Do we have $\sum_{n=1}^{\infty}\frac{\gcd\left(1+n!,1+n^{2}\right)}{n!}\stackrel{?}{=}e$?

Question

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.

Edit 01/11/2023 :

We can represent (I suggest to show it) that :

$\exists a,r,n,m,u>0$ such that :

$$n!+1=\lfloor(1-a^r)^{-m}\rfloor=P$$

And :

$$n^2+1=\lfloor(1-a^r)^{-u}\rfloor=Q$$

Then it seems we have if $a$ is irrational and $P,Q$ have the prime factorization :

$$P=\prod_{i=1}^{v}p^{b_i}_{a_i},Q=\prod_{i=1}^{N}p^{c_i}_{d_i}$$

If so $\exists a_k,b_k,c_k,d_k$ such that $p^{b_k}_{a_k}$ is not in the prime factorization of $Q$ and vice versa .

A simple progress :

If $p$ is a prime having as last digit a $9$ and let $(p-1)^2+1=M$ then if $M$ have $9$ representation as sum of two square and $M^2$ have $4$ pythagorean triples then the conclusion follows and the conjecture is true .To understand that try an example like $p=269$

it seems also $p=883$ works too $25<882,37<882,29×29<882$ .Build it is rather easily .

Remark see https://publications.ias.edu/deligne/paper/357 and Can we invert these analogous "Dirichlet" series for GCD / LCM convolution?

Some works :

If $n,n+2$ is a twin pairs then :

$$4(n-1)!+4+n=0\operatorname{mod}(n(n+2))$$

Or $\exists k$ such that :

$$4(n-1)!+4+n=kn(n+2)$$

Or :

$$4(n-1)!+4=kn^2+(2k-1)n$$

It's divisible by four next we compare with :

$$kn^2+(2k-1)n=4q((n-1)^2+1)$$

Or :

$$(k-4q)n^2+(2k-1+8q)n-8q=0$$

If the discriminant is not a perfect square then the roots are irrational and so there is no solution in $q$:

$$2(10k^2-6k+1)$$

Is never a perfect square using $k$ is odd and (optional) the Hasse-Minkowski theorem so it's true that $\gcd((n-1)!+1,(n-1)^2+1)=1$ when $n,n+2$ are prime numbers if $10k^2-6k+1,(n-1)^2+1$ have no common prime factor .

We can enlarge with $$f\left(x,y\right)=\gcd\left(\left(10\cdot\left(2\operatorname{floor}\left(x\right)+1\right)^{2}-6\left(2\operatorname{floor}\left(x\right)+1\right)+1\right),\left(\operatorname{floor}\left(y\right)\right)^{2}+1\right)$$ Let introduce :

$$f\left(x,y\right)=\gcd\left(10\left(2x+1\right)^{2}-6\left(2x+1\right)+1,y^2+1\right)$$

Conjecture :

Let $n>1$ be a natural number then :

$$f(n^3,n)=1$$ Or : $$f(n^3,n)=\prod_{i=1}^{m}p_{a_i}$$

Where $p_{a_j}$ is a prime number here distinct

As a factorial cannot be a perfect power greater than one we delete some cases

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I know that if $n-1$ finishes with a $8$ then $(n-1)^2+1$ divisible by $5$ and primes then I cannot go further

One example :

It's known that it can be write as a sum of square and squared as a sum of square too. The prime factorization of the number in the sum of square is not in $(n-1)^2+1$ but could be in $(n-1)!+1$ as the example $n=29$

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To be continued.

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In formalism it gives (all the unknow are positive integers and here $p,p+2$ are twin primes):

Theorem 1 :

Using Wilson's theorem for a prime $p$ Then $p$ is the smallest prime divisor of $(p-1)!+1$ if $p-1$ ends with a $8$ .

Theorem 2 :

If $p-1$ have as last digit $8$ then $(p-1)^2+1=P$ is divisible by $5$ .Then if $P=a^2+b^2,P^2=c^2+d^2$ then the integer factors in the prime decomposition of $a,b,c,d$ are not in the prime decomposition of $P$. Finally one of $a,b,c,d$ share a prime with $(p-1)!+1=K$ .This prime is the smallest divisor excepting $1$ of $K$.

Theorem 3 with condition of theorem 2 :

Recalling theorem 1 ,$p$ is the smallest divisor excepting $1$ of $K$. The smallest prime divisor of $K$ greater than $p$ is greater than the greater divisor of $P$ which is a prime number as divisor, if $P^2$ have $4$ representation as sum of two square .

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Some materials for a proof :

Let says with the conditions of theorem 1,2,3 on $K$ and let $K$ have $C_p$ prime factor distinct then it seems we have :

$$C_p<\ln\left(\left(p-\frac{3}{2}\right)\ln\left(p\right)-p-1+\frac{1}{2}\ln\left(2\pi\right)\right)+\frac{1}{\sqrt{p}}$$

Using Ramanujan-Hardy theorem and Binet's formula .

For $P$ which have $T_p$ prime factor distinct it seems we have for $p$ sufficiently large :

$$T_p<\ln\left(\ln\left(\left(p-1\right)^{2}+1\right)\right)+\frac{1}{2}\sqrt{\ln\left(\ln\left(\left(p-1\right)^{2}+1\right)\right)}$$

Reference :

https://www.jstor.org/stable/2305816?origin=crossref

Answer 1

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 $$\sum_{p = n^2 + 1, p> 10^6} \frac{2}{p - 1} < \sum_{n = 10^3}^\infty \frac{2}{n^2} \leq 0.01.$$

Answer 2

The easy way :

Lemma 1 :

Let $n>6$ be an integers and $k$ a positive integers such that $(n+1)^2+1$ is not a prime number then we can write: $1+(n(n+1)+1)!=1+k((n(n+1)+1)^2+1)$.

Straight proof of lemma 1 :

To show it use Fibonacci's identity $(n^2+1)((n+1)^2+1)=1+(n(n+1)+1)^2$, and remark the divisors are less than the square roots of $(n(n+1)+1)^2+1$ and use the fact that's not a prime numbers for $(n+1)^2+1$.

Then $$\gcd((n^2+n+1)!+1,(n^2+n+1)^2+1)=1$$ Where $(n+1)^2+1$ is not a prime number

To go further Cauchy's induction ? or from wikipedia source :

To be continued...