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Using the first 100,000 values of $\varphi(n)$ it seems that the following is true. Let $\mathcal A$ be a finite subset of $\mathbb{N}$, $\forall n\in \mathbb{N} \setminus \mathcal{A}$, $\displaystyle \frac{1}{\varphi(2n)} - \frac{1}{\varphi(2n+1)} \geqslant \frac{1}{n\ln n} frac{1}{2n\ln (2n)} $. Is this true? Is there a stronger lower bound? P.S.: I looked at Handbook of Number Theory I by Mitrinović and Sándor which has a lot of info about $\varphi (n)$ but it doesn't appear there. |
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Inequality with Euler's totientUsing the first 100,000 values of $\varphi(n)$ it seems that the following is true. Let $\mathcal A$ be a finite subset of $\mathbb{N}$, $\forall n\in \mathbb{N} \setminus \mathcal{A}$, $\displaystyle \frac{1}{\varphi(2n)} - \frac{1}{\varphi(2n+1)} \geqslant \frac{1}{n\ln n} $. Is this true? Is there a stronger lower bound? P.S.: I looked at Handbook of Number Theory I by Mitrinović and Sándor which has a lot of info about $\varphi (n)$ but it doesn't appear there.
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