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).