I would tend to think that the following has already been investigated.

But as implied from the title, I have no idea how to even start looking for it.

Let $P_n$ denote the sum of the squares of the prime factors of $n$ with multiplicity; i.e.

let $\ S\ $ be a finite set of primes. Let $\ e: S\rightarrow\{1\ 2\ ...\}\ $ represents positive integer exponents in $\ n\ :=\ \prod_{p\in S}\ p^{e(p)}.\ $ Then

$$P_n\ :=\ \sum_{p\in S}\ e(p)\cdot p^2$$

Let $A_n$ denote the following sequence (defined for every $n>1$):

$ A_n= \begin{cases} \sqrt{P_n} & \text{$\sqrt{P_n} \in\mathbb{N}$}\\ {P_n} & \text{$\sqrt{P_n}\not\in\mathbb{N}$}\\ \end{cases} $

A few examples:

  • $A_{5}=\sqrt{5^2}=5$
  • $A_{30}=2^2+3^2+5^2=38$
  • $A_{48}=\sqrt{2^2+2^2+2^2+2^2+3^2}=5$
  • $A_{60}=2^2+2^2+3^2+5^2=42$

Now, form a sequence by performing this operation repeatedly, beginning with any positive integer $k>1$, and taking the result at each step as the input to the next. My conjecture is that this process will eventually reach a number $n$ such that $A_n=n$, regardless of which positive integer $k$ is chosen initially.

For example, an initial value of $k=9$ yields the following process:

  • $A_{9}=3^2+3^2=18$
  • $A_{18}=2^2+3^2+3^2=22$
  • $A_{22}=2^2+11^2=125$
  • $A_{125}=5^2+5^2+5^2=75$
  • $A_{75}=3^2+5^2+5^2=59$
  • $A_{59}=\sqrt{59^2}=59$

It's easy to observe that $A_n=n$ for every prime number $n$ ("landing" on the 1st case).

In addition to that, $A_n=n$ also for the non-prime $n=27$ ("landing" on the 2nd case).

I am pretty sure that $A_n\neq{n}$ for any other value of $n$, but I do not know how to prove this.

It would be nice if someone suggested a proof, although it is not the primary question at hand.

In any case, this conjecture seems much more probable than the $3n+1$ conjecture, since the latter has to "land" on a power of $2$, while the former can "land" on any prime number (or $27$), which is significantly more likely to occur within any given range.

However, the process seems to be growing up to rather large figures for certain initial values of $k$, such as $30$ and $60$ (and many others), so it may be possible to refute this conjecture by showing an initial value of $k$ for which the process never terminates.

So at this point I can consider three different approaches:

  1. Prove that the process converges for every initial value
  2. Prove that the process diverges for some initial value
  3. Show that the process loops for some initial value

The first two options seem quite difficult, and I wouldn't even know where to start.

The third option is the easiest, but I find it very unlikely that such initial value exists.

So any ideas or points of reference to related researches will be highly appreciated.

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    $\begingroup$ I do not actually understand how $9-18-22-125-59$ is possible.$3$ is the only prime divisor of $9$ so $A_9$=3 $\endgroup$ – Konstantinos Gaitanas Jan 10 '15 at 12:54
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    $\begingroup$ @KonstantinosGaitanas: Prime factorization $\equiv$ non-distinct primes. $\endgroup$ – barak manos Jan 10 '15 at 12:55
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    $\begingroup$ Who said anything about downvoting for a typo? Why so sensitive to downvotes? $\endgroup$ – Gerry Myerson Jan 11 '15 at 5:10
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    $\begingroup$ "...and then getting down-votes without any comment explaining them...." You have one downvote on this question. Not "down-votes"; one downvote. Get over it. $\endgroup$ – Gerry Myerson Jan 11 '15 at 5:17
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    $\begingroup$ The question to me seems contrived. $P_n$ could be defined without multiplicities, or with a power higher than 2 (and $A_n$ then the corresponding root) or in any number of ways that sends a prime to itself and that seems likely to send infinitely many integers into primes. Unless this particular formulation arises from some other problem (eg. if $P_n$ were a well studied arithmetic function), why is it interesting? In itself it lacks the beauty and tantalizing mix of approachability and inaccessibility of the $3x+1$ conjecture. $\endgroup$ – Yaakov Baruch Jan 11 '15 at 7:42

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