11. If the sequence $\left\{a_{n}\right\}$, satisfies $a_{1}=2, a_{n} a_{n+1}=2^{n}(n \geqslant 1)$, then the general term formula for $a_{n}$ is $\qquad$ .
11. $a_{n}=\left\{\begin{array}{l}2^{\frac{n}{2}-1}, n \text { is even; } \\ 2^{\frac{n+1}{2}}, n \text { is odd. }\end{array}\right.$

Given $a_{n} \cdot a_{n+1}=2^{n}$, and $a_{n-1} \cdot a_{n}=2^{n-1}$, dividing the two equations yields $\frac{a_{n+1}}{a_{n-1}}=2$, hence $\left\{a_{2 n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2, $a_{2 n}=2^{n-1} ;\left\{a_{2 n-1}\right\}$ is a geometric sequence with the first term 2 and common ratio 2, $a_{2 n-1}=2^{n}$.