In this post we denote the Stirling numbers of the second kind as ${n\brace k}$. I present a variant of the problem showed in the penultimate paragraph of section **B33** of [1] (see also the cited bibliography of the book), for a different sequence from combinatorics. I don't know if next problem is the most interesting that one can evoke as a variant, but I'm curious about my question. Your comments are welcome.

Up to $3000$ the sequence of integers $1\leq n$ satisfying $$\gcd\left({2n\brace n},105\right)=1$$
are $1, 6, 762, 2520, 2521$ and $2526$ (see if you want my section *Details for the computational evidence and documentation*).

Question.Is it possible to determine if our equation $$\gcd\left({2n\brace n},105\right)=1$$ has finitely many or infinitely many solutions for integers $n\geq 1$?Many thanks.

If it is very difficult, mainly I am asking about what work can be done about the **Question**.

**Details for the computational evidence and documentation.** My code was written in Pari/GP. Search in Internet *Sage Cell Server* and paste next codes choosing as Language *GP*. I say this line

`for (n = 1, 1000, if(gcd(stirling(2*n, n, {flag = 2}),105)==1,print(n)))`

or

`for (n = 2000, 3000, if(gcd(stirling(2*n, n, {flag = 2}),105)==1,print(n)))`

You can find also a *Pari-GP reference card* from some universities, searching these words in Internet.

## References:

[1] Richard K. Guy, *Unsolved Problems in Number Theory*, Unsolved problems in Intuitive Mathematics Volume I, Second Edition, Springer-Verlag (1994).