(1)先将x1=109分成26+25+23+22+1从而得到1it-1it-2…i1i的值,然后根据x3=(1i1iit-1…i3i2)2进行求解即可;
(2)根据x1=22m+2+22m+1+22m+1则x1=(1i2m+1i2m…i1i)2,从而x2=(1ii2m+1i2m…i1)2,x3=(1i1ii2m+1i2m…i2)2,依此类推x2m+3=x1=(1i2m+1i2m…i1i)2,从而得到结论.
【解析】
(1)∵x1=109=26+25+23+22+1
∴x1=(1101101)2而x3=(1i1iit-1…i3i2)2=(1011011)2,
∴x3=26+24+23+21+1=91
(2)∵x1=22m+2+22m+1+22m+1
∴x1=(1i2m+1i2m…i1i)2而x2=(1ii2m+1i2m…i1)2,x3=(1i1ii2m+1i2m…i2)2,
当i跑到最后时移动了2m+2次,此时x2m+3=x1,
满足xn=x1(n∈N+且n≠1)的n的最小值为2m+3
故答案为:91、2m+3