Given a natural number $N\in\mathbb N$, we can construct another number as follows: Let $p_1\leq p_2\leq\ldots\leq p_n$ the prime factorisation of $N$, where duplicates can occur. Let $\pi$ be the prime-counting function and write $a\|b$ for the concatenation of digits in base 10, then let $\psi(N)=\pi(p_1)\|\pi(p_2)\|\cdots\|\pi(p_n)$.
I am interested in numbers $N$ such that $N=\psi(N)$. Examples are $2\cdot 7=14=1\|4=\pi(2)\|\pi(7)$ and $3\cdot709=2127=2\|127=\pi(3)\|\pi(709)$. I searched for numbers up to $10^6$, but couldn't find any more (unfortunately, my programming skills aren't good enough to search much further).
Interesting questions that I would like to know the answer to include
- Can you explicitely find any more such numbers?
- How many of these numbers are there?
- Can you construct bounds on the number of primes in the factorisation of $N$ or the size of $N$ depending on the number of primes in its factorisation?
- Can every prime occur as a factor in such a number?
- Feel free to point out any other interesting properties you find.
It might be the case that this has a really easy answer and I'm just stupid, but the only thing I managed to prove is that there are no other numbers other than the two above with $\psi(N)=N$ and $N$ has two prime factors, by using the inequalities $a\|b\leq (10a+1)b$ and $\pi(p)\leq\frac{1.3p}{\log p}$, thus finding bounds on $p_2$ in terms of $p_1$ and checking all remaining possibilities.
I am looking forward to all of your ideas and observations.