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Let $p(n)$ denote the largest prime factor of $n$. Prove that there are infinitely many $n$ such that $$p(n)<p(n+1)<p(n+2)<p(n+3)$$

if the three consecutive numbers,I can prove it,also can see three,erdos,but for four (or more) consecutive number maybe is old?and How do to?

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By standard conjectures (see Dickson's conjecture), there exist infinitely many $k$ such that $3k+1$, $20k+7$, $30k+11$, and $60k+23$ are all prime. Then the required inequalities hold for $n=60k+20$, since $p(n)=3k+1$, $p(n+1)=20k+7$, $p(n+2)=30k+11$, and $p(n+3)=60k+23$, and clearly $3k+1<20k+7<30k+11<60k+23$.

I don't know whether there is an unconditional proof. I note that Erdos and Pomerance were unable to prove the existence of infinitely many $n$ such that $p(n)>p(n+1)>p(n+2)$, so this could be a very hard problem.


Original, incorrect, answer follows:

By standard conjectures (see Dickson's conjecture), there exist infinitely many $k$ such that $3k+2$, $10k+7$, $15k+11$, and $30k+23$ are all prime. Then the required inequalities hold for $n=30k+20$, since $p(n)=3k+2$, $p(n+1)=10k+7$, $p(n+2)=15k+11$, and $p(n+3)=30k+23$, and clearly $3k+2<10k+7<15k+11<30k+23$.

I don't know whether there is an unconditional proof. I note that Erdos and Pomerance were unable to prove the existence of infinitely many $n$ such that $p(n)>p(n+1)>p(n+2)$, so this could be a very hard problem.

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  • $\begingroup$ I think one needs to be careful on $3k+2$ and $15k+11$. If $k$ is even, then $3k+2$ is not a prime, and if $k$ is odd then $15k+11$ is not a prime. That is, we need to check whether the given four linear forms are admissible. $\endgroup$ May 17, 2020 at 1:50
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    $\begingroup$ @SungjinKim im so confused. why does one need to be "careful"? this answer is just blatantly wrong. why does it have 2 upvotes $\endgroup$ May 17, 2020 at 2:21
  • $\begingroup$ @mathworker, I'll fix it. $n=60k+20$ should do. $\endgroup$ May 17, 2020 at 2:48
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    $\begingroup$ $n=12k+8$ works, too, with smaller numbers. It needs $n/4=3k+2$, $(n+1)/3=4k+3$, $(n+2)/2=6k+5$, and $n+3=12k+11$ to be simultaneously prime, which already happens for $k=1$. $\endgroup$ May 17, 2020 at 3:19
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    $\begingroup$ @GerryMyerson The problem of $p(n)>p(n+1)>p(n+2)$ for infinitely many $n$ is solved by Balog with counting function $\gg \sqrt x$, and by Tao and Teravainen at arxiv.org/pdf/1904.05096.pdf with counting function $\gg x$. $\endgroup$ May 20, 2020 at 16:49

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