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Jan 16, 2022 at 16:10 comment added Terry Tao @Wojowu Not that it is particularly helpful for this question, but primes of the form $2p+1$ where $p$ itself is prime are known as "safe primes". en.wikipedia.org/wiki/Safe_and_Sophie_Germain_primes
Jan 16, 2022 at 16:07 comment added Terry Tao For $n$ of the form $3^j \leq n \leq 4 \times 3^{j-1}$ for some $j \geq 2$ the sum is non-integral because $\varphi(k), k \leq n$ is divisible by $3^{j-1}$ only at $k=3^j$ and (possibly) at $k = 2 \times 3^{j-1}+1$ if the latter is prime, and in either case we can write $\sum_{k=1}^n 1/\varphi(k) = \frac{a}{b} + \frac{c}{2 \times 3^{j-1}}$ for some integers $a,b,c$ with $c \in \{1,2\}$ and $b$ not divisible by $3^{j-1}$. However this technique does not seem to extend well to much larger primes than $3$.
Jan 15, 2022 at 14:36 answer added Max Alekseyev timeline score: 7
Jan 15, 2022 at 11:01 comment added Wojowu There should be only finitely many such $n$. Assuming some quantitative version of Dickson's conjecture a version of Bertrand's postulate should hold for primes of the form $2p+1$, where $p$ itself is prime. We can then repeat the argument for harmonic series, noting that $1/\varphi(2p+1)$ would be the only term with $p$ in the denominator.
Jan 15, 2022 at 10:37 history asked annie CC BY-SA 4.0