Task 4. (20 points) A finite increasing sequence of natural numbers $a_{1}, a_{2}, \ldots, a_{n}(n \geq 3)$ is given, and for all $\kappa \leq n-2$ the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$ holds. The sequence must contain $a_{k}=2022$. Determine the maximum number of three-digit numbers, divisible by 4, that this sequence can contain.
# Solution.

Since it is necessary to find the largest number of three-digit numbers that are multiples of 4, the deviation between the members should be minimal. Note that the arithmetic
progression with a difference of $d=2$, defined by the formula $a_{k}=2 k$, satisfies the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$.

Indeed,

$$
a_{k}=2 k, a_{k+1}=2 k+2, a_{k+2}=2 k+4, \text { or by the formula }
$$

$$
a_{k+2}=3 a_{k+1}-2 a_{k}-2=3(2 k+2)-2 \cdot 2 k-2=2 k+4
$$

The sequence containing 2022: 4, 6,

This finite sequence can contain all even three-digit numbers from 100 to 999. Among them, the numbers divisible by 4 are: 100, 104, 108, 112, ..., 992, 925, 950, 996 - 25 in each of the nine hundreds, i.e., 225 numbers.

Answer. 225.