2
$\begingroup$

While I was exploring the squares in the sequence defined for integers $n\geq 1$ $$\prod_{k=1}^n(\varphi(k)+1),\tag{1}$$ where $\varphi(m)$ denotes the Euler's totient function I wondered a different question. I've considered the sequence $$\lambda+\prod_{k=1}^n(\varphi(k)+1),\tag{2}$$ for fixed (and small, positive or negative) integers $\lambda\neq 0$.

Question. Can you provide me an example of $\lambda$ for which a sequence of the type $(2)$ should have infinitely many square-free terms? That is an example and reasoning or heuristic to know why such sequence that I evoke has infinitely many terms without repeated factors (see this Wikipedia). Many thanks.

Only is required a heuristic, if the proof doesn't seem feasible. My belief is that in fact next conjecture holds, but I don't know how to prove (I've tested the first 60 terms).

Conjecture. For $n>1$, the integer $2+\prod_{k=1}^n(\varphi(k)+1)$ has no repeated prime factors (that is a square-free integer).

I don't know if this question (or the question in the first paragraph) was in the literature. I think that maybe there is an heuristic to show such evidence for some $\lambda\neq 0$. If this question was in the literature please refer it, and I am going to search and try to understand such propositions from those articles. Many thanks.

$\endgroup$
  • $\begingroup$ For the Question only is required a reasoning or heuristic that convince us that is reasonable that exists a sequence of the type $(2)$ with infinitely many terms without repeated prime factors. I've tested few terms of my Conjecture, if you find a counterexample please add it. $\endgroup$ – user243301 Feb 28 '18 at 19:09
  • $\begingroup$ If some user wants to explore the question about the squares in the sequence $\prod_{k=1}^n(\varphi(k)+1)$ (if it isn't studied in the literature) that he/she feels free to do it. $\endgroup$ – user243301 Feb 28 '18 at 19:17
  • 1
    $\begingroup$ According to my calculations, for no prime factor $q\le 10^5$ , the product can be divisible by $q^2$, which is a good indication that the conjecture is true. $\endgroup$ – Peter Mar 1 '18 at 21:58
  • 1
    $\begingroup$ Many thanks for such great effort @Peter now the conjecture seems more interesting. $\endgroup$ – user243301 Mar 1 '18 at 22:39
  • 1
    $\begingroup$ Upto $n=10^5$, $(1)$ is only a perfect square for $n=2$ and $n=4$, so probably the only perfect squares in the sequence are $4$ and $36$. $\endgroup$ – Peter Mar 3 '18 at 10:27
0
+50
$\begingroup$

The following PARI/GP - program searches for a counterexample. The zeros in the second column only mean that no counterexample has been found yet. If it finds a counterexample, it will post the numbers $q$ and $k$. This then means that $q^2$ divides the product upto $k$ plus $2$.

? gef=0;forprime(q=1,10^6,if(Mod(q,10^4)==1,print(q,"  ",gef));p=2;k=1;while(k<q
,k=k+1;p=lift(Mod(p*(eulerphi(k)+1),q^2));if(p==q^2-2,gef=1;print(q,"  ",k))))
70001  0
90001  0
150001  0
160001  0
180001  0
270001  0
280001  0
390001  0

As you can see , no prime factor less than $390\ 000$ can do the job. So, there is a good chance that there is none. But this is only a heuristic, a counterexample is still well possible.

$\endgroup$
  • $\begingroup$ Thank you very much, it is a great help and draw attention about the conjecture and Question. Just about the penultimate of your commments, is that this conjecture is of all us, I just did a question and my few experiments, I was lucky this time with this post. $\endgroup$ – user243301 Mar 2 '18 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy