5
$\begingroup$

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 $p=10^9$

$\endgroup$
  • 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
  • 2
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ $$p=4776*10^{2979}-8441$$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.