We write 1s at the two endpoints of a segment. Then, step by step, we bisect the resulting segments and write the sum of the numbers at the two ends of the segments to the midpoint. Repeating the bisection of the segments one million times, how many times will the number $1978$ appear among the written numbers?
Consider the numbers $a$ and $b$ (in this order) as neighbors if, during the process, a segment is formed where $a$ is at the left endpoint and $b$ is at the right endpoint. For example, according to Figure 1, 1 and 2, 3 and 2, and 3 and 5 are all neighbors. First, we will show that neighboring numbers are relatively prime. We do this using complete induction on their sum.

During the process, we write the sum of positive integers over each point of the segment (Figure 1). Thus, if $a$ and $b$ are neighbors and $a + b \leq 2$, then only $a = b = 1$ is possible, and these are indeed relatively prime.

![](https://cdn.mathpix.com/cropped/2024_05_02_eae9c7eb50e1d8f393a8g-1.jpg?height=881&width=1848&top_left_y=443&top_left_x=152)

Now, suppose that $a$ and $b$ are neighbors and appeared in the $i$-th step. This can only happen in two ways: either $a > b$ and in the $(i-1)$-th step, $a - b$ and $b$ were neighbors (Figure 2a), or $a < b$ and in the $(i-1)$-th step, $a$ and $b - a$ were neighbors (Figure 2b). If $a + b > 2$, then $a$ and $b$ are different, for example, $a > b$. In this case, $a - b$ and $b$ are relatively prime, and their sum $(a - b) + b < a + b$. By our induction hypothesis, $a - b$ and $b$ are neighbors, and during the process, the midpoint of the segment they define is assigned the sum $a$, so $a$ and $b$ are also neighbors. The case $a < b$ can be handled similarly.

The number 1978 appears as many times as there are ways to express it as the sum of relatively prime numbers (of course, $a + b$ and $b + a$ are considered different representations). This is equal to the number of integers less than 1978 that are relatively prime to it, which is 924. After the millionth division, only numbers greater than a million can be placed at the endpoints of the segment, so all 924 pairs will appear among the previously written numbers.

Therefore, the number 1978 will appear 924 times among the written numbers.