After the change of the year I realized, as everyone did, that $2018=2\times1009$ and of course $1009$ is a prime number. $2017$ is also a prime number. Furthermore $2019=3\times 673$ and $673$ is also a prime number. So here is the conjecture.

For every number $n\in \mathbb{N}$ there exists a prime number $p$ such that

$p+1=2\times p_1$,

$p+2=3\times p_2$,

$p+3=4\times p_3$,

$p+4=5\times p_4$,

. . .

$p+n=(n+1)\times p_n$

where $p_1,p_2,p_3,.....,p_n$ are some prime numbers.

One can notice that this is an optimal situation in the sense that in every 2 numbers at least one should be divided by $2$ in every 3 numbers at least one should be divided by $3$ and in every n numbers at least one should be divided by $n$. Of course the conjecture is too hard in the sense that it implies the open conjecture that there are infinitely many primes such that $2p-1$ is also prime, or you can ask the same question for the opposite direction meaning:

For every number $n\in \mathbb{N}$ there exists a prime number $p$ such that

$p-1=2\times p_1$,

$p-2=3\times p_2$,

. . .

$p-n=(n+1)\times p_n$

where $p_1,p_2,p_3,.....,p_n$ are some prime numbers that implies that the Sophie Germain Primes are infinite, or both directions simultaneously,etc.

I don't know if this is a known question, I tried to find related ones but I didn't made it.


Thinking to this direction one can make much more general conjectures that implies these ones. Par example:

For any number $n\in \mathbb{N}$ and for every prime $p'<n$ and every power $i$ for each one of them there exists a $p$ such that

every number $p+1,...,p+n$ is divided by:

1) some(or none) primes $p'<n$ to some powers $i$ in any legal way meaning that none of this ways should force $p'|p$ or for some $n'$ $n'|p+j, n'|p+k$ and $n'$ does not divide $k-j$ or some $n''$ not to divide $n''$ consecutive numbers.

2) at most one more prime number.

Of course this would imply much more than the twin prime conjecture...

approaching the general philosophy that "With prime numbers (or natural numbers in general) everything that is possible to happen happens somewhere"

Happy New Year!

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    $\begingroup$ Surely this is a special case of the (notorious) prime $k$-tuples conjecture, and surely no easier to settle than the general case. $\endgroup$ – Gerry Myerson Jan 2 '18 at 21:57
  • $\begingroup$ @Gerry Myerson: the generalization implies the K-tuples conjecture, I cant see how the opposite happens $\endgroup$ – Asterios Gkantzounis Jan 2 '18 at 22:03
  • $\begingroup$ So this seems to be a generalization also of k-tuples, not a special case $\endgroup$ – Asterios Gkantzounis Jan 2 '18 at 22:30
  • $\begingroup$ Sorry, I was referring to the questions above the generalization. $\endgroup$ – Gerry Myerson Jan 3 '18 at 4:27

Let $L = \text{lcm}(2,\ldots, n+1)$. Then $p = Lm -1$ satisfies your conjecture for $n$ if $Lm/(k+1)-1$ for $k = 0 \ldots n$ are all primes. Dickson's conjecture implies that this is the case for infinitely many $m$.

See OEIS sequence A078502.

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  • $\begingroup$ Very interesting. What about the generalization? $\endgroup$ – Asterios Gkantzounis Jan 3 '18 at 5:04
  • $\begingroup$ seems like the generalization is equivalent to Dickson's conjecture $\endgroup$ – Asterios Gkantzounis Jan 3 '18 at 10:42

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