a) We call [i]admissible sequence[/i] a sequence of 4 even digits in which no digits appears more than two times. Find the number of admissible sequences.

b) For each integer $ n\geq 2$ we denote $ d_n$ the number of possibilities of completing with even digits an array with $ n$ rows and 4 columns, such that 

(1) any row is an admissible sequence; (2) the sequence 2, 0, 0, 8 appears exactly ones in the array.

Find the values of $ n$ for which the number $ \frac {d_{n\plus{}1}}{d_n}$ is an integer.
### Part (a)
We need to find the number of admissible sequences of 4 even digits where no digit appears more than twice. The set of even digits is \( A = \{0, 2, 4, 6, 8\} \).

1. **Sequences with all distinct digits:**
   - We choose 4 distinct digits from the set \( A \). The number of ways to choose 4 digits out of 5 is \( \binom{5}{4} = 5 \).
   - Each choice of 4 digits can be permuted in \( 4! = 24 \) ways.
   - Therefore, the total number of sequences with all distinct digits is:
     \[
     5 \times 24 = 120
     \]

2. **Sequences with exactly two digits appearing twice:**
   - Choose 2 digits from the set \( A \) to appear twice. The number of ways to choose 2 digits out of 5 is \( \binom{5}{2} = 10 \).
   - Arrange these 4 digits (two pairs) in \( \frac{4!}{2!2!} = 6 \) ways.
   - Therefore, the total number of sequences with exactly two digits appearing twice is:
     \[
     10 \times 6 = 60
     \]

3. **Sequences with one digit appearing twice and two other distinct digits:**
   - Choose 1 digit to appear twice and 2 other distinct digits from the remaining 4 digits. The number of ways to choose 1 digit out of 5 is \( \binom{5}{1} = 5 \).
   - The number of ways to choose 2 digits out of the remaining 4 is \( \binom{4}{2} = 6 \).
   - Arrange these 4 digits (one pair and two singles) in \( \frac{4!}{2!} = 12 \) ways.
   - Therefore, the total number of sequences with one digit appearing twice and two other distinct digits is:
     \[
     5 \times 6 \times 12 = 360
     \]

Adding all these cases together, the total number of admissible sequences is:
\[
120 + 60 + 360 = 540
\]

### Part (b)
We need to find the values of \( n \) for which \( \frac{d_{n+1}}{d_n} \) is an integer, where \( d_n \) is the number of ways to fill an \( n \times 4 \) array with admissible sequences such that the sequence \( 2, 0, 0, 8 \) appears exactly once.

1. **Fix the sequence \( 2, 0, 0, 8 \) in one of the \( n \) rows:**
   - There are \( n \) ways to choose the row for the sequence \( 2, 0, 0, 8 \).

2. **Fill the remaining \( n-1 \) rows with admissible sequences:**
   - Each of the remaining \( n-1 \) rows can be filled in \( 540 \) ways (from part (a)).

Therefore, the total number of ways to fill the array is:
\[
d_n = n \times 540^{n-1}
\]

3. **Calculate \( \frac{d_{n+1}}{d_n} \):**
   \[
   \frac{d_{n+1}}{d_n} = \frac{(n+1) \times 540^n}{n \times 540^{n-1}} = \frac{(n+1) \times 540^n}{n \times 540^{n-1}} = \frac{(n+1) \times 540}{n} = 540 + \frac{540}{n}
   \]

For \( \frac{d_{n+1}}{d_n} \) to be an integer, \( \frac{540}{n} \) must be an integer. Therefore, \( n \) must be a divisor of 540.

The divisors of 540 are:
\[
\{1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 45, 54, 60, 90, 108, 135, 180, 270, 540\}
\]

Since \( n \geq 2 \), the valid values of \( n \) are:
\[
\{2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 45, 54, 60, 90, 108, 135, 180, 270, 540\}
\]

The final answer is \( \boxed{ \{2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 45, 54, 60, 90, 108, 135, 180, 270, 540\} } \)