Do we have $\sum_{n=1}^{\infty}\frac{\gcd\left(1+n!,1+n^{2}\right)}{n!}\stackrel{?}{=}e$? I try to find a balanced exercise on it with Desmos.
It seems we have:
$$\sum_{n=1}^{\infty}\frac{\gcd\left(1+n!,1+n^{2}\right)}{n!}\overset?=e$$
It seems non trivial as you can remark in replacing the square by a cube.
At first glance we can think it's problem with hidden rearrrangement theorem. Keeping in mind that I try the Riemann theorem on sequence and rearrangemnt without success. I keep in mind also the Wilson's theorem but cannot progress.
Notice I check the result with Wolfram Alpha.
How to (dis)prove it?
Thanks in advance for the help.
Side notes :
It's a strange idea but why calculus couldn't be here I mean ,and I think it's stupid but let's try that :
$$\gcd\left(\operatorname{floor}\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\right)+1,n^{2}+1\right)=^?1,n\in N^+$$
I think it's not very useful but why not ? I think that because it's really not the same problem but as Leibniz said knowledge is like path between some islands...
Ps: It's the floor function .
As second notes see also Bounds on the difference between the polylogarithm with negative base and the gamma function
Edit 04/02/2023
There is also where $x$ is a prime :
$$\gcd(x!-x^x,1+x^2)=1$$
Then I was thinking to Green-Tao theorem but nothing consistent at all.
 A: Here is a reason why this problem would probably be very hard to solve.
Suppose $(n^2 + 1, n! + 1) = d$. Then any prime divisor of $d$ must be a prime greater than $n$, so $d$ must be a prime greater than $n$ and equivalent to $1$ modulo $4$. Thus, we are looking for a prime $p \equiv 1 \bmod{4}$ and an $n < p$ such that $p | n^2 + 1$ and $p | n! + 1$.
Look at this from the perspective of a random $p \equiv 1 \bmod{4}$. It is well known that there exactly two choices for $n$, $a_p$ and $b_p = p - a_p$, such that $p | a_p^2 + 1$. Given this, as far as I can tell, there is nothing definite we can say about $a_p!$ and $b_p!$ modulo $p$, other than the fact that they are not divisible by $p$. Thus, we can heuristically assume they are equally likely to be equidistributed in $(\mathbb{Z} / p\mathbb{Z})^*$ modulo $p$. Therefore, there is about $2 / (p - 1)$ chance that $p$ satisfies the desired relation.
Taking union over $p$, I conclude that in an interval $[A,B]$, roughly
$$\sum_{p \in[A,B], p \equiv 1 \bmod 4} \frac{2}{p - 1}$$
$p$ would satisfy the desired constraint. We know that
$$\sum_{p \in[A,B], p \equiv 1 \bmod 4} \frac{2}{p - 1} \approx (\log\log B - \log\log A).$$
Solving for this greater than $1$(i.e. at least one solution exists), we get
$$B \geq A^e.$$
@Peter has kindly verified that no solution less than $2 \times 10^6$ exists. Thus, if I were to gamble, I would bet that a solution will eventually pop up, but it is around
$$(2 \times 10^6)^e \approx 10^{17}$$
which is far beyond the computational ability of my computer.
A few more words: the reason I believe we can say nothing about $a_p!$ modulo $p$ is because number theorists have not figured out how to deal with factorials beyond Wilson's theorem. For example, seemingly innocent problems like Brocard's problem remain wide open.
P.S. If you were to search for primes of the form $p = n^2 + 1$ only, I would bet no solution exists. The reason is that we can show
$$\sum_{p = n^2 + 1, p> 10^6} \frac{2}{p - 1} < \sum_{n = 10^3}^\infty \frac{2}{n^2} \leq 0.01.$$
