可以证明由n个1与n个2构成的2n位数11.122...2等于由n个3构成的n位数与n-1个3和1个4构成的n位数的乘积.即
11.122...2=33...3*33..34
11.122...2=(1/9)(99..9)*10^n+(2/9)(99..9)
=(1/9)(10^n-1)*10^n+(2/9)(10^n-1)
=(1/9)(10^n-1)(10^n+2)
=(1/3)(10^n-1)*(1/3)(10^n-1+3)
=(1/3)(99..9)*(1/3)(99..9+3)
=33..33*33..34