Given a fixed $n\in \mathbb{Z}^+$, what is the value of the sum
$$f(n) = \sum_{i=0}^{n-2}~n \!\!\!\!\pmod{n - i}?$$
So far, I have only been able to get a lower bound using triangular numbers. Let $a = \lfloor\frac{n}{2}\rfloor.$ Then we have the equality
$$\sum_{i=0}^{a} n \!\!\!\pmod{n-i} = \sum_{i=0}^{a} i = T_{a}$$
So $T_{a}$ is a very rough lower bound, as it is a truncation of half the sum.
Consider generating functions, odd numbers will give a contribution of $\frac{x^3}{1-x^2}$ , modulo 3 will give $\frac{x^4+2x^5}{1-x^3}$ and so on ... we have
\begin{eqnarray*} \sum_{i=2}^{\infty} \frac{x^{i+1}}{(1-x^{i})} \sum_{j=1}^{i-1} jx^{j-1} \end{eqnarray*} Now invert the sums & replace $\frac{1}{(1-x^i)}$ by the geometric sum $\sum_{k=1}^{\infty} x^{ik}$ \begin{eqnarray*} \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} jx^{j-1} \sum_{i=j+1}^{\infty} x^{ik} \end{eqnarray*} The inner sum is geometric ... we have \begin{eqnarray*} \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} jx^{j-1} \frac{x^{(j+1)k}}{(1-x^{k})} \end{eqnarray*} Now perform the $j$ sum ... we have \begin{eqnarray*} \sum_{k=1}^{\infty} \frac{x^{2k+1}}{(1-x^{k})(1-x^{k+1})^2} \end{eqnarray*} So the value you want is the coefficient of $x^n$ in the sum above.