Arithmetic progression and most significant digits in different bases

Given a number $x \geq 3$, let $b(x) \in \{0,1\}$ be the second most significant digit (bit) of its binary representation, and $t(x)\in \{1,2\}$ the most significant digit of its ternary representation.

In other words: $x = 2^n + b(x) 2^{n-1} + ... = t(x) 3^m + ...$ for some $n,m \geq 1$

Let $A$ be the set of $K \geq 0$ such that there exist integers $x \geq 3, y \geq 1$ and

$$b(x + i y) + t(x + i y) = 1 \text{ for all } 0 \leq i \leq K$$

Informally the longest (integer) arithmetic progression that can be formed with numbers having $0$ as the second most significant digit of its binary representation and $1$ as the most significant digit of its ternary representation.

Is $A$ finite?
Do exist $x,y$ such that $b(x + i y) + t(x + i y) = 1 \text{ for all } i$?
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No , A is infinite, as there are infinitely many intervals of increasing length of numbers with such representation. Eventually the mth interval has more than 2^m many consecutive integers. – The Masked Avenger Mar 5 '14 at 1:14

Thanks, it confirms my idea (use $4 \cdot 3^{2m}+2i$); but I'm not an expert; do you have a suggestion about how to formally prove the second claim (there are not progressions of infinite length)? – Marzio De Biasi Mar 5 '14 at 13:18