A $2021$ problem: $20\sim 21$ and $43\times 47$ Notice that $2021$ is a concatenation of consecutive integers:  $20\sim 21$
Also $2021$ is a product of consecutive primes: $43\times 47$.
What is the next number with both of these properties?
$24502451$ is close, $4943\times 4957$  but $4951$ is in between.
$2484224843$ is close, $49831\times 49853$ but $49843$ is in between.
$715353612\sim 715353607$ isn't quite there but is $845785793\times 845785799$
Are there any other numbers with the $2021$ property?
 A: If $p$ is prime and $q$ is the next prime and $pq$ has $2n$ digits, the probability that $pq$ is the concatenation of consecutive integers is heuristically about $10^{-n}$.  For $pq$ to have $2n$ digits (i.e. $10^{2n-1} \le pq < 10^{2n}$), $p$ must be between about $10^{n-1/2}$ and $10^n$, and there are approximately $$\frac{10^{n}}{n \ln 10} - \frac{10^{n-1/2}}{(n-1/2) \ln 10} \sim \frac{10 - \sqrt{10}}{10 \ln(10)} \frac{10^n}{n}$$
such primes.  Thus we expect about $0.297/n$ examples with $2n$ digits.  Since the harmonic series diverges, there should be infinitely many, but the next one could be quite large and hard to find.  I've checked by brute force that there are no further examples with up to $16$ digits.
EDIT: To expand on my comment above:
The concatenation of $y$ and $y+1$, where $y+1$ has $n$ digits,
is $(10^n+1) y + 1$, and thus $\equiv 1 \mod (10^n+1)$.
Given an even positive integer $d$ and positive integer $n$, you can solve the equation $x(x+d) \equiv 1 \mod (10^n+1)$, and then check in each solution
that $10^{2n} > x(x+d) \ge 10^{2n-1}$ and $x$ is prime and $x+d$ is the next prime.  For $n \le 36$ and $d \le 10000$ the only solutions found are
$n=2$, $d=4$, $x = 43$, $x(x+d) = 2021$ and that amazing
$n=36$, $d=2$, $x=891077215721081784886888257701070827$, $x(x+d)=794018604377235322848433897872605582794018604377235322848433897872605583$.
A: I would have just commented but I don't have sufficient reputation. A quick brute-force approach in bases 2 to 16 is here and shows solutions where the second part of the concatenation contains leading zeros. Hence 249950~00249951 = 4999493*4999507.
