13. Find the value of $10 \operatorname{ctg}(\operatorname{arcctg} 3+\operatorname{arcctg} 7$ $+\operatorname{arcctg} 13+\operatorname{arcctg} 21)$.
$\begin{array}{l}\text { Let } f(n)=1+n+n^{2}, \text { it is easy to know that } \\ f(1)=3, f(2)=7, f(3) \\ =13, f(4)=21. \\ \text { It is easy to prove that } \\ \operatorname{arcctg}\left(1+n+n^{2}\right) \\ =\operatorname{arctg}(n+1)-\operatorname{arctg} n. \\ \text { Therefore, } \operatorname{arcctg} 3=\operatorname{arctg} 2-\operatorname{arctg} 1, \\ \operatorname{arcctg} 7=\operatorname{arctg} 3-\operatorname{arctg} 2, \\ \operatorname{arcctg} 13=\operatorname{arctg} 4-\operatorname{arctg} 3, \\ \text { arctg } 21=\operatorname{arctg} 5-\operatorname{arctg} 4. \\ \text { Adding the four equations, and let } \alpha=\operatorname{arcctg} 3+\operatorname{arcctg} 7 \\ +\operatorname{arcctg} 13+\operatorname{arcctg} 21, \text { we get: } \\ \alpha=\operatorname{arctg} 5-\operatorname{arctg} 1. \\ \therefore \operatorname{tg} \alpha=\frac{5-1}{1+5 \cdot 1}=\frac{4}{6}, \\ \operatorname{ctg} \alpha=\frac{6}{4}=\frac{3}{2}. \\ \therefore \quad 10 \operatorname{ctg} \alpha=10 \operatorname{ctg}(\operatorname{arcctg} 3 \\ +\operatorname{arcctg} 7+\operatorname{arcctg} 13+\operatorname{arcctg} 21) \\ =15. \\\end{array}$