# Strange (or stupid) arithmetic derivation

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?

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I suspect $p^{p^p}$ and so on will need several steps before hitting a cycle so the answer to the second question should be no. For example $3^{27}$ takes $10$ steps before hitting a fixed point. –  Felipe Voloch Sep 21 '13 at 16:34
Note that this differs from the usual arithmetic derivative en.wikipedia.org/wiki/Arithmetic_derivative. –  Martin Brandenburg Sep 22 '13 at 11:29
Why wouldn't you make it distributive over products of primes: seems a bit more natural? –  Lev Borisov Sep 22 '13 at 11:54
A similar question is mathoverflow.net/questions/25974/calculus-on-rationals. –  Roland Bacher Feb 1 at 23:15

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.

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How do you ensure that after the first $n$ steps the chain doesn't go back to some $f^k(p^n)$, $k<n$ ? –  BS. Sep 22 '13 at 7:14
I don't see either how $p>n^n$ ensures that, but it can be done. You just consider the sequence without powers of $p$: $$1→n→f(n)(n−1)→f(f(n)(n−1))(n−2)→...$$ and you choose a prime greater than every element in it. Thank you Mark Sapir, for the elegant solution. (Also by observing $f(n)<n^2$ one could obtain a bound on $p$, but this isn't enough to prove that $n^n$ is enough.) –  Daniel Soltész Sep 22 '13 at 8:58
I replaced $n^n$ by $n^{2^n}$. Hope it is enough. I first wanted to write simply $p\gg n$ (which is clearly enough for what I wrote). But @BS's comment is in fact relevant. One needs to prove also that the chain does never produces an extra power of $p$. I have to think about it. –  Mark Sapir Sep 22 '13 at 11:31
@BS: I changed the answer again. $n^n$ may not be enough but there is a bound anyway (although we may never know what it is). –  Mark Sapir Sep 22 '13 at 11:50
@MarkSapir : nice trick! –  BS. Sep 22 '13 at 12:11

Here's an elementary observation inspired by Felipe's insightful comment.

Let's attack a simpler version of this problem and try to engineer long cycles when the starting point is $n = p^{k_0}$ for some $k_0 > 1$. Starting with such an $n$, we immediately see that $f(n) = k_0p^{k_0-1}$ and so we try to establish some control over the prime factors of $k_0$. In order to keep things as simple as possible, we might want $k_0$ to again be a power of $p$, say $k_0 = p^{k_1}$ so that we are left with $f(n) = p^{k_1 + k_0 -1}$. So long as $k_1 > 1$, we are guaranteed $f(n) \neq n$.

Proceeding in this fashion, it seems as though we are after a sequence $k_j$ of integers which satisfy the following property: denoting the partial sums as $K_j = \sum_{0 \leq i < j}k_j$, we want $$k_j = p^{K_j - j}.$$

In order to ensure a cycle of length $L+1$, we would want a contiguous subsequence $k_j, \ldots , k_{j+L}$ so that the set of shifted partial sums $$\{K_{j+i} - (j+i) \mid 0 \leq i \leq L\}$$ has cardinality $L+1$. I'm not sure how one goes about finding such a sequence $k_j$, but hopefully there are number theory wizards out there who know such things...

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I don't understand this answer either! If $k_0=p^{k_1}$ then except for some trivial cases, it cannot be that $k_0+k_1-1$ is a power of $p$. So I don't understand what proceeding in this fashion" means. –  Lucia Sep 21 '13 at 22:52

While working on the first question, i managed to get another solution to the second, different from Mark's. Consider the formal product: $$P_0=\prod_{p_i \, prime} p_{i}^{p_i}.$$ Note that it is the fixed point of the formal derivation. Then the formal derivative of $2P_0$ is $f(2P_0)=3P_0$. In general it is true that $f^{(k)}(2P_0)=c_k f^{(k-1)}(2P_0)$ for a $c_k>1$. Indeed as every exponent is at least as big as its base. Thus lets formally derive $2P_0$, k-times. Then there are only finitely many primes which had their exponents changed during the process. Removing every other prime from $2P_0$ we obtain a finite product which has its first $k$ derivatives strictly increasing.

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Suppose that a number $n$ needs $K$ steps to get to a cycle.Let $Q$ be a prime number greater than all the $a_i$ 's and all the $p_i$ 's in $$n=\prod_{i=1}^{k}p_i^{\alpha_i}$$ Then $nQ^{Q^K}$ will need at least $K+1$ steps to fall into a cycle because $Q^{Q^K}$ clearly will not "mix" with what is going on with $n$ and it needs at least $K$ steps to fall to cycle.
This means that such a fixed number $K$ does not exist.Very nice question
Did you suddenly switch from $n$ to $x$ after the equation? If not, what is $x$? –  Vidit Nanda Sep 21 '13 at 18:29
What do you mean by not "mix"? $$f(f(Q^Q^K))=f(Q^{Q^K+K-1})=Q^{Q^K+K-2}(Q^k+K-1)$$ this may count as mixing, but i am not sure. Anyway, i like the approach! –  Daniel Soltész Sep 21 '13 at 19:57
I don't understand this answer. The statement that $Q^Q^K$ will clearly not mix seems vague and incorrect, as pointed out by the OP. There also seems to be an assumption that everything must cycle down to $1$ which also is false (e.g. consider $p^p$). –  Lucia Sep 21 '13 at 22:49
As lucia pointed out, not every number drops to one. Also $2^2^6$ gets to a fixed point in $5$ steps so $Q^Q^K$ doesn't need more than $K$ steps to get to a cycle. –  Daniel Soltész Sep 21 '13 at 22:59