A $2021$ problem: $20\sim 21$ and $43\times 47$

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 $n=36$, $d=2$, $x=891077215721081784886888257701070827$, $x(x+d)=794018604377235322848433897872605582794018604377235322848433897872605583$.

I've found two more solutions, bringing the total to 4 solutions. I've posted code here.

$794018604377235322848433897872605582\sim794018604377235322848433897872605583 = 891077215721081784886888257701070827\times891077215721081784886888257701070829$

$2518711810848159770018909254809359591672377471484881441744436703324716\sim2518711810848159770018909254809359591672377471484881441744436703324717 = 5018676928083894672666012088036109843105301546773725102790665815794437\times5018676928083894672666012088036109843105301546773725102790665815794441$

$353879205744237011544616255111782082608671961515039134082358165448687146\sim353879205744237011544616255111782082608671961515039134082358165448687147 = 594877471202462845078583461328011525336167267541426222873827376039101347\times594877471202462845078583461328011525336167267541426222873827376039101401$

I found more twin prime solutions. For n = 396, d = 2 I found four; the smallest is x =

434219772837481616940726338933362452273916097301492635627291443512051997247961750150908315825206335025681182252954355463934702107964728269885096262688644034853639395077615105799339919571601786192665103151170947581386505769710635828116973131647706180301379657813220432413064536727826883282252811784121256116486385454859849292857777667719016881251956101283374338871503089431823276421261037909974359.

For n = 420, d= 2 I found x =

797632045320122442922746848370123218495083238021539262463793196558953236606673684986633691999805212011226776035396567202878987657452660659814874176096052099397205337415190293836597252999499063444222715671072742933640380698030812583452694556061309433747142801949662038045758937332990610602377203706320829077950947557183280793831442171456664369310820656043310649088552521802135329739094023210425721483954126295715977820771.

I wrote a Java program to find modular square roots using known factorizations of $10^n+1$ at https://stdkmd.net/nrr/repunit/10001.htm

The n = 396 solution shown may not be the second smallest twin prime solution because I did not analyse $10^n+1$ with unknown factorization (eg: 323, 392).

Note that all these solutions involve only probable primes.