The idea here is you want to have no carries. Thus, we want the placement of the $i$th nonzero digit to be at least twice the placement of the $i-1$ th. Now, we see that if $$a_{n} = \sum_{i=1}^{n-1} 10^{2^{i}}.$$Notice that $T(a_{n})=n,$ while $$T(a_{n}^{2}) = T(\sum_{i=2^{j}+2^{k}} 10^{i}) = 1 \cdot n^{2} = n^{2}.$$Thus, we are done.