# (Unsolved) In this infinite sequence, no term is a prime: prove/disprove.

$343,~ 34343, ~3434343, ~343434343,\ldots$

$\begin{array}\\ \color{Red}{343} &\color{Red}{: 7^3}\\ 34343 &: 61\times 563\\ \color{green}{3434343} &\color{green}{: 3\times 11^2\times 9461}\\ \color{red}{343434343} &\color{Red}{: 7\times 521\times 94169}\\ 34343434343 &: 47\times 79\times 9249511\\ \color{green}{3434343434343} &\color{green}{: 3^2\times 19\times 29\times 67\times10336531}\\ \color{red}{343434343434343} &\color{Red}{: 7\times 151\times 324914232199}\\ 34343434343434343 &: 5638147\times 6091262669 \end{array}$

Update: The numbers in black are,

$$F_n = \frac{34\times10^{6n-1}-43}{99}$$

and $F_n$ is composite for $n<1667$ (user Uncountable) and $n<3101$ (user A.P.).

• It's easy to prove that one third of the numbers are divisible by 3, and another third are divisible by 7. – gnasher729 Feb 21 '15 at 23:23
• It's only the black ones that are of interest. – Brian M. Scott Feb 21 '15 at 23:25
• Where did you find the problem?????? – Will Jagy Feb 21 '15 at 23:41
• The black numbers are of the form $(34\times10^n+43)/99$ where $n$ is of the form $6r-1$. That doesn't solve the problem, but it gives you a more convenient form for the numbers. – Gerry Myerson Feb 21 '15 at 23:45
• It's worth noting that since this sequence is exponential, it satisfies a linear recurrence relation with constant coefficients, and so some of the tools that have been applied to those sequences might be applicable here. See math.stackexchange.com/questions/1090319/… for an example. – Steven Stadnicki Feb 22 '15 at 0:22

We have $343434 = 2 \times 3 \times 7 \times 13 \times 17 \times 37$. $34343$ is not divisible by any of these numbers. Therefore the subsequence $34343$, $34343 + 343434 \times 10^5$, $34343 + 343434 \times 10^5 \times (10^0 + 10^6)$, $34343 + 343434 \times 10^5 \times (10^0 + 10^6 + 10^{12})$, $34343 + 343434 \times 10^5 \times (10^0 + 10^6 + 10^{12} + 10^{18})$ etc. consists of numbers which are not divisible by $2, 3, 5, 7, 13, 17, \quad \text{or}\quad 37$.
Heuristically, the chance of a random number $N$ being prime is $1 / \ln N$. Having $7$ small primes excluded as possible factors increases the chances by a factor $(2/1)(3/2)(5/4)(7/6)(13/12)(17/16)(37/36) ≈ 5.1757$.
The numbers are around $3.4343 \times 10^{5+6k}$ with $k = 0, 1, 2, 3,\ldots$ The natural logarithm is about $13.8155k + 10.4442$. So the chance that each of the numbers is a prime is about $5.1757 / (13.8155k + 10.4442)$. The expected number of primes among the numbers for $k = 0$ to $n$ is about $0.3746 \cdot\ln (n) + 0.4013$. For $n = 1,666$ the expected number of primes is about $3.1803$; that's the range that Uncountable checked. So finding no primes is slightly unlucky, but not that unlikely.
For a $50\%$ chance of finding a prime, the expected number of primes needs to increase by $\ln 2 ≈ 0.6931$, so $\ln n$ needs to be increased by $0.6931/0.3746 ≈ 1.8502$, $n$ needs to be multiplied by $6.36$. So there's a $50\%$ chance of finding a prime with up to $63,600$ digits; then a $50\%$ chance for a prime with up to $404,000$ digits and so on.
Of course all that is just heuristic. If it is correct, then a prime will almost certainly exist. Finding a probable prime might be very hard. If checking a range that gives a $50\%$ chance fails, the next range giving a $50\%$ chance is $6.36$ times larger. Miller-Rabin test grows more than quadratic with the number of digits, and there are more numbers to test, so the next range takes more than $6.36^3 = 257$ times longer.
• Can you look at this similar question and give a comment why $7373737373737\dots$ has prime values for $n=3,\,7,\,95,\,422,\dots$. – Tito Piezas III Apr 17 '15 at 5:37