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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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    $\begingroup$ 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. $\endgroup$
    – The Masked Avenger
    Commented Mar 5, 2014 at 1:14

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The problem is a little harder than it seems at first glance. Pick m large, say m > 8. There are 3k=3^m numbers with first ternary digit 1 and m other ternary digits. Suppose 0<= a < 3^m is smallest such that 3k +a= 3*2^l. Then 2^(l+2)=4k +4a/3, so there will be either at least k/2 numbers before 3k+a or after 4k+ 4a/3 which are part of the 3k numbers in ternary which begin 10 in binary. The claim in the comment above that A is infinite follows, while the claim about the mth interval does not, although there will be an interval of length > 2^m for every m.

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    $\begingroup$ Indeed k/2 can be replaced by something like 6k/7. In any case there will be at least 2^m consecutive numbers of the desired form for all m not too small. $\endgroup$
    – The Masked Avenger
    Commented Mar 5, 2014 at 4:07
  • $\begingroup$ 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)? $\endgroup$
    – Marzio De Biasi
    Commented Mar 5, 2014 at 13:18
  • $\begingroup$ Certainly. Given an arithmetic progression with first term a and common difference d, choose m so that 3^m > a+d. There will be k such that 2*3^m < a+kd < 3*3^m. Figure how to find k. $\endgroup$
    – The Masked Avenger
    Commented Mar 5, 2014 at 15:12

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