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 $p=10^9$

  • 2
    $\begingroup$ 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 ? $\endgroup$ – Roddy MacPhee Apr 21 at 10:57
  • $\begingroup$ 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. $\endgroup$ – Roddy MacPhee Apr 21 at 21:19
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    $\begingroup$ The same $10^9$ claim is made at primes.utm.edu/curios/page.php?short=479 and attributed to Galliani. $\endgroup$ – Gerry Myerson Apr 23 at 4:35
  • $\begingroup$ Now posted to (but closed on) MO, mathoverflow.net/questions/330066/… $\endgroup$ – Gerry Myerson Apr 26 at 23:48

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.

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

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