# Arithmetic progressions that generate an infinite number of powers of 2?

For an arithmetic progression of the form $a_i = ki, i \in \mathbb{N}$, the question is trivial - if $k$ is a power of 2, then the progression will generate an infinite number of powers of 2, and no powers of 2 for any other case.

I'm having trouble figuring out the more general $a_0 \neq 0$ case. What form does the progression $a_0 + k\mathbb{N}$ have to take to generate an infinite number of powers of 2? I.e., if an arithmetic progression generates an infinite number of powers of 2, what will the relationship between $a_0$ and $k$ look like?

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Fix a common difference $d$. Take the $d$ arithmetic progressions $A_0,A_1,\ldots, A_{d-1}$ all with common difference $d$ with the first term for $A_i$ chosen as $i$. They partition the set of positive integers. As they are finite in number at least one of them should contain infinitely many powers of 2.
Wow, that was pretty straightforward. Is there any way to determine which sequences $A_i$ will contain infinitely many powers of 2 for a fixed difference $d$? – xjtian Apr 28 '14 at 0:43
If $i$ and $d$ have some common factor $k>2$, then every element the corresponding A.P. $A_i$ will be divisible by that $k$, and can never be a power of $2$. So choose $i$ such that $\gcd(i,d)=1$ or $2$. This cuts down the search space. – P Vanchinathan Apr 28 '14 at 2:36