Interesting. First off, we can assume any expansion containing a 1 is less than ideal, because it would be better to combine into something else. With that in mind, let’s pick an arbitrary (but larger than 6) number, and list the possible expansions. How about 24?
Two-number:
22, 2
21, 3
20, 4
19, 5
…
12, 12
The largest of these is 12, 12. This is the typical Calculus/Pre-calculus problem about enclosing a maximum area with a specific amount of fencing.
Three-number:
20, 2, 2 = 80
19, 3, 2 = 114
18, 4, 2 = 144
18, 3, 3 = 162
17, 5, 2 = 170
17, 4, 3 = 204
16, 6, 2 = 192
16, 5, 3 = 240
16, 4, 4 = 256
…
The general pattern is that the small numbers are getting bigger and the big numbers are getting smaller. And the product is (in general) getting bigger. Therefore, I hypothesize that the ideal (three-number) solution is when all three numbers are equal, which is
8, 8, 8 = 512
In general, (I’m guessing) we want all the numbers to be the same or as close to the same as possible. IF this is true (a big if) then…
Say we started with the number x, and assume it’s divisible by just about anything, just for sake of argument. To make all the numbers equal, the possibilities are:
x/2, x/2 = x[sup]2[/sup]/4
x/3, x/3, x/3 = x[sup]3[/sup]/27
x/4, x/4, x/4, x/4 = x[sup]4[/sup]/256
x/n, x/n, x/n, x/n, x/n, x/n … (out to n numbers) = (x[sup]n[/sup])/(n[sup]n[/sup])
I guess the question is then: At what value of n is (x[sup]n[/sup])/(n[sup]n[/sup]) maximized?
This is of course assuming that letting all the numbers be the same is a good idea. Which I’m not sure if it is.