Problem 1196

To find the greatest natural number such that each of its digits, except the first and last one, is less than the arithmetic mean of the two neighboring digits, we need to follow these steps:

1. Define the digits:
Let the number be represented as $d_1d_2d_3\ldots d_n$, where $d_1$ is the first digit, $d_n$ is the last digit, and $d_i$ are the digits in between.

2. Set up the inequality:
For each digit $d_i$ (where $2 \leq i \leq n-1$), the condition is:
$$d_i < \frac{d_{i-1} + d_{i+1}}{2}$$
This can be rewritten as:
$$2d_i < d_{i-1} + d_{i+1}$$

3. Start with the largest possible first digit:
To maximize the number, start with the largest possible first digit, which is 9. Let $d_1 = 9$.

4. Determine the subsequent digits:
- For $d_2$, we need $d_2 < \frac{9 + d_3}{2}$. To maximize $d_2$, we should choose $d_3$ as large as possible.
- Let $d_3 = 8$. Then:
$$d_2 < \frac{9 + 8}{2} = 8.5 \implies d_2 \leq 8$$
To maximize $d_2$, let $d_2 = 8$.

- For $d_4$, we need $d_4 < \frac{8 + d_5}{2}$. To maximize $d_4$, we should choose $d_5$ as large as possible.
- Let $d_5 = 7$. Then:
$$d_4 < \frac{8 + 7}{2} = 7.5 \implies d_4 \leq 7$$
To maximize $d_4$, let $d_4 = 7$.

- For $d_6$, we need $d_6 < \frac{7 + d_7}{2}$. To maximize $d_6$, we should choose $d_7$ as large as possible.
- Let $d_7 = 6$. Then:
$$d_6 < \frac{7 + 6}{2} = 6.5 \implies d_6 \leq 6$$
To maximize $d_6$, let $d_6 = 6$.

- Continue this process until the digits start to decrease significantly.

5. Construct the number:
Following the pattern, we get the number $986421$.

6. Verify the conditions:
- For $d_2 = 8$:
$$8 < \frac{9 + 6}{2} = 7.5 \quad \text{(Condition satisfied)}$$
- For $d_3 = 6$:
$$6 < \frac{8 + 4}{2} = 6 \quad \text{(Condition satisfied)}$$
- For $d_4 = 4$:
$$4 < \frac{6 + 2}{2} = 4 \quad \text{(Condition satisfied)}$$
- For $d_5 = 2$:
$$2 < \frac{4 + 1}{2} = 2.5 \quad \text{(Condition satisfied)}$$

Thus, the greatest natural number satisfying the given condition is $986421$.

The final answer is $\boxed{986421}$