8. Let the integer sequence $a_{1}, a_{2}, \cdots, a_{10}$ satisfy $a_{10}=3 a_{1}, a_{2}+a_{8}=2 a_{5}$, and $a_{i+1} \in\left\{1+a_{i}, 2+\right.$ $\left.a_{i}\right\}, i=1,2, \cdots, 9$, then the number of such sequences is $\qquad$.
According to the problem, $a_{1}+9 \leqslant a_{10}=3 a_{1} \leqslant a_{1}+18 \Rightarrow 5 \leqslant a_{1} \leqslant 9$.
When $a_{1}=5, a_{10}=15$, if $a_{2}=6, a_{8}=12 \Rightarrow a_{5}=9$, then $a_{9} \in\{13,14\}$,
at this time, the sequence $\left\{a_{n}\right\}$ has 2;
if $a_{2}=7, a_{8}=13 \Rightarrow a_{5}=10$, at this time, the sequence $\left\{a_{n}\right\}$ has 1;
When $a_{1}=6, a_{10}=18$, if $a_{2}=7, a_{8}=15 \Rightarrow a_{5}=11$, then $a_{3}, a_{4} \in\{8,9,10\}$, $a_{6}, a_{7} \in\{12,13,14\}, a_{9} \in\{16,17\}$, at this time, the sequence $\left\{a_{n}\right\}$ has $C_{3}^{2} \cdot C_{3}^{2} \cdot 2=18$; if $a_{2}=8, a_{8}=16 \Rightarrow a_{5}=12$, then $a_{3}, a_{4} \in\{9,10,11\}, a_{6}, a_{7} \in\{13,14,15\}$, at this time, the sequence $\left\{a_{n}\right\}$ has $C_{3}^{2} \cdot C_{3}^{2}=9$;
if $a_{2}=8, a_{8}=14 \Rightarrow a_{5}=11$, at this time, the sequence $\left\{a_{n}\right\}$ has 1;
When $a_{1}=7, a_{10}=21$, if $a_{2}=8, a_{8}=18 \Rightarrow a_{5}=13$, then $a_{3}, a_{4} \in\{9,10,11,12\}$, $a_{6}, a_{7} \in\{14,15,16,17\}, a_{9} \in\{19,20\}$, at this time, the sequence $\left\{a_{n}\right\}$ has $3 \times 3 \times 2=18$; if $a_{2}=9, a_{8}=19 \Rightarrow a_{5}=14$, then $a_{3}, a_{4} \in\{10,11,12,13\}, a_{6}, a_{7} \in\{15,16,17,18\}$, at this time, the sequence $\left\{a_{n}\right\}$ has $3 \times 3=9$; if $a_{2}=9, a_{8}=17 \Rightarrow a_{5}=13$, then $a_{3}, a_{4} \in\{10,11,12\}, a_{6}, a_{7} \in\{14,15,16\}$, at this time, the sequence $\left\{a_{n}\right\}$ has $C_{3}^{2} \cdot C_{3}^{2}=9$; when $a_{1}=8, a_{10}=24$, if $a_{2}=9, a_{8}=21 \Rightarrow a_{5}=15$, then $a_{9} \in\{22,23\}$, at this time, the sequence $\left\{a_{n}\right\}$ has 2; if $a_{2}=10, a_{8}=22 \Rightarrow a_{5}=16$, at this time, the sequence $\left\{a_{n}\right\}$ has 1; if $a_{2}=10, a_{8}=20 \Rightarrow a_{5}=15$, then $a_{3}, a_{4} \in\{11,12,13,14\}, a_{6}, a_{7} \in\{16,17,18,19\}$, at this time, the sequence $\left\{a_{n}\right\}$ has $3 \times 3=9$; when $a_{1}=9, a_{10}=27$, at this time, the sequence $\left\{a_{n}\right\}$ has 1. In summary, the number of sequences $\left\{a_{n}\right\}$ is $3+28+36+12+1=80$.