China-TST-A

Proof: First, we prove that $\{T_n\}_{n=k}^{\infty}$ is unbounded. If not, let $M = \max_{n \ge k}\{T_n\}$, then for any $n > 2M$, from $T_{n-1} \le M$ and $T_{n-2} \le M$, we have $T_n = T_{n-1} + T_{n-2}$. By the boundedness of $T_n$, the only possibility is $T_{n-1} = T_{n-2} = 0$, which inductively leads to $T_{k+1} = T_k = 0$, a contradiction!

Let $m_0$ be the number of digits of $k$. We further prove that for any $m > m_0$, there exists $n$ such that $10^{m-1} \le T_n < 10^m$. By the unboundedness of $T_n$, there exists a minimal $T_n \ge 10^{m-1}$, then $n > k+1$ and $T_{n-1}, T_{n-2} < 10^{m-1}$, thus $T_n \le T_{n-1} + T_{n-2} < 2 \cdot 10^{m-1} < 10^m$.

For convenience, denote $p \star q$ as the number obtained by concatenating positive integer $p$ with non-negative integer $q$. By contradiction, assume $0.T_k \star T_{k+1} \star \dots = 0.a_1 \dots a_s b_1 \dots b_t b_1 \dots b_t \dots$ (i.e., this repeating decimal has ultimate period length $t$), where $a_i, b_j \in \{0, 1, \dots, 9\}$. Take $n$ sufficiently large so that $T_n$ appears to the right of $a_s$, then for any $r \ge n$, $T_r \star T_{r+1} \star T_{r+2} \dots$ is a number with period $wt$, where $w$ is any positive integer. Take $m = \nu t$ with $\nu$ sufficiently large so that $10^{m-1} > T_i$ for all $i \le n$, then there exists a minimal $\ell$ such that $T_\ell$ has exactly $m$ digits, clearly $\ell > n$.

If $T_{\ell-1}$ has more than $m$ digits, then there must exist $j < \ell - 1$ where $T_j$ has $m-1$ digits, leading to some number before $T_\ell$ having $m$ digits, a contradiction!

Clearly $k \ne m$, so $k < m$. But $m$ is the period length, thus the last $k$ digits of $T_{\ell-1}$ and $T_\ell$ must be identical, which implies $T_{\ell-2}$ has at least $m$ digits. Therefore, there exists a number before $T_{\ell-1}$ with $m$ digits, another contradiction!

In conclusion, the assumption by contradiction is invalid, and the original proposition is proved. □