Connecting Euler Numbers of the Second Kind and Unsigned Stirling Numbers of the First Kind.

Question

The notation $ \left\langle\left\langle n\atop m\right\rangle\right\rangle$ denotes the Eulerian numbers of the second kind or order. Similarly, ${n \brack m}$ denotes the unsigned Stirling numbers of the first kind. The following relation between these two categories of numbers is widely known: that $${n \brack n-k}=\sum_{j=0}^{k-1}{\binom{n+j}{2k}\left\langle\left\langle k\atop j\right\rangle\right\rangle}$$ I am wondering if there is a way to rewrite the above summation so that the summation indices are in terms of $n$ and $k$. Something like this: $${n \brack n-k}=\sum_{j=k}^{n-1}{f(?, ?)\left\langle\left\langle ?\atop ?\right\rangle\right\rangle},$$ where $f(?, ?)$ is a function of of any of $n, k, j$, whichever they may be. Thank you.

Answer 1

Well, if we let $u=k+j$ (with $k\gt 0$), we have $${n \brack n-k}=\sum_{u=k}^{2k-1}{{n+u-k\choose 2k}\left\langle\left\langle k\atop u-k\right\rangle\right\rangle},$$ if $n \ge 2k$, the upper limit for the index in the sum can be set to $n-1$ because if $u\ge 2k$, then $\left\langle\left\langle k\atop u-k\right\rangle\right\rangle=0$

and if $n \lt 2k$, then we can suppose $j \ge 2k-n$, otherwise ${n+j\choose 2k}=0$ and $${n \brack n-k}=\sum_{j=2k-n}^{k-1}{{n+j\choose 2k}\left\langle\left\langle k\atop j\right\rangle\right\rangle},$$

we let $u= n-k+j $ so that $ k\le u\le n-1$ and $${n \brack n-k}=\sum_{u=k}^{n-1}{{u+k\choose 2k}\left\langle\left\langle k\atop u+k-n\right\rangle\right\rangle},$$

putting everything together we have

$${n \brack n-k}=\sum_{u=k}^{n-1}{{u+(n-k)\choose 2k}\left\langle\left\langle k\atop u-k\right\rangle\right\rangle}$$ or $${n \brack n-k}=\sum_{u=k}^{n-1}{{u+k\choose 2k}\left\langle\left\langle k\atop u-(n-k)\right\rangle\right\rangle},$$ depending on whether $n\ge 2k$ or $n \lt 2k$, respectively.