Let us consider the following operation on positive integers: $$n=\prod\_{i=1}^{k}p\_i^{\alpha\_i} \qquad f(n):= \prod\_{i=1}^{k}\alpha\_ip\_i^{\alpha\_i-1}$$ (Is it true that if we apply this operation to any integer multilpe times, it will eventually get into a finite cycle?) Is there a constant $K$ such that any integer will fall into a cycle after $K$ steps?


**Edit4:** We managed to settle affirmatively the question of Mark Sapir, whether a cycle of arbitrary length exists: <http://www.math.bme.hu/~kovacsi/Pub/arithmetic_derivation_v04.pdf>


**Edit3:** I proposed two questions (in retrospect, it was a minor mistake), one of them was answered. To appreciate this, i accept Mark Sapir's answer, and alter the original text by putting the unanswered stuff into parentheses. Making the answered one the main question.


**Edit2:** István Kovács pointed out that there is a nice formula for $f(n)$ using the 'number of divisors' function: $$ f(n):= d \left( \frac{n}{ \prod\_{i=1}^{n}p\_i } \right) \frac{n}{ \prod\_{i=1}^{n}p\_i } $$ from which it fillows that for any $\varepsilon >0 , \quad f(n)=o(n^{1+\varepsilon})$.


I think that the answer to the first question is yes, but to the second no. We tested the first $10000$ integers and every integer fell into a cycle after at most $6$ steps.


**Edit:** @MarkSapir proved that the answer to the second question is no. His proof raises the (third) question: How long can such a cycle be?



Strange (or stupid) arithmetic derivation
I will show that the answer to the second question is "no". Note that if the answer to the first question is "no", we are done. Hence assume that the answer is "yes" and for every number $n$, the chain eventually turns into a cycle. Take any number $n$ and consider a sequence of numbers $n, f(n)(n-1), f(f(n)(n-1))(n-2),...$, that is $a\_1=n, a\_{m+1}=f(a\_m)(n-m)$, for every $m=1,...,n-1$. Let $A=a\_n$. Let $p$ be a prime that is bigger than any number that occurs in the chain for $A$. Consider $p^n$. Then the chain for $p^n$ looks like $p^n\to a\_1p^{n-1}\to ...\to a\_n\to...$ and the chain for $A=a\_n$ follows. That chain will never hit the first $n$ of the numbers in the chain for $p^n$, hence the chain for $p^n$ does not go into a cycle before the step number $n$. Since $n$ was arbitrary, we are done. This answers the second question. The first question seems more difficult. 


 **Edit.** As @DanielSoltész pointed out, I answered a harder question about bounding the length of a pre-cycle in the chain $m\to f(m)\to\ldots$. If we want to show that there is no bound for the number of different elements in a chain, then assuming $p>n^{2^n-1}$ is enough. This leads to another reasonable question about bounding the lengths of cycles $m\to...\to m$. That question is open. 


