# Is there another pair of consecutive primes with this property?

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 $$p=10^9$$

• I'm guessing you see the restrictions to even digit starts for p, same digit length for q, 1 mod 6values of p, mapping to 5 mod 6 values of q, and 5 mod 6 vslues of p mapping to 1 mod 6 values for q ? – Roddy MacPhee Apr 21 at 10:57
• is the multiplier on 9 predictable on reversal of a number ? if so we might be able to figure out mod 7. and therefore mod 42. – Roddy MacPhee Apr 21 at 21:19
• The same $10^9$ claim is made at primes.utm.edu/curios/page.php?short=479 and attributed to Galliani. – Gerry Myerson Apr 23 at 4:35
• Now posted to (but closed on) MO, mathoverflow.net/questions/330066/… – Gerry Myerson Apr 26 at 23:48

## 1 Answer

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.

• Good job! (+1) But you did not find a solution with $r(p)=2q$ , right ? – Peter Apr 28 at 7:21
• @Peter: No. I've found only nearly consecutive pairs like $(4989999999999999999799, 4989999999999999999947)$ with one other prime in between. – Max Alekseyev Apr 28 at 10:01
• $$p=4776*10^{2979}-8441$$ solves the original problem. Thank you Max, I used your idea to find it ! – Peter Apr 29 at 18:05
• @Peter: Good catch! I did not bother looking at exponents greater than 1000. – Max Alekseyev Apr 29 at 20:03