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Topic: z^z^z^...
Replies: 14   Last Post: Mar 23, 2000 5:00 PM

 Messages: [ Previous | Next ]
 FinalFntsy Posts: 35 Registered: 12/8/04
Re: z^z^z^...
Posted: Mar 23, 2000 1:07 AM

Message:

In Donald Knuth's notation x^(x^(x^(...^x))) = x^^n (assuming
the tower of exponents has n levels). In a recent post I pointed
out an amazing result concerning the modular arithmetic version
of this problem: if a, m are positive integers, m >= 2, then
as n -> infinity a^^n _always_ converges modulo m (i.e., it is
eventually constant mod m). Has anybody found this result before?

Response:

This is trivial to prove by fixing a and inducting of m. m=2 is trivial.
Looking at the multiplicative orbit of a modulo m we see that it has k < m
elements in it. By induction, a^^n stabilizes mod k, and thus one step later
it stabelized mod m. (The case of a=2 was a problem on the USA Mathematical
Olympiad sometime in recent history, and the proof for a=2 easily extends to
arbitrary a).

Date Subject Author
3/21/00 Jonathan Hoyle
3/21/00 Johannes H Andersen
3/22/00 Jonathan W. Hoyle
3/22/00 David C. Ullrich
3/22/00 Jonathan Hoyle
3/22/00 denis-feldmann
3/22/00 John Bailey
3/23/00 Jonathan Hoyle
3/22/00 G. A. Edgar
3/22/00 Jonathan Hoyle
3/22/00 Miguel A. Lerma
3/23/00 FinalFntsy
3/22/00 Ioannis Galidakis
3/23/00 John Bailey
3/23/00 Ioannis Galidakis