16. Below is a "right-angled triangular array":
$\frac{1}{4}$,
$\frac{1}{2}, \quad \frac{1}{4}$
$\frac{3}{4}, \frac{3}{8}, \frac{3}{16}$
Each column of numbers forms an arithmetic sequence, and from the third row onwards, each row of numbers forms a geometric sequence with the same common ratio. Let the number in the $i$-th row and $j$-th column be $a_{i j}\left(i \geqslant j, i, j \in \mathbf{N}_{+}\right)$, then $a_{n 3}=$ $\qquad$
According to the problem, the common difference of each column $d=\frac{1}{4}$, the common ratio of each row $q=\frac{1}{2}$, then the first term of the $n$-th row $a_{n 1}=\frac{n}{4}$, so $a_{n 3}=\frac{n}{16}\left(n \in \mathbf{N}_{+}\right)$.