If $\{x\}$ denotes the fractional part of $x$ and for $n\in \mathbb{N}$ we define the set $A_{n}=\left\{k \in \mathbb{N}: \{\frac{n}{k}\} \geq \frac{1}{2} \right\}$. Then how can I try to calculate the sum: $ S_{n}=\sum_{k \in A_{n}} \varphi(k)$ where $\varphi$ is the Euler's totient function. I've tried calculating the sum by hand for integers $1\leq n\leq 10$ and noticed that $S_{n} = n^{2}$ for these, but other than this, I have not found any other patterns.
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I dk. I suggest you try $S_n$ for some larger $n.$ E.g. $ A_{17}={2,3,9,10,11}$ and $S_{17}=1+2+6+4+10 =23.$ – DanielWainfleet Nov 05 '17 at 05:54
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Well if I am not wrong, $A_{17}$ = {2, 3, 6, 9, 10, 11, 18,...,34} – Rikka Nov 05 '17 at 06:00
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Sorry. I forgot that $k>n$ is allowed. Did you calculate $S_{17}$? – DanielWainfleet Nov 05 '17 at 06:03
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That is really a lot of calculation, I only tried until 10 and I believe the statement is true. – Rikka Nov 05 '17 at 06:05
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Well I had tried 1+2+2+6+4+10+6+18+8+12+10+22+8+20+12+18+12+28+8+30+16+20+16 = 289 = $17^{2}$ – Rikka Nov 05 '17 at 06:12
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Since hand calculation is exhausting, I typed some Mathematica code and found it to be true for $n \le 1000$. But a computer verification is hardly substantive enough for an answer. – silvascientist Nov 05 '17 at 06:17
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@MonstrousMoonshiner. True.But it's a good motivation for suspecting that there may be an elementary proof. – DanielWainfleet Nov 05 '17 at 06:28
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@DanielWainfleet True. But I have no idea what sort of proof strategy one might employ for this. – silvascientist Nov 05 '17 at 06:29
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With some more work I have been able to verify the conjecture in C++ for the first 10,000 cases. Pretty sure this thing is correct. – silvascientist Nov 05 '17 at 07:45