Divisors from a Prime factorization

[CuCN]

Again, you don't need to separate the factors into three different groups. You can treat them all as (2^x)(3^y)(5^z) where x, y, and z might be 0.

Just wondering, how familiar are you with competition-level math problem solving?

[Savage]

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

[I would then need to continue to do this with 2,5 growing and 3,5 growing, then finally all 3 growing and that is where I need help]

===> here the goal is actually list every possible factor (with ... of course, can't list an infinite group)
writing it as "(2^x)(3^y)(5^z) where x, y, and z might be 0" only gives a 'general' sense of every factor

Double's can be listed in this manner (previous post) quite simply

[Lost in Nowhere]

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

[I would then need to continue to do this with 2,5 growing and 3,5 growing, then finally all 3 growing and that is where I need help]

===> here the goal is actually list every possible factor (with ... of course, can't list an infinite group)
writing it as "(2^x)(3^y)(5^z) where x, y, and z might be 0" only gives a 'general' sense of every factor

Double's can be listed in this manner (previous post) quite simply

If you want to extend it so that it has all 3 factors, just repeat it like this:

2⁰3⁰5⁰, 2¹3⁰5⁰, 2²3⁰5⁰,..., 2^a3⁰5⁰
2⁰3¹5⁰, 2¹3¹5⁰, 2²3¹5⁰,...,2^a3¹5⁰
.
.
.
2⁰3^b5⁰,..........................,2^a3^b5⁰

2⁰3⁰5¹, 2¹3⁰5¹, 2²3⁰5¹,..., 2^a3⁰5²
2⁰3¹5², 2¹3¹5², 2²3¹5²,...,2^a3¹5²
.
.
.
2⁰3^b5²,..........................,2^a3^b5²

.
.
.

2⁰3⁰5^c, 2¹3⁰5^c, 2²3⁰5^c,..., 2^a3⁰5^c
2⁰3¹5^c, 2¹3¹5^c, 2²3¹5^c,...,2^a3¹5^c
.
.
.
2⁰3^b5^c,..........................,2^a3^b5^c

For finite a, b, c, etc., this works in exactly the same way as CuCN's notation, except that instead of one long list, this is a bunch of shorter lists.
(hopefully I've fixed all of the errors that I had in it at first...)

[Savage]

Where are the single 5 digits factors in your list

Also, I was just asking for a list extending when 2,3,5 are all raised to some power n>=1 (you have included 0, but doing doubles and singles should be kept separate for readability purposes)

[Lost in Nowhere]

The factors of 5 are in the top-left of each block.
I see no reason to overcomplicate the format for the purpose of readability, when it would only give seemingly marginal improvements. At least to me, this format seems like it's the simplest, especially when you increase the number of factors.
But that's just me. Since you asked:

Spoiler for Hidden:

1
~~~
2¹, 2², ... 2^a
~~~
3¹, 3², ... 3^b
~~~
5¹, 5², ... 5^c
~~~
2¹3¹, 2¹3¹, ... 2^a3¹
2¹3², 2¹3², ... 2^a3²
.
.
.
2¹3^b, 2¹3^b, ... 2^a3^b
~~~
2¹3¹, 2¹3¹, ... 2^a3¹
2¹3², 2¹3², ... 2^a3²
.
.
.
2¹3^b, 2¹3^b, ... 2^a3^b
~~~
2¹5¹, 2¹5¹, ... 2^a5¹
2¹5², 2¹5², ... 2^a5²
.
.
.
2¹5^b, 2¹5^b, ... 2^a5^b
~~~
3¹5¹, 3¹5¹, ... 3^a5¹
3¹5², 3¹5², ... 3^a5²
.
.
.
3¹5^b, 3¹5^b, ... 3^a5^b
~~~
2¹3¹5¹, 2¹3¹5¹, ... 2^a3¹5¹
2¹3²5¹, 2¹3²5¹, ... 2^a3²5¹
.
.
.
2¹3^b5¹, 2¹3^b5¹, ... 2^a3^b5¹

2¹3¹5², 2¹3¹5², ... 2^a3¹5²
2¹3²5², 2¹3²5², ... 2^a3²5²
.
.
.
2¹3^b5², 2¹3^b5², ... 2^a3^b5²

.
.
.

2¹3¹5^c, 2¹3¹5^c, ... 2^a3¹5^c
2¹3²5^c, 2¹3²5^c, ... 2^a3²5^c
.
.
.
2¹3^b5^c, 2¹3^b5^c, ... 2^a3^b5^c

[andretimpa]

If you look closely, that's exactly what CuCN was talking about.

Also, to list all factors there is no correspondence that I think you'd find simple, but they all revolve on listing (x,y,z) for the values where x, y AND z are small in some sense and then working their way to infinity.

For example, you can start listing (using CuCN's method for example), all triples such that x,y,z < n at first for n=1, then 2, 3, 4, ... However, when you list the case n=4, for example, you must omit the cases that were enumerated up to n=3 to avoid repetitions.