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 $b(x + i y) + t(x + i y) = 1 \text{ for all } i$?
> 



Arithmetic progression and most significant digits in different bases
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.


