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Does there exist $x\in\mathbb{R}$ such that $\lfloor 10^nx\rfloor$ is a prime number for all $n\in\mathbb{N}$?

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    $\begingroup$ 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? $\endgroup$ Jul 30, 2023 at 16:40
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    $\begingroup$ All the integer parts will be right-truncatable primes of which there are finitely many, so the answer is no. $\endgroup$
    – Wojowu
    Jul 30, 2023 at 16:58
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    $\begingroup$ No, we can choose $E(x)$ a prime number of more of one digit $\endgroup$
    – Dattier
    Jul 30, 2023 at 17:19
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    $\begingroup$ @GHfromMO Wojowu's comment doesn't technically answer the question, since the starting point of the sequence needn't be a single digit. $\endgroup$ Jul 31, 2023 at 0:29
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    $\begingroup$ 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 $1$. $\endgroup$ Jul 31, 2023 at 12:37

2 Answers 2

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(This is an extended comment.) There couldn't be anything special about base 10, could there?

Notation: Given two positive integers $m,n$, let $m\oplus n$ be the integer that results from prepending the digits of $m$ to the left of the digits of $n$ (in whatever base we are considering).

Base 2: Here we can only append digits $0$ and $1$. We can't append a $0$ digit to a prime and have it remain prime, since the result would be divisible by $2$. So, we can only repeatedly append $1$. But such a process cannot always result in a prime. Indeed, $p\oplus \underbrace{111\cdots 1}_{p-1\text{ times}}$ is divisible by $p$, when $p$ is prime.

Base 3: Again, we can ignore the digit $0$, due to 3-adic considerations. We can also ignore the digit $1$, due to 2-adic considerations. This only leaves us with the ability to append the digit $2$ repeatedly, which leads to a similar contradiction as at the very end of the base 2 case.

Base 4: We can ignore the digits $0$ and $2$, due to 2-adic considerations. Looking 3-adically, we see that $n\oplus 1\equiv n+1\pmod{3}$ and $n\oplus 3\equiv n\pmod{3}$. Thus, the digit $1$ can be appended only finitely many times (before we reach a number that is $0\pmod{3}$), and hence eventually we must only see the digit $3$ used. That leads to a contradiction as before.

Base 5: We can ignore the digit $0$. Also, due to 2-adic considerations, we can ignore the digits $1$ and $3$. This leaves the digits $2$ and $4$. If $n\equiv 1\pmod{3}$, then $n\oplus 2\equiv 1\pmod{3}$ while $n\oplus 4\equiv 0\pmod{3}$. Thus, for numbers that are $1\pmod{3}$, we can only append the digits $2$ repeatedly, and this leads to a contradiction. Similar considerations apply to numbers that are $2\pmod{3}$.

Base 6: We can ignore the digits $0,2,3,4$, leaving only $1,5$. Now, looking 5-adically, we see that $n\oplus 1\equiv n+1\pmod{5}$ while $n\oplus 5\equiv n\pmod{5}$. So, the digit $1$ can be appended only finitely many times, leaving us to eventually repeat only the digit $5$, which leads to a contradiction.

Base 7: We can ignore the digit $0$. Due to 2-adic considerations, we can ignore $1,3,5$. This leaves digits $2,4,6$. Suppose that we have reached a prime $p\equiv 1\pmod{3}$. Then $p\oplus 2\equiv 0\pmod{3}$, $p\oplus 4\equiv 2\pmod{3}$, and $p\oplus 6\equiv 1\pmod{3}$. Similar considerations apply when $p\equiv 2\pmod{3}$. So, what has to happen is that we append some number of $6$'s, then $4$, then some number of $6$'s, then $2$, then some number of $6$'s, then $4$ again, etc...

Looking 5-adically, we can get some additional restrictions on the sequence of appended digits. At this point, additional $p$-adic considerations, for primes $p\geq 11$, give additional restrictions, but they don't seem to solve the problem.

Base 10: A similar analysis, using 2-adic, 3-adic, and 5-adic considerations, leads to eventually only having use of the digits $3$ and $9$. Moreover, $p$-adic considerations, for primes $p\geq 7$, give limitations on the allowable sequences of $3$'s and $9$'s. Ultimately, I believe there is no local obstruction (but I could be wrong). This base seems more tractable than for bases 7,8,9.

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A few comments:

a) this problem was mentioned in 2004 edition of R.Guy's book "Unsolved Problems in Number Theory". The author's comment indicated that at the time the solution was not known.

b) In 1997 Gerry Myerson reformulated original question about infinite sequence of right-extendable (truncatable) primes in the form presented here. He also mentioned that it was easily solved in bases 2 through 6.

c) In 2007 the record-holding sequence was only 14 numbers long.

It certainly seems that this problem (about infinite sequence) is still unsolved. The fact that we can only add digits 3 and 9 is pretty obvious because digits 1 and 7 cannot be used more than twice (modulo 3 argument proves that) - therefore we can assume that the sequence begins after the last time those digits were added.

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