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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 $[10^k,10^{k+1}]$.

MJD
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mathreadler
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    How far have you gotten in considering this question? Please do include such work in your post. – amWhy Jun 30 '17 at 17:26
  • @amWhy: I have so far only had some initial ideas about how to maybe use the prime number theorem. – mathreadler Jun 30 '17 at 17:38
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    Did you see this concept of intertwined prime elsewhere? Or did you come up with it yourself? – RitterSport Jun 30 '17 at 17:41
  • As far as I can tell I came up with it. But I have read an awful lot of "programming challenges" with prime math questions when I was younger so it is possible that I am mistaken and actually read it somewhere. – mathreadler Jun 30 '17 at 17:44
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    You be interested in https://en.wikipedia.org/wiki/Permutable_prime and https://en.wikipedia.org/wiki/Truncatable_prime and references therein. – RitterSport Jun 30 '17 at 17:58
  • There is also this question, about primes that stay prime if you delete digits. There is a heuristic about why there should only be finitely many. Your intertwined primes are similar: They are primes that remain prime if you delete certain digits. So maybe you can bootstrap of the answer. https://math.stackexchange.com/questions/33094/deleting-any-digit-yields-a-prime-is-there-a-name-for-this – RitterSport Jun 30 '17 at 18:00
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    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 – Hagen von Eitzen Jun 30 '17 at 18:06
  • Some mildy larger examples: $100000007\oplus 100000259$, $1000000007\oplus 1000000349$, $10000000019\oplus 10000000601$, $100000000003\oplus 100000000069$, $1000000000061\oplus 1000000000163$, $1000000000000223\oplus 1000000000001063$, $100000000000000000151\oplus 100000000000000000547$, $1000000000000000000000000000271\oplus 1000000000000000000000000000529$, $100000000000000000000000000000000000000000000000151\oplus 100000000000000000000000000000000000000000000001249$ -- no end in sight – Hagen von Eitzen Jun 30 '17 at 18:20
  • One thing we can say, is that since, $1+2\equiv 0 \pmod 3$, All pairs of primes that work, will have digital sums, that are same modular remainder mod 3. –  Jun 30 '17 at 18:22
  • MathReadler Adding what you added to your post, plus your active participation in the comments: Thumbs up! Thanks for the added information! – amWhy Jun 30 '17 at 18:45
  • The number of intertwined primes with $k$ digits, $3\le k\le9$ is: $$25,\ 86,\ 504,\ 2741,\ 17769,\ 112863,\ 807657$$ – Julián Aguirre Jun 30 '17 at 21:44

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