7. The integer sequence $\left\{a_{n}\right\}$, has $a_{1}=1, a_{2}=2, a_{n+2}=5 a_{n+1}+a_{n}$, then $\left[\frac{a_{2}}{a_{1}}\right]\left\{\left[\frac{a_{3}}{a_{2}}\right\}\left\{\frac{a_{4}}{a_{3}}\right\} \cdots\left\{\left[\left\{\frac{a_{2025}}{a_{2024}}\right\}\left[\frac{a_{2024}}{a_{2}}\right]\right.\right.\right.$ is

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7. The integer sequence $\left\{a_{n}\right\}$, has $a_{1}=1, a_{2}=2, a_{n+2}=5 a_{n+1}+a_{n}$, then $\left[\frac{a_{2}}{a_{1}}\right]\left\{\left[\frac{a_{3}}{a_{2}}\right\}\left\{\frac{a_{4}}{a_{3}}\right\} \cdots\left\{\left[\left\{\frac{a_{2025}}{a_{2024}}\right\}\left[\frac{a_{2024}}{a_{2}}\right]\right.\right.\right.$ is one
Answer: 1
$5 a_{n+1}<a_{n+2}=5 a_{n+1}+a_{n}<6 a_{n+1}$, so $\left\{\frac{a_{n+2}}{a_{n+1}}\right\}=\frac{a_{n+2}}{a_{n+1}}-5=\frac{a_{n}}{a_{n+1}}$
By the periodicity of $\left\{a_{n}\right\}$ modulo 2, we know that $a_{2024}$ is even.
From the above two points, the answer is 1.