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Oct 6, 2018 at 11:05 vote accept Woett
Apr 13, 2017 at 12:57 history edited CommunityBot
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Mar 31, 2011 at 20:30 comment added GH from MO $z=\max(17,b)$ is fine, I was lazy typing and had to fit my message in a bounded box :-) Also I made no attempt to optimize any of my estimates. Perhaps I should add that the original problem naturally reduces to square-free numbers as the density of numbers with a square divisor exceeding $u^2$ is less than $1/u$ (think of $u$ as a large constant). Actually, this is how I was led to my solution.
Mar 31, 2011 at 19:53 comment added Woett Thanks a lot! Not that it matters much, but isn't it for large $b$ better (in the sense that, when $b$ is actually known, we get a better lower bound on the number of good integers) to choose $z$ to be $\max{17, b}$ instead of $> 5b$?
Mar 31, 2011 at 19:20 comment added GH from MO Here is how to deduce the statement for any residue class $a \pmod{b}$ using my construction. Choose $z > 5b$, then the construction yields a set of good integers $x < n \leq 2x$ coprime with $b$, which is of positive lower density. This set intersects at least one reduced residue class $c \pmod{b}$ with positive lower density. Multiply this intersection with the residue $1 \leq a\bar c \leq b$, where $\bar c$ stands for multiplicative inverse mod $b$. For $x > b^2$ the result is a set of good integers $x < n \leq 2bx$ in the residue class $a \pmod{b}$, which is of positive lower density.
Mar 31, 2011 at 17:51 comment added GH from MO Woett, I believe my proof can be easily adapted to residue classes, but I am lazy to check all the details.
Mar 31, 2011 at 17:37 history edited Woett CC BY-SA 2.5
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Mar 31, 2011 at 15:45 answer added GH from MO timeline score: 4
Mar 31, 2011 at 13:40 history asked Woett CC BY-SA 2.5