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In this post we denote the Stirling numbers of the second kind as ${r\brace s}$ and we consider the proposal to ask if the equation of the title has infinitely many solutions $${n\brace a}{m\brace b}={k\brace c}\tag{1}$$ with the additional requirements $2\leq a$, $2\leq b$ and $2\leq c$, and $a+1<n$, $b+1<m$ and $c+1<k$ for the integers $m,n,k,a,b$ and $c$ in our equation.

The Stirling numbers of the second kind have a good mathematical content in combinatorics.

The additional requirements are due to make a more interesting question.

Example. A solution is $(n,a;m,b;k;c)=(7,2;6,4;13,2)$ since

$${7\brace 2}\cdot{6\brace 4}=63\cdot65=4095={13\brace 2}.$$ A table for some of first few solutions is given below.

Question. Has the equation $${n\brace a}{m\brace b}={k\brace c}$$ infinitely many solutions $(n,a;m,b;k;c)$ for positive integers $m,n,k,a,b$ and $c$ satisfying the conditions $2\leq a$, $2\leq b$, $2\leq c$, $a+1<n$, $b+1<m$ and $c+1<k$? Many thanks.

With a Pari/GP, this line of code that you can to run in the web Sage Cell Server, choose GP as Language and press Evaluate,

for(n=2, 20, for(m=2, 20, for(k=2,20, for(a=2,20, for(b=2, 20, for(c=2, 20, if((a+1<n)&&(b+1<m)&&(c+1<k)&&stirling(n,a,{flag=2})*stirling(m,b,{flag=2})==stirling(k,c,{flag=2}),print(n))))))))

one can to get examples of solutions denoted as $(n,m,k;a,b,c)$. The implementarion of the Stirling number of the second kind is from an user's guide to Pari/GP of the Université de Bordeaux.

The table of solutions that I've calculated and that thus I know is the following:

$$\begin{array} {|r|r|r|r|r|r|} \hline n&a&m&b&k&c \\ \hline 5&3&18&4&10&5\\ \hline 5&3&18&16&17&14 \\ \hline 6&4&7&2&13&2\\ \hline 7&2&6&4&13&2\\ \hline 8&4&5&3&10&5\\ \hline 18&16&5&3&17&14 \\ \hline \end{array}$$

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  • $\begingroup$ All, feel free to add comments about the mathematical content of my Question, and in particular add or remove more suitable tags. $\endgroup$
    – user142929
    Sep 10, 2019 at 9:00

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