China-TST-2025A

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 $0.T_k\star T_{k+1}\star \cdots =0.a_1\dots a_s{b}_1\dots b_t{b}_1\dots b_t\dots$ (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. □