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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Sep 25, 2015 at 2:42 vote accept Brad Graham
Sep 24, 2015 at 3:26 vote accept Brad Graham
Sep 25, 2015 at 2:42
Sep 24, 2015 at 3:17 comment added Gerhard Paseman If you choose x to be a larger primorial, you can get n (say n+6=x) not far from x to be squarefree, and you will run into the same problem again. I suggest you look through your proof and use these choices of x and n to find where the proof breaks. Gerhard "Always, Always Use Test Cases" Paseman, 2015.09.23
Sep 24, 2015 at 3:12 comment added Brad Graham Ps i have a feeling my results works for $x,n$ squarefree - whose example my result was extrapolated from. I need to think more carefully about non-squarefree numbers.
Sep 24, 2015 at 3:06 history edited Gerhard Paseman CC BY-SA 3.0
improve readability through shorter paragraphing
Sep 24, 2015 at 3:06 comment added Brad Graham For $x=210$ and $n=208$, $gcd(x,n)=2$, $x'=105$ and $n'=104$. So $V \in [0,e]$ where $e$ is very small. But we need $\Lambda(x,n)=47$ which the formula doesn't give. Hmmm. Ps typing from phone hence premature comment.
Sep 24, 2015 at 2:57 comment added Gerhard Paseman I made a mistake: $\frac{x'-n'}{x'n'}$ differs from the result I use by a factor of 2 for my choices of $x$ and $n$. However, the result when corrected still shows no integer in the range, and the claim is still false. Gerhard "Correct Up To Small Error" Paseman, 2015.09.23
Sep 24, 2015 at 2:52 history answered Gerhard Paseman CC BY-SA 3.0