The first four terms of an infinite sequence $S$ of decimal digits are $1$, $9$, $8$, $2$, and succeeding terms are given by the final digit in the sum of the four immediately preceding terms. Thus $S$ begins $1$, $9$, $8$, $2$, $0$, $9$, $9$, $0$, $8$, $6$, $3$, $7$, $4$, $\cdots$. Do the digits $3$, $0$,  $4$,  $4$ ever come up consecutively in $S$?
1. **Understanding the Sequence Generation**:
   The sequence \( S \) is defined such that each term after the first four is the final digit of the sum of the four immediately preceding terms. The first four terms are given as \( 1, 9, 8, 2 \).

2. **Generating the Sequence**:
   Let's generate the next few terms to understand the pattern:
   \[
   \begin{align*}
   S_5 &= (1 + 9 + 8 + 2) \mod 10 = 20 \mod 10 = 0, \\
   S_6 &= (9 + 8 + 2 + 0) \mod 10 = 19 \mod 10 = 9, \\
   S_7 &= (8 + 2 + 0 + 9) \mod 10 = 19 \mod 10 = 9, \\
   S_8 &= (2 + 0 + 9 + 9) \mod 10 = 20 \mod 10 = 0, \\
   S_9 &= (0 + 9 + 9 + 0) \mod 10 = 18 \mod 10 = 8, \\
   S_{10} &= (9 + 9 + 0 + 8) \mod 10 = 26 \mod 10 = 6, \\
   S_{11} &= (9 + 0 + 8 + 6) \mod 10 = 23 \mod 10 = 3, \\
   S_{12} &= (0 + 8 + 6 + 3) \mod 10 = 17 \mod 10 = 7, \\
   S_{13} &= (8 + 6 + 3 + 7) \mod 10 = 24 \mod 10 = 4, \\
   \end{align*}
   and so on.

3. **Checking for Repetition**:
   Since there are only \( 10^4 = 10000 \) possible combinations of four consecutive digits, the sequence must eventually repeat. This is due to the Pigeonhole Principle, which states that if more items are put into fewer containers than there are items, at least one container must contain more than one item.

4. **Finding the Specific Sequence**:
   We need to check if the sequence \( 3, 0, 4, 4 \) appears. To do this, we can work backwards from the given sequence \( 1, 9, 8, 2 \) to see if \( 3, 0, 4, 4 \) can be the preceding four terms.

5. **Calculating Backwards**:
   \[
   \begin{align*}
   x_4 + 1 + 9 + 8 &\equiv 2 \pmod{10} \implies x_4 + 18 \equiv 2 \pmod{10} \implies x_4 \equiv 4 \pmod{10}, \\
   x_3 + 4 + 1 + 9 &\equiv 8 \pmod{10} \implies x_3 + 14 \equiv 8 \pmod{10} \implies x_3 \equiv 4 \pmod{10}, \\
   x_2 + 4 + 4 + 1 &\equiv 9 \pmod{10} \implies x_2 + 9 \equiv 9 \pmod{10} \implies x_2 \equiv 0 \pmod{10}, \\
   x_1 + 0 + 4 + 4 &\equiv 1 \pmod{10} \implies x_1 + 8 \equiv 1 \pmod{10} \implies x_1 \equiv 3 \pmod{10}.
   \end{align*}
   \]
   Thus, the four terms preceding \( 1, 9, 8, 2 \) are \( 3, 0, 4, 4 \).

6. **Conclusion**:
   Since we have shown that \( 3, 0, 4, 4 \) precedes \( 1, 9, 8, 2 \) in the sequence, and the sequence is periodic, the digits \( 3, 0, 4, 4 \) do indeed appear consecutively in \( S \).

The final answer is \( \boxed{ 3, 0, 4, 4 } \).