^0)(3^1)(5^2)=75
7->(0,1,3)->(2^0)(3^1)(5^3)=375
8->(0,2,0)->(2^0)(3^2)(5^0)=9
...
15->(0,3,3)->(2^0)(3^3)(5^3)=3375
16->(1,0,0)->(2^1)(3^0)(5^0)=2
17->(1,0,1)->(2^1)(3^0)(5^1)=10
...
62->(3,3,2)->(2^3)(3^3)(5^2)=5400
63->(3,3,3)->(2^3)(3^3)(5^3)=27000

Thank you very much...that helps quite a bit.
Now, referring to andre's post regarding no restrictions on a,b,c I am still confused on your process in both your system and how you applied it to CuCN's notation.

You start off with (0,0,0), (0,0,1), (0,1,0), (1,0,0), then do (2,0,0)...but then you next say (1,1,0)...why do this step next as opposed to say...after (1,0,0) go to (0,1,1)?
Again, I am just trying to understand the system so I don't get confused if I were to list 100 different sets of (a,b,c) and lost count (that would suck)

The link with CuCN's notation is that (x,y,z) -> (2^x)(3^y)(5^z)

I added a picture about what I meant with covering triangles and pyramids



Note that you first go through all the (x,y,z) such that x+y+z = 1, then x+y+z = 2, x+y+z = 3, etc. when you do it like this. But if you want to restrict the exponents CuCN gave the simplest answer imo.

[Savage]

What about this method for list notation:

2, 2^2, 2^3,...,2^a
3, 3^2,...,3^b
5,...,5^c

Then for doubles: follow the pattern of exponent growth:

2*3, 2²*3,...2^a3
2*3², 2²3²,..., 2^a3²
.
.
.
2*3^b, 2²3^b,..., 2^a3^b

and then do the same for 2*5 and 3*5

For triples: would something like this be better off for a list?

2*3*5, 2²3*5, 2³3*5,..., 2^a3*5
2*3²5, 2²3²*5, 2³3²*5,...,2^a3²*5
.
.
.
2*3^b5,..........................,2^a3^b5

[CuCN]

What about this method for list notation:

2, 2^2, 2^3,...,2^a
3, 3^2,...,3^b
5,...,5^c

The for doubles: follow the pattern of exponent growth:

2*3, (2^2*3), (2*3^2), (2^2*3^2), (2^3*3), (2*3^3), (2^3*3^3),...,(2^a*3^b)
2*5 (follow the same pattern putting raise 2 once, then raise 5 once, then raise both together)
3*5 (same as above)

For triples:

2*3*5, (2^2*3*5), (2*3^2*5), (2*3*5^2), (2^2*3^2*5), (2^2*3*5^2), (2*3^2*5^2)

question: would it be easier (nicer looking) to go:

2*3*5, 2^2*3*5, 2^3*3*5,...,2^a*3*5
           2*3^2*5, 2*3^3*5,...,2*3^b*5
           now we just raise 5 until 2*3*5^c

how would you suggest I list the rest in a nice listed way?

This seems more complicated than it needs to be. If you rewrite 2^x as (2^x)(3^0)(5^0) all the factors become analogous and then can be placed in correspondence with the triples (x,y,z).

[Savage]

Just edited the post
Can you see if that makes more sense