This seems familiar

We change the notation (for reasons apparent later on) from $f(n,k)$ to $f(x,y)$ .

Make the following observations:

• Observation 1: $f(x,y+1) = \sum^x_{a_{y}=1}\left(\sum^{a_{y}}_{a_{y-1}} \ldots \right)$ $= \sum^x_{a_y=1} f(a_y, y)$ $f(x,y+1) = \sum^x_{i=1} f(i, y) \tag{1}$

• Observation 2: $f(x,1) = \sum^x_{a_0=1} a_0^2$ From $(1)$ , we also know that $f(x,1) = \sum^x_{i=1} f(i, 0)$ $\Rightarrow f(x,0) = x^2 \tag{2}$

From $(1)$ we also conclude that $f(x,y+1) = f(x-1,y+1) + f(x,y) \tag{3}$

We also extend the function $f(x,y)$ to $y=-1,-2,-3$ . This is not strictly necessary but it allows us to spot a pattern. Next, we construct a table using equations $(2)$ and $(3)$ for $1 \leq x \leq 3$ and $-3 \leq y \leq 3$ .

$1 \quad 7 \quad 27 \quad 77 \\ 1 \quad 6 \quad 20 \quad 50 \\ 1 \quad 5 \quad 14 \quad 30 \\ 1 \quad 4 \quad 9 \quad 16 \\ 1 \quad 3 \quad 5 \quad 7 \\ 1 \quad 2 \quad 2 \quad 2 \\ 1 \quad 1 \quad 0 \quad 0$

Does this seem familiar? Well, maybe not. What if we rotated it?

$1\\ 1\quad 1\\ 0\quad 2 \quad 1\\ 0\quad 2 \quad 3 \quad 1\\ 0\quad 2 \quad 5 \quad 4 \quad 1\\ 0 \quad 2 \quad 7 \quad 9 \quad 5 \quad 1\\ \\ 0 \quad 2 \quad 9 \quad 16 \quad 14 \quad 6 \quad 1\\ \cdots$

Compare this to Pascal's Triangle: $1\\ 1\quad 1\\ 1\quad 2 \quad 1\\ 1\quad 3 \quad 3 \quad 1\\ 1\quad 4 \quad 6 \quad 4 \quad 1\\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1\\ 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \\ \cdots$

The only difference is that there is a $0$ instead of a $1$ in the third row. We notice that by comparing $0,2,7,16, \ldots$ and $1,4,10,20, \ldots$ the differences are $1,2,3,4$ . In general, the $f(x,y)$ diagonal is the corresponding Pascal diagonal $-$ the second diagonal away from it.

The Pascal equivalent of columns and rows can be found in terms of $n = x+y+2, r = y+3$ .

Hence $f(x,y) = \binom{n}{r} - \binom{n-2}{r}$ $f(x,y) = \binom{x+y+2}{y+3}-\binom{x+y}{y+3}$

$f(5,2016) = \dfrac{2020\times 2021 \times 2022 \times 2023}{24} - \dfrac{2020 \times 2021}{2} \mod{2016}$ $= \boxed{193}$