The Stirling numbers of the first kind $\left[\begin{array}{c}n\\r\end{array}\right],$ where $n$ and $r$ are natural numbers, are named after James Stirling (1692 - 1770). According to the corresponding recursive formula, they form a triangular scheme, in analogy to the Pascal's triangle for binomial coefficients. For the first $10$ values of $n$ this scheme is
\[\begin{array}{r|rrrrrrrrrr} n&\left[\begin{array}{c}n\\0\end{array}\right]&\left[\begin{array}{c}n\\1\end{array}\right]&\left[\begin{array}{c}n\\2\end{array}\right]&\left[\begin{array}{c}n\\3\end{array}\right]&\left[\begin{array}{c}n\\4\end{array}\right]&\left[\begin{array}{c}n\\5\end{array}\right]&\left[\begin{array}{c}n\\6\end{array}\right]&\left[\begin{array}{c}n\\7\end{array}\right]&\left[\begin{array}{c}n\\8\end{array}\right]&\left[\begin{array}{c}n\\9\end{array}\right]&\left[\begin{array}{c}n\\10\end{array}\right]\\ \hline 0&1\\ 1&&1\\ 2&&1&1\\ 3&&2&3&1\\ 4&&6&11&6&1\\ 5&&24&50&35&10&1\\ 6&&120&274&225&85&15&1\\ 7&&720&1764&1624&735&175&21&1\\ 8&&5040&13068&13132&6769&1960&322&28&1\\ 9&&40320&109584&118124&67284&22449&4536&546&36&1\\ 10&&362880&1026576&1172700&723680&269325&63273&9450&870&45&1\\ \end{array}\]
Note that empty entries in this table are actually \(0\)'s.
