Quote:
Originally Posted by CRGreathouse
That cuts it down to a little more than
\[n^{n! + (n1)! + (n2)! + n3}/2.\]
You can do better: reassign the variables so the first one you use is 1, the first one you use other than that is 2, and so on. This cuts it to about
\[n^{n! + (n1)! + (n2)! + n3}/n! \approx n^{n! + (n1)! + (n2)!  3}.\]
But these numbers are huge, we need to do much, much better.

Only patterns I see is that when the current lengths known n<6 are divided by 3 we get 1 mod 4, 3 mod 4, 1 mod 4, 3 mod 4. Wonder if that holds.