I came across the following curious problem while playing around with my calculator.

Take any positive integer $n$; for this example we'll use $216$. Create a sequence as follows:

  • Factor $n$ into its prime factors, listing smaller factors first and expanding exponents; $216=2\times2\times2\times3\times3\times3$
  • Create the next number by concatenating the first $d$ digits of its prime factors, where $d$ is the number of digits in $n$; $2\times2\times2\times3\times3\times3\to222$
  • Repeat until...? $222\to2\times3\times37\to233\to233\to\cdots$

The first thing that is easy to show is that you will always be able to create a $d$-digit number at any step.

The sequence obviously stops once it hits a prime number. But can it stop for any other reason? It turns out that the first number whose sequence does not end in a prime number is $333$, a composite which miraculously generates itself: $333=3\times3\times37\to333$.

A quick computer program found several other numbers that exhibit this property:

  • $2255\to5114\to2255\to\cdots$ a loop!
  • $22222\to24127\to23104\to22222\to\cdots$ a longer loop!
  • $22564\to22564\to\cdots$
  • $31111\to53587\to41130\to23354\to21167\to61347\to31111\to\cdots$ !!!
  • $210526\to210526\to\cdots$
  • $252310\to252310\to\cdots$
  • $1143241\to1143241\to\cdots$
  • $3331233\to3331233\to\cdots$
  • $3710027\to3710027\to\cdots$

Update: up to 150 million, I found

  • $219371601\to367109140\to225183554\to219371601\to\cdots$

I'll call these numbers, whose sequence does not end in a prime, self-factoring numbers. As far I've tested, these are all the self-factoring numbers below forty million.

This topic may have no depth to explore (it may just be a fun mathematical coincidence), but I would still like to pose the following questions:

  • Are these self-factoring numbers related in any way? Can we prove any property about these numbers at all?
  • Is there any property that might help speed up a computer search past the brute force approach?
  • Are there infinitely many such self-factoring numbers? Could there be a constructive proof?
  • $\begingroup$ Interesting question! One idea: let $X_d$ be the set of $d$ digit natural numbers and $\phi: \mathbb{N} \rightarrow \mathbb{N}$ be the transformation you've described. Then let $P^n_d = \bigcup_{i = n}^\infty \phi^i[X_d]$. The question you are asking is basically if $\bigcap_{i=1}^\infty P^i_d$ contains any non-primes for any $d$. $\endgroup$ – Bruce Zheng Oct 7 '14 at 21:13

Similar to $a_0=210526$ is $$a_3=210526315789473684210526315789473684210526315789473684210526$$ where $$a_k=2\cdot\frac{2\cdot10^{18k+6}-3}{19}=210\dots526=2\cdot10\dots5263.$$ In each case $a_k$ is twice a prime. I have checked all the $a_k/2$ for $k=0,\dots,745$ for primality, and not found any further primes (according to GAP 4.7).

Another possible source of similar examples (i.e. where the number self-factors to itself by being twice a prime) is the numbers \begin{align*} b_k&=2\cdot\frac{2\cdot10^{18k+1}-1}{19}=210\dots842=2\cdot10\dots8421, \end{align*} though $b_0=2$ is prime, and $b_k/2$ is composite for $k=1,\dots,739$. Stopping the decimal expansion of $4/19$ anywhere else cannot yield a number that is twice a prime. For example the numbers \begin{align*} 2\cdot\frac{2\cdot10^{18k+11}-9}{19}&=210\dots578=2\cdot10\dots5789 \end{align*} are all multiples of 11, and the numbers \begin{align*} 2\cdot\frac{2\cdot10^{18k+13}-7}{19}&=210\dots894=2\cdot10\dots8947 \end{align*} are all multiples of 13. (For $p=11$ and $p=13$, $p\mid 1001$, so, modulo $p$, $1000=-1$, so $10^6=1$ and $10^{18k}=1$. Thus, if $p$ divides any member of the above sequence, $p$ divides them all.)


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