13.正项等比数列{an}\{a_{n}\}{an}中,a2023=a2022+2a2021a_{2023}=a_{2022}+2a_{2021}a2023=a2022+2a2021,若aman=16a12a_{m}a_{n}=16{a}_{1}^{2}aman=16a12,则1m+4n\frac{1}{m}+\frac{4}{n}m1+n4的最小值等于((( )))