Konagin S.

Find $x_{1000}$, if $x_{1}=4, x_{2}=6$, and for any natural $n \geq 3$, $x_{n}$ is the smallest composite number greater than $2 x_{n-1}-x_{n-2}$.
We will prove by induction that $x_{n}=1 / 2 n(n+3)$.

Base. For $n=3,4$, the formula is correct: $2 x_{2}-x_{1}=8$, so $x_{3}=9$; $2 x_{3}-x_{2}=12$, so $x_{4}=14$.

Induction step. $2 x_{n}-x_{n-1}=2 \cdot 1 / 2 n(n+3)-1 / 2(n-1)(n+2)=1 / 2(n+1)(n+4)-1$. By the condition, $x_{n+1}$ is the first composite number greater than

$1 / 2(n+1)(n+4)-1$. But the number $1 / 2(n+1)(n+4)$ is composite. Indeed, if $n$ is odd, then $1 / 2(n+1)(n+4)=$ $(n+4) \cdot 1 / 2(n+1)$. Each of the factors is an integer greater than 2. Similarly, the case of even $n$ is considered. Thus, $x_{n+1}=1 / 2(n+1)(n+4)$.

Substituting $n=1000$, we get $x_{1000}=500 \cdot 1003$.

## Answer

501500.