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Let $\sqrt 7=\sum_{i=0}^\infty a_i 7^{-i}, 0\le a_i \le 6$ be the expansion of $\sqrt 7$ in the base $7$.

I am curious about the following question: Is there a $K\in \mathbb{N}$ such that for any $n\ge K,$ there exists an $ i\in [n, 2n] $ for which $ a_i\ne 6.$

Let $\sqrt 7=\sum_{i=0}^\infty a_i 7^{-i}, 0\le a_i \le 6$ be the expansion of $\sqrt 7$ in base $7$.

I am curious about the following question: Is there a $K\in \mathbb{N}$ such that for any $n\ge K$, there exists an $ i\in (n, 2n] $ for which $ a_i\ne 6$?

Let $\sqrt 7=\sum_{i=0}^\infty a_i 7^{-i}, 0\le a_i \le 6$ be the expansion of $\sqrt 7$ in the base $7$.

I am curious about the following result: is there a $K\in \mathbb{N}$ such that for any $n\ge K,$ there exists an $ i\in [n, 2n] $ for which $ a_i\ne 6.$

Let $\sqrt 7=\sum_{i=0}^\infty a_i 7^{-i}, 0\le a_i \le 6$ be the expansion of $\sqrt 7$ in the base $7$.

I am curious about the following question: Is there a $K\in \mathbb{N}$ such that for any $n\ge K,$ there exists an $ i\in [n, 2n] $ for which $ a_i\ne 6.$

Let $\sqrt 7=\sum_{i=0}^\infty a_i 7^{-i}, 0\le a_i \le 6$ be the expantion of $\sqrt 7$ in the base $7$.

I am curious about the following result: is there a $K\in \mathbb{N}$ such that for any $n\ge K,$ there exists an $ i\in [n, 2n] $ for which $ a_i\ne 6.$

Let $\sqrt 7=\sum_{i=0}^\infty a_i 7^{-i}, 0\le a_i \le 6$ be the expansion of $\sqrt 7$ in the base $7$.

I am curious about the following result: is there a $K\in \mathbb{N}$ such that for any $n\ge K,$ there exists an $ i\in [n, 2n] $ for which $ a_i\ne 6.$