2. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-b\right|<\frac{1}{125}$ is what?
2. $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$, it can be known that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore, $\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250>3^{5}$, from which we can deduce that $n-1>5, n>6$, hence $n \geqslant 7$, so the smallest integer $n=7$.