Timeline for Does this number exist?
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 6, 2023 at 22:35 | comment | added | Zach Teitler | @DenisShatrov At most twice. 3, 31, and 311 are prime. | |
| Oct 6, 2023 at 20:15 | answer | added | JimT | timeline score: 4 | |
| Aug 1, 2023 at 19:24 | answer | added | Pace Nielsen | timeline score: 13 | |
| Aug 1, 2023 at 16:07 | comment | added | Dattier | No, it's not the same question, the question here talk about a real $x$ with $\lfloor x\times 10^n\rfloor$ is prime with all $n$ natural integer, and not $\lfloor x^n \rfloor$. | |
| Aug 1, 2023 at 13:15 | comment | added | David E Speyer | @DanielAsimov That question has been asked on this site before, but no one answered it mathoverflow.net/questions/436518 . | |
| Aug 1, 2023 at 8:44 | comment | added | Yaakov Baruch | @DanielAsimov It's easy to see, without knowing $C$, that $p_{n+2}\in \left( \frac{p_{n+1}^2}{p_n+1}, \frac{(p_{n+1}+1)^2}{p_n}\right)$ and that once the size of $p_n$ is large enough the width of this interval is less than something like $2C^2+5$, and therefore the probability of each branch continuing becomes less than $1$ (again, for $\log(p_n)$ larger that $2C^2+5$). So with probability $1$ all branches will eventually die out. | |
| Aug 1, 2023 at 2:12 | comment | added | Daniel Asimov | Does there even exist any constant C > 1 and an x > 0 such that the sequence floor(C^n x) is prime for all n in ℤ+ ? | |
| Jul 31, 2023 at 12:37 | comment | added | Yaakov Baruch | Euristically the answer seems a clear no. Say that the first prime in the sequence has length $m$ (so about $10^m/(m\log(10))$ possibilities for it) and we try to add a digit, we have a probability of $\frac{10}{\log(10^m)}$ to successfully land on a prime again, so we see that once $m$ is larger than $2$ the expected number of live branches decreases exponentially and soon will come to an end with probability $1$. | |
| Jul 31, 2023 at 11:01 | comment | added | Dattier | $E$ is the function integer part, hence $E(1.92)=1$ | |
| Jul 31, 2023 at 10:56 | comment | added | მამუკა ჯიბლაძე | @Dattier What is $E(x)$? | |
| Jul 31, 2023 at 0:38 | comment | added | GH from MO | @PaceNielsen I see. Thank you. | |
| Jul 31, 2023 at 0:29 | comment | added | Pace Nielsen | @GHfromMO Wojowu's comment doesn't technically answer the question, since the starting point of the sequence needn't be a single digit. | |
| Jul 31, 2023 at 0:01 | comment | added | GH from MO | @Wojowu Please turn your comment into an answer, so that this question can be closed. | |
| Jul 30, 2023 at 23:48 | history | edited | GH from MO | CC BY-SA 4.0 |
deleted 39 characters in body
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| Jul 30, 2023 at 22:24 | comment | added | Denis Shatrov | Actually, to prevent divisibility by 3, digits $1, 7$ can be added at most once. | |
| Jul 30, 2023 at 22:19 | comment | added | Juan Moreno | Equivalently, as any prime number greater than $5$ ends in $1$, $3$, $7$ or $9$, can we add a combination of these digits to some prime number digits to always obtain a greater prime number? | |
| Jul 30, 2023 at 22:14 | comment | added | Pietro Majer | So every prime is the root of a "decimal tree of primes", e.g. from the root 43 there are three branches, to 431, to 433, and to 439. Then 431 has no sons; from 433 a couple of branches, to 4337 and to 4339; from 439 just one to 4391. Are all these tree finite? | |
| Jul 30, 2023 at 22:01 | comment | added | Nick S | @CarloBeenakker I think that the answer to the weaker question is yes. If $k$ is any positive integer, the PNT implies that for $n$ large enough we get $$\frac{p_{n+1}}{p_{n}}< \frac{k+1}{k}$$. This means that for $m$ large enough there exists a prime $p$ such that $k\cdot 10^{m} < p < (k+1) \cdot 10^{m}$. It is trivial now to constrcut some $x$ with the property from your comment. | |
| Jul 30, 2023 at 20:49 | review | Close votes | |||
| Jul 31, 2023 at 19:35 | |||||
| Jul 30, 2023 at 17:19 | comment | added | Dattier | No, we can choose $E(x)$ a prime number of more of one digit | |
| Jul 30, 2023 at 17:14 | comment | added | Wojowu | It's a tedious but easy computational task. | |
| Jul 30, 2023 at 16:58 | comment | added | Wojowu | All the integer parts will be right-truncatable primes of which there are finitely many, so the answer is no. | |
| Jul 30, 2023 at 16:40 | comment | added | Gerald Edgar | Like this ... $2$ is prime, $29$ is prime, $293$ is prime, $2939$ is prime; can we continue this indefinitely? For this start, no, at $29399999$ we get stuck. But what about other starting values? | |
| Jul 30, 2023 at 16:18 | comment | added | Carlo Beenakker | A weaker question would be if it is possible that this set contains an infinite subset of prime numbers. I think this is an open problem. | |
| Jul 30, 2023 at 16:08 | history | asked | Dattier | CC BY-SA 4.0 |