Conjecture :
Let $x<0,n>2$ ,$n$ an integer such that $|x|<(n!+1)(n^2+1)$ then it seems that :
$$1/(x/(n!+1))!-1/(x/(n^2+1))!=0$$
Have no integer solution in $x$ where :
$$x!=\Gamma(x+1)$$
Or : $$\Gamma(x/(n!+1)+1)=\Gamma(x/(n^2+1)+1)$$
Based on empirical trial it seems true for low $n$ .It generates a lot of almost integer and the calculus of the factorial is delicate for large value .It recall me a bit the LFT by these empirical facts .
Context :
Do we have $\sum_{n=1}^{\infty}\frac{\gcd\left(1+n!,1+n^{2}\right)}{n!}\stackrel{?}{=}e$?
The goal is to show the two arithmetical sequences doesn't match and then conclude for the $\gcd$.As similarities is the Green-Tao theorem.
Some other information :
There exists $x<0$:
$$1/2+1/2×(2-1/x!^{x!})^2-1/2(x!(1-x!))^2-20(x!(1-x!))^4-127(x!(1-x!))^6+\cdots=f(x!)$$
$$f(z)=z^z-z^2(1-z)^2-20z^4(1-z)^4-21z^6(1-z)^6-33z^8(1-z)^8-33z^{10}(1-z)^{10}-66z^{12}(1-z)^{12}-249z^{14}(1-z)^{14}+\cdots$$
Question :
Is there an argument in the sense of the conjecture or a counter-example ?
