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  • $13$ is prime
  • $\frac {14}2$ is prime
  • $\frac {15}3$ is prime

The same thing also holds for $2017$, and I've found cases where this holds even up to $6$, others have shown it works even up to $10$. My question:

Is there a way to show that there will always exist a prime number such that $p$, $\frac{p+1}{2}, \frac{p+2}{3}, \frac{p+3}{4},\dots, \frac{p+n}{n+1}$ are all prime numbers?

What is the largest value of $n$ for which this is possible?

abiessu
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Ryan Cole
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    An interesting consideration: the numbers $N!+2,N!+3,N!+4,\dots, N!+N$ will each be respectively divisible by $2,3,4,\dots,N$. – abiessu May 12 '18 at 03:00
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    It seems probable albeit quite difficult to even show that $p, \frac{p+1}{2}$, and $\frac{p+2}{3}$ would be prime for infinitely many primes $p$. – JavaMan May 12 '18 at 03:05
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    I believe it is not even proven that there are infinitely many such primes that $p+1\over2$ is a prime. – Ivan Neretin May 12 '18 at 05:41
  • @ivan: with $q=(p+1)/2$ and thus $p=2q-1$ : aren't $p,q$ defining the Sophie-Germain-primes? – Gottfried Helms May 12 '18 at 06:47
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    @GottfriedHelms No, a Sophie germain prime is a prime $p$, such that $2p+1$ is prime as well (not $2p-1$) – Peter May 12 '18 at 06:49
  • Ah,yes, I see. Anyway - I'd be surprised if it were easier to show infinititude of $p,q$ pairs than that of SG-primes... – Gottfried Helms May 12 '18 at 07:04
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    If the Bunyakovsky conjecture is true, then for every $n$, such a prime can be found. To see that, define $f(x):=lcm(2,3,\cdots ,n)x+1$. If we find $x$, such that $\frac{f(x)+j}{j+1}$ is prime for $j=0,1,\cdots,n-1$, then we have found a solution. Assuming the Bunyakovsky conjecture, there are even infinite many solutions for each $n$. – Peter May 12 '18 at 07:21
  • However, such solutions are difficult to find in practice. Even for $n=10$, the smallest example seems to be quite large. Dana Jacobsen is an expert for such questions, maybe he helps you. – Peter May 12 '18 at 07:26
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    For those who are interested, a solution for $n=10$ is $$2\ 394\ 196\ 081\ 201$$ – Peter May 12 '18 at 14:51
  • With the notation in the question however, we only have $n=9$. But the last denominator is $10$ – Peter May 12 '18 at 16:18

1 Answers1

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Below is just a proof of a link between $p$ and $n$, thus, not an answer.

Primes larger than $5$ can be partitioned into 2 subsets of type $6n+1$ and $6n+5$. It's easy to see that $p+1=6n+5+1=6(n+1) \Rightarrow \frac{p+1}{2}=3(n+1)$ which is not a prime. So, we can discard $6n+5$ class of primes. Particularly, $13=6\cdot2+1$.

We will be looking at the primes $p=6n+1$. We can write $6n=2^{k}n_1$, where $n_1$ is odd, i.e. $$p=2^{k}n_1+1 \tag{1}$$ and $n_1>1$, since there is $3$ in $6n$.

Proposition $$n<2^k-1 \tag{2}$$

Easy to see from: $$q_1=\frac{p+1}{2}$$ $$q_2=\frac{p+2}{3}$$ $$...$$ $$q_{2^k-1}=\frac{p+2^k-1}{2^k}\overset{(1)}{=}\frac{2^{k}n_1+1+2^k-1}{2^k}=n_1+1$$ But $n_1$ is odd, thus $q_{2^k-1}=n_1+1$ is even and, thus, definitely not prime.

rtybase
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