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Peter

D 6

**Posted:** Fri May 25, 2007 2:24 am

ZetaX

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.

**Posted:** Fri May 25, 2007 2:24 am