04/03/2010, 12:00 PM
(04/03/2010, 12:18 AM)Stereotomy Wrote: \( 7^{7^{7}} \text{mod} 5 = 3 \)
Is true as well. In fact, thinking about it, the numbers I tried out with this all had b>a. Perhaps that's an additional condition that either b > a or m, n > 1?
There is an article which proves that \( {^n a} \) (which is \( a{\uparrow}^2 n \) in Knuth's arrow notation) finally will be constant for \( n\to\infty \) mod any \( M \), see this thread.
