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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).

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  • $\begingroup$ It is welcome, that if some professor or user of this site MathOverflow wants to study/explore some of my problems posted in this site, that he/she can do it. Also I am especially interested in knowing how to improve the mathematical content of the problems that I propose, thus feel free to add comments. $\endgroup$ – user142929 Aug 14 at 19:45

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