An increasing arithmetic sequence with infinitely many terms is determined as follows.  A single die is thrown and the number that appears is taken as the first term. The die is thrown again and the second number that appears is taken as the common difference between each pair of consecutive terms.  Determine with proof how many of the 36 possible sequences formed in this way contain at least one perfect square.
1. **Understanding the Problem:**
   We need to determine how many of the 36 possible sequences formed by rolling a die twice contain at least one perfect square. The first roll determines the first term \(a\) of the arithmetic sequence, and the second roll determines the common difference \(d\).

2. **Analyzing the Common Differences:**
   We need to check for each possible common difference \(d\) (from 1 to 6) and see if there exists at least one perfect square in the sequence for each possible first term \(a\) (from 1 to 6).

3. **Case \(d = 1\):**
   - The sequence is \(a, a+1, a+2, \ldots\).
   - Since \(d = 1\), all integers are included in the sequence.
   - Therefore, all perfect squares are included for every value of \(a\).
   - Hence, all 6 sequences (one for each \(a\)) contain at least one perfect square.

4. **Case \(d = 2\):**
   - The sequence is \(a, a+2, a+4, \ldots\).
   - If \(a\) is even, then all even perfect squares are included.
   - If \(a\) is odd, then all odd perfect squares are included.
   - Therefore, all 6 sequences (one for each \(a\)) contain at least one perfect square.

5. **Case \(d = 3\):**
   - The sequence is \(a, a+3, a+6, \ldots\).
   - All perfect squares are \(\equiv 0, 1 \pmod{3}\).
   - Therefore, \(a\) must be \(0, 1, 3, 4 \pmod{3}\).
   - This corresponds to \(a = 1, 3, 4, 6\).
   - Hence, 4 sequences contain at least one perfect square.

6. **Case \(d = 4\):**
   - The sequence is \(a, a+4, a+8, \ldots\).
   - All perfect squares are \(\equiv 0, 1 \pmod{4}\).
   - Therefore, \(a\) must be \(0, 1 \pmod{4}\).
   - This corresponds to \(a = 1, 4, 5\).
   - Hence, 3 sequences contain at least one perfect square.

7. **Case \(d = 5\):**
   - The sequence is \(a, a+5, a+10, \ldots\).
   - All perfect squares are \(\equiv 0, 1, 4 \pmod{5}\).
   - Therefore, \(a\) must be \(0, 1, 4 \pmod{5}\).
   - This corresponds to \(a = 1, 4, 5, 6\).
   - Hence, 4 sequences contain at least one perfect square.

8. **Case \(d = 6\):**
   - The sequence is \(a, a+6, a+12, \ldots\).
   - As with \(d = 3\), all perfect squares are \(\equiv 0, 1 \pmod{3}\).
   - Therefore, \(a\) must be \(0, 1, 3, 4 \pmod{3}\).
   - This corresponds to \(a = 1, 3, 4, 6\).
   - Hence, 4 sequences contain at least one perfect square.

9. **Summing Up:**
   - For \(d = 1\): 6 sequences
   - For \(d = 2\): 6 sequences
   - For \(d = 3\): 4 sequences
   - For \(d = 4\): 3 sequences
   - For \(d = 5\): 4 sequences
   - For \(d = 6\): 4 sequences

   Total number of sequences containing at least one perfect square:
   \[
   6 + 6 + 4 + 3 + 4 + 4 = 27
   \]

The final answer is \(\boxed{27}\)