4．数列{an}\{a_{n}\}{an}满足an+1={2an,0⩽an<12,2an−1,12⩽an<1,{a}_{n+1}=\left\{\begin{array}{l}2{a}_{n},0\leqslant {a}_{n}<\frac{1}{2},\\ 2{a}_{n}-1,\frac{1}{2}\leqslant {a}_{n}<1,\end{array}\right.an+1={2an,0⩽an<21,2an−1,21⩽an<1,若a1=25{a}_{1}=\frac{2}{5}a1=52，则a2023a_{2023}a2023等于((( )))