# there are infinitely many $n$ such that $p(n)<p(n+1)<p(n+2)<p(n+3).$

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

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？

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.

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.
• 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. May 17, 2020 at 1:50
• @mathworker, I'll fix it. $n=60k+20$ should do. May 17, 2020 at 2:48
• $n=12k+8$ 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 $k=1$. May 17, 2020 at 3:19
• @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 $\gg x$. May 20, 2020 at 16:49