∵T(
k−1
5) −T(
k−2
5)成的数列为1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1…,k=1,2,3,4,5,…一一代入计算得数列{xn}为1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,…即{xn}是以5为周期的周期数列,从而有x5n+1=1,x5n+2=2,x5n+3=3,x5n+4=4,x5n=5.n∈N*.
数列{yn}为1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,…即yn的重复规律是y5n+k=n,0≤k<5.∴由题意可知第2011棵树种植点的坐标应为(1,202).
故答案为(1,202)