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We can call a prime intertwined if it consists of two smaller primes having their digits alternating in it, for example:

$$1433 = 13 \oplus 43\\ 1873 = 17 \oplus 83$$

If we let $\oplus$ mean this alternating operation.

Is there likely to exist a largest intertwined prime?


Own work

I have considered the prime distribution from the prime number theorem

$$\pi(N) = \frac{N}{\log(N)}$$

Maybe it would be possible to consider probabilities two prime numbers of $k$ digits each forming a prime number of $2k$ digits taking into account approximately how many prime numbers there are on each interval $[10^k,10^{k+1}]$.

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    $\begingroup$ How far have you gotten in considering this question? Please do include such work in your post. $\endgroup$ – Namaste Jun 30 '17 at 17:26
  • $\begingroup$ @amWhy: I have so far only had some initial ideas about how to maybe use the prime number theorem. $\endgroup$ – mathreadler Jun 30 '17 at 17:38
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    $\begingroup$ Did you see this concept of intertwined prime elsewhere? Or did you come up with it yourself? $\endgroup$ – RitterSport Jun 30 '17 at 17:41
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    $\begingroup$ You be interested in en.wikipedia.org/wiki/Permutable_prime and en.wikipedia.org/wiki/Truncatable_prime and references therein. $\endgroup$ – RitterSport Jun 30 '17 at 17:58
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    $\begingroup$ Heuristics: Starting from one of the $\sim \frac{10^{2n}}{2n}$ primes with $2n$ digits, the two $n$-digit "twine summands" are prime with probability $\sim\frac 1n$ each. So $\sim\frac{10^{2n}}{2n^3}\gg1$ primes should succeed in that range. -- Or taking pairs from the $\sim\frac{10^n}n$ primes with $n$ digits, we can $\oplus$ them to $\sim\frac{10^{2n}}{n^2}$ longer numbers, each being prime with probability $\frac1{2n}$; this leads to the same estimate $\sim\frac{10^{2n}}{2n^3}$. -- so the answer is "probably works for all $n$", but that's similar to what we know about Goldbach's conjecture $\endgroup$ – Hagen von Eitzen Jun 30 '17 at 18:06

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