View Single Post
Old 2018-02-02, 03:25   #20
science_man_88's Avatar
"Forget I exist"
Jul 2009

838410 Posts

Originally Posted by CRGreathouse View Post
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.
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.

Last fiddled with by science_man_88 on 2018-02-02 at 03:34
science_man_88 is offline