We define, for $k\in\mathbb{N}$, the sequence $\left(S_{k,n}\right)_{n\in\mathbb{N}}$: $$S_{k,1}=k,\;\;\; S_{k,n+1}=p_1q_1\cdots p_mq_m \text{ (written out in decimal)}$$ Where $p_1^{q_1}*\cdots *p_m^{q_m}$, with $p_1<\cdots<p_m$ is the ordered prime factorisation of $S_{n}$.

For example: $$S_{28,1}=28$$ Now since $28=2^2*7^1$: $$S_{28,2}=2271$$ Since $2271=3^1*757^1$: $$S_{28,3}=317571$$ Etcetera.

We see that, at least in this particular case, the sequence grows rapidly. This turns out to be a general phenomenon. However, sometimes the sequence can decrease: $$S_{512,1}=512=2^9\;\;\;\;\; S_{512,2}=29$$ Motivated by this, I am wondering wether or not there exists a $k\in\mathbb{N}$ such that the sequence $\left(S_{k,n}\right)_{n\in\mathbb{N}}$ has cycles (i.e. is periodic).

By doing computer research, I have found that there exist no cycles which have highest element $\leq 10^7$. My conjecture is that no cycles exist for any $k$, however I am not sure where to start a proof.

For further investigation we might also consider the problem in other number systems (with a different base). Is there any base for which there are cycles?

EDIT: The answer to the last question is yes. Thanks to Joffan we have found bases in which there even exist fixed points.

Some of my further attempt at fixed points:

Note that in order to find fixed points in base $x$, we will need to find an $n\in\mathbb{N}$ such that there exist prime numbers $p_1<\cdots<p_n$ and positive integers $q_1,\cdots,q_n$ satisfying:

$$p_1^{q_1}*\cdots*p_n^{q_n}=p_1x^{2n-1}+q_1x^{2n-2}+\cdots+p_nx+q_n$$ We are particularly interested in the case $x=10$.

Maybe it is easier to first solve the equation for $x$. For example, for $n=1$ the equation is: $$p^q=px+q$$ Which gives us $x=p^{q-1}-\frac{q}{p}$. So we would like to find a prime number $p$ and a positive integer $q$ such that $p^{q-1}-\frac{q}{p}=10$ (probably no luck here).

For $n=2$ we get: $$ p_1^{q_1}p_2^{q_2}=p_1x^3+q_1x^2+p_2x+q_2$$ Which gives us some nastier solutions for $x$, but will maybe give us more chance to end up with the magical $10$ since we have more $p_i$s and $q_i$s to choose from.

Maybe someone can even find a proof that there exist no fixed points for base $10$, which of course will also be a big step. Please do not hesitate to post any suggestions that come to mind.

  • $\begingroup$ The base 10 representation makes your iterated representation somewhat "meaningless" in terms of the preceding prime factorization. The fact that you are appending the powers after the primes makes it even more somewhat "meaningless". I think this is probably a very, very hard question to attack with theoretical math, and so probably the only hope is that if there is a cycle that doesn't start with too large of a number, or have too long a cycle length, then a computer program will find it. $\endgroup$ – user2566092 Feb 19 '15 at 22:23
  • $\begingroup$ Thank you for your feedback. I agree that this sequence might not be that "meaningful" from a theoretical point of view, especially because of the dependence of the number system used. However, I don't feel like that should stop us. If you don't like base $10$, you can always try to work in another base. Maybe someone with better programming skills than me will find a way to easily check for sequences in other number systems, and it would be pretty cool if we actually found one! I don't care about the meaning, I'm just interested in some fun maths. Any ideas will help! Many thanks in advance. $\endgroup$ – Uncountable Feb 19 '15 at 22:39
  • $\begingroup$ Are you familiar with the $3n + 1$ conjecture? That deals with stuff that is seemingly even more mathematically "meaningful" than your sequences but Paul Erdos famously said that he believed the problem was perhaps beyond current mathematics. $\endgroup$ – user2566092 Feb 19 '15 at 22:44
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    $\begingroup$ Fairly trivially, perhaps, we can make $k^k = kk$ in base $n=k^{k-1}-1$ - for example, $3^3 = 33$ base $8$. $\endgroup$ – Joffan Feb 20 '15 at 0:26
  • $\begingroup$ $Joffan Very nice! I hadn't thought of that yet. Well done, sir. $\endgroup$ – Uncountable Feb 20 '15 at 0:29

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