most recent 30 from http://mathoverflow.net 2013-05-20T05:48:40Z http://mathoverflow.net/feeds/question/100187 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100187/is-it-true-that-the-sum-of-a-specific-floor-function-of-a-prime-1 Larry Freeman 2012-06-20T22:52:58Z 2012-06-20T23:08:57Z

<p>I noticed that for primes $p \le 109$, the following seems to be true:</p> <p>$\sum_{i | p\#}^{p\#} \lfloor{\frac{p}{i}\mu(i)}\rfloor = 1$</p> <p>where $\mu(i)$ is the Mobius function.</p> <p>For example: </p> <p>$\frac{2}{1} + \frac{2}{2}(-1) = 1$</p> <p>$\frac{3}{1} + \lfloor\frac{3}{2}(-1)\rfloor + \frac{3}{3}(-1) + \lfloor\frac{3}{6}\rfloor = 1$</p> <p>and so on...</p> <p>I verified this up to $p=109$ using a simple java application.</p> <p>I might be making a mistake in my code or my thinking. This seems like a very surprising result to me.</p> <p>Is it correct? If it is, does it stop being true for some prime? Could anyone help me to understand this result.</p> <p>Thanks very much,</p> <p>-Larry</p>

http://mathoverflow.net/questions/100187/is-it-true-that-the-sum-of-a-specific-floor-function-of-a-prime-1/100189#100189 David Cohen 2012-06-20T23:08:57Z 2012-06-20T23:08:57Z

<p>Assuming that $p\text{#}$ means the product over primes $\prod_{q\leq p}q$, then it is clear that $$\sum_{i\leq p} \mu(i)[\frac{p}{i}] = \sum_{i | p\text{#}} \mu(i)[\frac{p}{i}].$$</p> <p>But this formula is very well known: $$\sum_{d\leq n} \mu(d)[\frac{n}{d}]=\sum_{k\leq n}\sum_{d | k} \mu(d)=1+0+0+\ldots$$</p>