Five. (20 points) The three sides of a triangle are positive integers in an arithmetic sequence, and the longest side is no greater than the positive integer $n$. The number of such triangles is denoted as $f(n)$ (congruent triangles are counted as one). Write down $f(1), f(2), f(3), f(4)$, and then find the formula for $f(n)$.

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When $n=1,2,3$, there are only equilateral triangles with a common difference of 0, $f(1)=1, f(2)=2, f(3)=3$.

When $n=4$, in addition to the equilateral triangles with a common difference of 0 and side lengths of $1,2,3,4$, there are also integer-sided triangles with a common difference of 1, $(2,3,4)$, giving $f(4)=5$.

In general, the number of integer-sided triangles with side lengths not exceeding $n-1$ and forming an arithmetic sequence is $f(n-1)$. When the maximum side length increases to $n$, the newly added triangles have side lengths that can be set as $n-2m, n-m, n$, where $m (m \geqslant 0)$ is the common difference. By the triangle inequality, the sum of the two shorter sides must be greater than the longest side, so
$$
(n-2m) + (n-m) > n, \quad 3m < n.
$$

Since $m$ and $n$ are integers, we have $3m \leqslant n-1, \quad m \leqslant \left[\frac{n-1}{3}\right]$. Here $\left[\frac{n-1}{3}\right]$ denotes the greatest integer not exceeding $\frac{n-1}{3}$.
When $m$ takes all values from $0, 1, 2, \cdots, \left[\frac{n-1}{3}\right]$, each type of triangle with a common difference of $m$ adds one more triangle to the existing count. A total of $\left[\frac{n-1}{3}\right] + 1$ triangles are added, leading to the recurrence relation:
$$
\begin{array}{l}
f(n) = f(n-1) + \left[\frac{n-1}{3}\right] + 1 \\
= f(n-2) + \left[\frac{n-2}{3}\right] + 1 + \left[\frac{n-1}{3}\right] + 1 = \cdots \\
= f(1) + \left[\frac{1}{3}\right] + 1 + \left[\frac{2}{3}\right] + 1 + \cdots + \left[\frac{n-1}{3}\right] + 1 \\
= n + \left[\frac{1}{3}\right] + \left[\frac{2}{3}\right] + \cdots + \left[\frac{n-1}{3}\right] \\
= \left\{\begin{array}{ll}
n + 3[1 + 2 + \cdots + (k-1) + k, & n = 3k + 1 \text{ when, } \\
n + 3[1 + 2 + \cdots + (k-1) + k + k, & n = 3k + 2 \text{ when, } \\
n + 3(1 + 2 + \cdots + k), & n = 3k + 3 \text{ when }
\end{array}\right. \\
= \left\{\begin{array}{ll}
n + \frac{3k(k+1)}{2} - 2k, & n = 3k + 1 \text{ when, } \\
n + \frac{3k(k+1)}{2} - k, & n = 3k + 2 \text{ when, } \\
n + \frac{3k(k+1)}{2}, & n = 3k + 3 \text{ when }
\end{array}\right. \\
= \left\{\begin{array}{ll}
n + \frac{(n-1)(n-2)}{6}, & 3 \nmid n \text{ when, } \\
n + \frac{n(n-3)}{6}, & 3 \mid n \text{ when }
\end{array}\right. \\
= \left\{\begin{array}{ll}
\frac{(n+1)(n+2)}{6}, & 3 \nmid n \text{ when, } \\
\frac{n(n+3)}{6}, & 3 \mid n \text{ when. }
\end{array}\right.
\end{array}
$$

Analysis: Directly using
$$
\sum_{k=1}^{n-1} \left[\frac{k}{3}\right] = \left\{\begin{array}{ll}
\frac{(n-1)(n-2)}{6}, & 3 \nmid n \text{ when, } \\
\frac{n(n-3)}{6}, & 3 \mid n \text{ when. }
\end{array}\right.
$$

can save the intermediate steps.