# On $\sigma(n)=\sigma \left( \left \lfloor{e^{H_n-\gamma}}\right \rfloor \right)$, for integers $n\geq 1$

Let for integers $n\geq 1$ the sum of divisors function $\sigma(n)=\sum_{d\mid n}d$, and we denote with $H_n=1+1/2+\ldots+1/n$ the nth harmonic number.

Euler's summation formula states that $$H_n=\log n+\gamma+O \left( \frac{1}{n} \right)$$ as $n$ tends to infinity, where $\gamma$ is the Euler-Mascheroni constant. Thus by exponentiation one has as $n\to\infty$ $$e^{H_n}=ne^\gamma e^{O(\frac{1}{n})}.$$ That is $$n \approx \frac{e^{H_n}}{e^{\gamma}}$$ holds as $n\to\infty$.

Let the floor function $\left \lfloor{x}\right \rfloor=x-\{x\}$, where thus $\{x\}$ is the fractional part.

Question. Is it possible to determine rigurously for which positive integers $m$ one can writes $$\sigma(m)=\sigma \left( \left \lfloor{e^{H_m-\gamma}}\right \rfloor \right)?$$ Thanks in advance.

I don't know if it is easy to answer previous question. I am interested in different exponentials that one can to evaluate for the sum of divisor function. A different example is $e^{\psi(m)}$, where $\psi(m):=\sum_{k\leq m}\Lambda(k)$ with $\Lambda(k)$ the von Mangoldt function thus $\psi(m)$ is the second Chebyshev function and $e^{\psi(m)}$ is integer for any integer $m\geq 1$, and has mathematical sense the particular value $f(e^{\psi(m)})$ for any arithmetical function $f$, thus also for the sum of divisor function.

• This seems to bring in in a rather artificial way the sum of divisors function $\sigma$ into the problem of determining when $$m\leqslant e^{H_m+\gamma}<m+1.$$ To solve this, note that, for every $m$, $$m< e^{H_m+\gamma}<me^{1/2m}$$ hence the inequality you are interested in holds as soon as $e^{1/2m}\leqslant (m+1)/m$, that is, for every $m\geqslant1$. – Did Aug 9 '16 at 8:50
• I am agree with you, very thanks much for your opinion and calculations @Did – user243301 Aug 9 '16 at 8:58