add Worpitzky's identity

Author: SuprDewd

comprog.pdf

@@ -843,7 +843,7 @@ \section{Other Algorithms}
         $\sum_{i=1}^n i^2 = n(n+1)(2n+1)/6$ & $\sum_{i=1}^n i^3 = n^2(n+1)^2/4$ \\
         $!n = n\times!(n-1)+(-1)^n$ & $!n = (n-1)(!(n-1)+!(n-2))$ \\
         $\sum_{i=1}^n \binom{n}{i} F_i = F_{2n}$ & $\sum_{i} \binom{n-i}{i} = F_{n+1}$ \\
-        $\sum_{k=0}^n \binom{k}{m} = \binom{n+1}{m+1}$ & $x^k = \sum_{i=0}^k i!\stirling{k}{i}\binom{x}{i}$ \\
+        $\sum_{k=0}^n \binom{k}{m} = \binom{n+1}{m+1}$ & $x^k = \sum_{i=0}^k i!\stirling{k}{i}\binom{x}{i} = \sum_{i=0}^k \left\langle {k \atop i} \right\rangle\binom{x+i}{k}$ \\
 
         $a\equiv b\pmod{x,y} \Rightarrow a\equiv b\pmod{\lcm(x,y)}$ & $\sum_{d|n} \phi(d) = n$ \\
         $ac\equiv bc\pmod{m} \Rightarrow a\equiv b\pmod{\frac{m}{\gcd(c,m)}}$ & $(\sum_{d|n} \sigma_0(d))^2 = \sum_{d|n} \sigma_0(d)^3$ \\