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Let $p$ be a fixed prime number. We define the ring of Witt vectors $W(R)$ for any commutative ring $R$ as follows:

For every ring morphism $R \rightarrow R'$ the induced morphism $W(R) \rightarrow W(R')$ is an ring morphism.

The maps $w_n : W(R) \rightarrow R$, given by $\underline{b}=(b_0,b_1,\ldots) \mapsto b_0^{p^n}+p b_1^{p^{n-1}}+\ldots+p^n b_n$, are ring morphisms.

We define Frobenius $^{F} : W(R) \rightarrow W(R)$ by functoriality in $R$ and the relations, for all $n\geq 0$ and $x \in W(R)$: $w_n (^{F} x) = w_{n+1} (x)$.

In particular, for $R = \mathbb{Z}$ we can compute the Frobenius on $W(R)$ directly. Now let $p =2$ (we could also take other primes, but I made only calculations on this case): Regard the sequence $x = (2,3,5,7,\ldots) \in W(R)$ and let us apply the Frobenius on $x$. We then obtain $$^{F} x = (a_0, a_1, \ldots)$$ with $$a_0 = 10 = 2 \cdot 5,$$ $$a_1 = -23,$$ $$a_2 = -2621,$$ $$a_3 = -5 \cdot 3198913,$$ $$a_4 = -11 \cdot 19 \cdot 107 \cdot 34195794911,$$ $$a_5 = -19 \cdot 180738889676329198795378497899,$$ where we denoted the prime factorization of each $a_n$.

There is one conjecture which arises (even though not formulated as a precise mathematical statement): is it true that $a_n$ factors into a very large prime number times very small primes (compared to $a_n$)?

Unfortunately, I could not test it for more $a_n$, since my PC ran into difficulties. If this would be true to some extent then it could be imaginable to get a kind of prime number generator in the following way: compute $a_n$ for some "large" $n$. Then we can divide $a_n$ by all small primes (when we know that there is some upper limit to their value) and then we get a very large prime number remaining. I did not test it for other primes or the iterated Frobenius, but there could be arise similar questions.

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I honestly don't understand this last part. What do you mean by "divide $a_n$ by small primes" and then ending up with a large prime number? For instance, in the example you gave what are you planning on dividing by? –  Matt Oct 14 '11 at 15:29
    
@Matt: I think tobias suggesting something like: there is a function $f$ such that $a_n$ factors as $pu$, where $p$ is prime, and $u< f(n)$, and that furthermore $a_n$ grows much more quickly than $f(n)$ as $n\to \infty$. –  Charles Rezk Oct 14 '11 at 17:59
    
Yes, Charles ist right. This was my idea. –  tobias Oct 15 '11 at 5:46
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