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Post Posted: May 24, 2007, 5:24 pm • # 1 


Show that, for any fixed integer \,n \geq 1,\, the sequence 2, \; 2^{2}, \; 2^{2^{2}}, \; 2^{2^{2^{2}}}, \cdots \pmod{n} is eventually constant.
 
 
Post Posted: May 24, 2007, 5:24 pm • # 2 


Write a_n=\underbrace{2^{2^{...^2}}}_{n \text{ times}}, then a_{n+1}=2^{a_n}.

Induction on n (one could also use some straightforward way, but I think this way's nicer, and it kills topic 139, too):

We show that it is constant beginning from the n-1-th term (or earlier):

n=1 is clear.
For any n>1, we write n=2^m n^\prime with n^\prime odd.
The sequence a_k clearly gets constantly 0 \mod 2^m for all k \geq m \leq n-1. So we are left to prove the same \mod n^\prime.

By induction, the sequence gets constant \mod \tilde n := \varphi(n^\prime) < n^\prime \leq n. Thus there is c such that for all k \geq \tilde n-1 we have a_k \equiv c \mod \varphi(n^\prime).
This gives a_{k+1} = 2^{a_k} \equiv 2^c \mod n^\prime by the theorem of Euler-Fermat, meaning nothing else than constantness of the sequence for all k>\bar n \leq n-1, our result.
 
 
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