Is there another pair of consecutive primes with this property?

Question

Denote $$r(n)$$ to be the number that occurs if we reverse the digits of $n$

Suppose, $\ (p,q)\ $ is a pair of consecutive primes.

The only prime $p$ with the property $$r(p)=2q$$ I found is $\ p=479\ $.

Is there another prime $p$ with the given property ?

For the opposite equation, namely $$r(q)=2p$$ I did not find yet a single example.

I checked both equations upto $p=10^9$

Answer 1

Here are two pairs of consecutive primes $(p,q)$ with $r(q)=2p$: $$p=4574\cdot 10^{123} - 3123,\quad q = 4574\cdot 10^{123} - 2581$$ and $$p=494\cdot 10^{213} - 303,\quad q = 494\cdot 10^{213} - 211.$$

Background. The difference between consecutive primes is much smaller than the primes (e.g., see Cramér's conjecture), but there are not so many patterns for numbers $(p,q)$ with $r(p)=2q$ or $r(q)=2p$ with small difference $q-p$. Furthermore, some of these patterns produce numbers with small factors, and thus they cannot deliver primes. Below I describe patterns for the differences below $100$ that can potentially produce prime pairs.

The most simple and attractive pattern for $r(q)=2p$ with difference $2$ is $p = 5\cdot 10^n - 3$ and $q = 5\cdot 10^n - 1$ with $n\geq 3$. As soon as these $p$ and $q$ are both prime, we are guaranteed that they are consecutive as prime twins. Unfortunately, if such prime twins exist, $n$ would be very large as can be seen from the sequences A103003 and A056712 lacking small common terms.

Next possible prime difference in increasing order are

• $28$ given by $p = 48\cdot 10^n - 41$ and $q = 48\cdot 10^n - 13$ with $r(p)=2q$ for all $n\geq 2$.
• $32$ given by $p=454\cdot 10^n - 323$ and $q= 454\cdot 10^n - 291$ with $r(q)=2p$ for all $n\geq 3$.
• $58$ given by $p = 493\cdot 10^n - 411$ and $q=493\cdot 10^n - 353$ with $r(p)=2q$ for all $n\geq 3$.
• $62$ given by $p=474\cdot 10^n - 313$ and $q=474\cdot 10^n - 251$ with $r(q)=2p$ for all $n\geq 3$.
• $92$ given by $p = 494\cdot 10^n - 303$ and $q = 494\cdot 10^n - 211$ with $r(q)=2p$ for all $n\geq 3$.

I've quickly tested these patterns for $n\leq 1000$ and for the last one found the second pair of consecutive primes given at the top.

UPDATE. I've also made a more extensive search over larger differences and found another pair (coming first at the top) having difference 542.