Let $n$ be a positive integer. A sequence $(a, b, c)$ of $a, b, c \in \{1, 2, . . . , 2n\}$ is called [i]joke [/i] if its shortest term is odd and if only that smallest term, or no term, is repeated. For example, the sequences $(4, 5, 3)$ and $(3, 8, 3)$ are jokes, but $(3, 2, 7)$ and $(3, 8, 8)$ are not. Determine the number of joke sequences in terms of $n$.
1. **Identify the smallest term in the joke sequence:**
   The smallest term in a joke sequence can be expressed as \(2n - 2k + 1\), where \(k = 1, 2, \ldots, n\). This ensures that the smallest term is odd.

2. **Case 1: All terms are the same:**
   The sequence is \(\{2n - 2k + 1, 2n - 2k + 1, 2n - 2k + 1\}\).
   - There is only 1 possible joke sequence in this case.

3. **Case 2: Two terms are the same, and one term is different:**
   The sequence is a permutation of \(\{2n - 2k + 1, 2n - 2k + 1, a\}\), where \(a \in \{2n - 2k + 2, 2n - 2k + 3, \ldots, 2n\}\).
   - The number of choices for \(a\) is \(2n - (2n - 2k + 2) + 1 = 2k - 1\).
   - Since there are 3 permutations of \(\{2n - 2k + 1, 2n - 2k + 1, a\}\), the total number of joke sequences in this case is \(3(2k - 1) = 6k - 3\).

4. **Case 3: All terms are different:**
   The sequence is a permutation of \(\{2n - 2k + 1, a, b\}\), where \(2n - 2k + 2 \le a < b \le 2n\).
   - The number of ways to choose \(a\) and \(b\) is \(\binom{2k - 1}{2} = \frac{(2k - 1)(2k - 2)}{2} = k(2k - 1) - k = 2k^2 - 3k + 1\).
   - Since there are 6 permutations of \(\{2n - 2k + 1, a, b\}\), the total number of joke sequences in this case is \(6(2k^2 - 3k + 1) = 12k^2 - 18k + 6\).

5. **Sum the number of joke sequences for each \(k\):**
   - For each \(k\), the total number of joke sequences is:
     \[
     1 + (6k - 3) + (12k^2 - 18k + 6) = 12k^2 - 12k + 4
     \]
   - Summing this expression for \(k = 1\) to \(n\):
     \[
     \sum_{k=1}^n (12k^2 - 12k + 4)
     \]

6. **Simplify the summation:**
   - The summation can be broken down as follows:
     \[
     \sum_{k=1}^n (12k^2 - 12k + 4) = 12 \sum_{k=1}^n k^2 - 12 \sum_{k=1}^n k + 4 \sum_{k=1}^n 1
     \]
   - Using the formulas for the sums of squares and integers:
     \[
     \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^n k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^n 1 = n
     \]
   - Substituting these into the expression:
     \[
     12 \left(\frac{n(n+1)(2n+1)}{6}\right) - 12 \left(\frac{n(n+1)}{2}\right) + 4n
     \]
   - Simplifying each term:
     \[
     12 \left(\frac{n(n+1)(2n+1)}{6}\right) = 2n(n+1)(2n+1)
     \]
     \[
     12 \left(\frac{n(n+1)}{2}\right) = 6n(n+1)
     \]
     \[
     4n = 4n
     \]
   - Combining these:
     \[
     2n(n+1)(2n+1) - 6n(n+1) + 4n
     \]
   - Factoring out \(2n\):
     \[
     2n \left((n+1)(2n+1) - 3(n+1) + 2\right)
     \]
     \[
     2n \left(2n^2 + 3n + 1 - 3n - 3 + 2\right)
     \]
     \[
     2n \left(2n^2 - 2\right)
     \]
     \[
     4n^3 - 4n
     \]

The final answer is \(\boxed{4n^3}\)