14.已知数列{an}\{a_{n}\}{an}的前nnn项和为SnS_{n}Sn,a1=1a_{1}=1a1=1且nSn+1=(n+2)SnnS_{n+1}=(n+2)S_{n}nSn+1=(n+2)Sn.(1)求{an}\{a_{n}\}{an}的通项公式;(2)bkb_{k}bk为满足k⩽an⩽2kk\leqslant a_{n}\leqslant 2^{k}k⩽an⩽2k的ana_{n}an的个数,求使b1+b2+…+bk>2023b_{1}+b_{2}+\ldots +b_{k}>2023b1+b2+…+bk>2023成立的最小正整数kkk的值.