Quote:
Originally Posted by CRGreathouse
That cuts it down to a little more than
\[n^{n! + (n-1)! + (n-2)! + n-3}/2.\]
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
\[n^{n! + (n-1)! + (n-2)! + n-3}/n! \approx n^{n! + (n-1)! + (n-2)! - 3}.\]
But these numbers are huge, we need to do much, much better.
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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.