We are given a sequence where the first term is 5. The rules for generating the sequence are as follows:

• If a term is even, divide it by 2.
• If a term is odd, multiply it by 3 and add 1.

The first three terms are provided: 5, 16, and 8. Let's generate more terms following these rules, and find the smallest $n$ such that the sum of the first $n$ terms is a four-digit number.

Generating the Sequence:

1. The first term is 5.
   • Since 5 is odd, we apply the rule $3 \times 5 + 1 = 16$.
2. The second term is 16.
   • Since 16 is even, we divide by 2 to get $16 / 2 = 8$.
3. The third term is 8.
   • Since 8 is even, divide by 2 to get $8 / 2 = 4$.
4. The fourth term is 4.
   • Since 4 is even, divide by 2 to get $4 / 2 = 2$.
5. The fifth term is 2.
   • Since 2 is even, divide by 2 to get $2 / 2 = 1$.
6. The sixth term is 1.
   • Since 1 is odd, apply the rule $3 \times 1 + 1 = 4$.

At this point, we see that the sequence starts to repeat (since we reached 4 again, which will generate the same terms we already saw).

Sequence:

$5, 16, 8, 4, 2, 1, 4, 2, 1, \dots$

This is a repeating cycle starting at the 4th term.

Calculating the Sum:

We now calculate the sum of the sequence and determine the smallest $n$ for which the sum is a four-digit number.

The sum of the first 6 terms: $5 + 16 + 8 + 4 + 2 + 1 = 36$

Since the sequence starts repeating after the 6th term, the next terms are $4, 2, 1$ repeating in a cycle. The sum of one cycle (terms 4, 2, and 1) is: $4 + 2 + 1 = 7$

Now, let's continue adding cycles to reach a sum of at least 1000.

Sum after $n = 6$ terms:

$36$

Adding the next cycle of $4 + 2 + 1$: $36 + 7 = 43 \quad (n = 9)$ Adding another cycle: $43 + 7 = 50 \quad (n = 12)$ Keep adding cycles until we reach a sum of 1000:

\text{Sum for n = 144 terms: } 36 + (7 \times terms= Total four digits} term ( cycle qusts during four Digits}.