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## Triangle of the Stirling Numbers of the Second Kind

The Stirling numbers of the second 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

$\begin{array}{r|rrrrrrrrrr} n&\left\{\begin{array}{c}n\\0\end{array}\right\}&\left\{\begin{array}{c}n\\1\end{array}\right\}&\left\{\begin{array}{c}n\\2\end{array}\right\}&\left\{\begin{array}{c}n\\3\end{array}\right\}&\left\{\begin{array}{c}n\\4\end{array}\right\}&\left\{\begin{array}{c}n\\5\end{array}\right\}&\left\{\begin{array}{c}n\\6\end{array}\right\}&\left\{\begin{array}{c}n\\7\end{array}\right\}&\left\{\begin{array}{c}n\\8\end{array}\right\}&\left\{\begin{array}{c}n\\9\end{array}\right\}&\left\{\begin{array}{c}n\\10\end{array}\right\}\\ \hline 0&1\\ 1&&1\\ 2&&1&1\\ 3&&1&3&1\\ 4&&1&7&6&1\\ 5&&1&15&25&10&1\\ 6&&1&31&90&65&15&1\\ 7&&1&63&301&350&140&21&1\\ 8&&1&127&966&1701&1050&266&28&1\\ 9&&1&255&3025&7770&6951&2646&462&36&1\\ 10&&1&511&9330&34105&42525&22827&5880&750&45&1\\ \end{array}$

Note that empty entries in this table are actually $$0$$’s.

| | | | created: 2020-04-04 12:29:03 | modified: 2020-04-10 18:36:53 | by: bookofproofs | references: [8404]