Square-free integers in the sequence $\lambda+\prod_{k=1}^n(\varphi(k)+1)$, where $\lambda\neq 0$ is integer

Question

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.

Answer 1

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.