22.数列{an}\{a_{n}\}{an}的前nnn项和为SnS_{n}Sn,且a1=1a_{1}=1a1=1,an+1−2an=n+1a_{n+1}-2a_{n}=n+1an+1−2an=n+1,则满足Sn>2048S_{n}>2048Sn>2048的最小的自然数nnn的值为 [ 10 ].