It has been proven that $$\limsup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n}=e^\gamma$$ where $\sigma(n)$ is the sum of divisors and $\gamma$ is Euler's constant ($0.5772…$). The way I understand it is that for every $\epsilon$, there is some $b$, so that when $n>b$, $(e^\gamma + \epsilon)n \ln\ln n>\sigma(n)$. Is there a way to find such smallest $b$ for every $\epsilon$ to make above inequality true (that is, for any smaller $b$, the inequality wouldn't hold)? If there is not a way to find the exact smallest $b$, is there at least a way to find a "tight" estimate/upper bound (any smaller estimate/upper bound wouldn't work) on such smallest $b$?
Question about Certain Upper Bounds of the Sum of Divisors Function
divisor-sumupper-lower-bounds
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Partial answer: Your function $C:\ \Bbb{N}\ \longrightarrow\ \Bbb{N}$ can also be described as follows: $$C(n)=\begin{cases}\varphi(n)&\text{ if $n$ is even}\\\sigma(n)&\text{ if $n$ is odd}\end{cases}.$$ Of course $\varphi(n)$ is even for every $n\geq3$, so for even $n$ we get $\varphi^{k}(n)\leq2$ for sufficiently large $k$. As $\varphi(2)=\varphi(1)=1$ this shows that every even number ends up at $1$.
Similarly $\sigma(n)$ is even for every odd $n$ that is not a perfect square, and so every odd number $n$ that is not a perfect square also ends up at $1$. So the question that remains is:
Is there an odd perfect square $n$ such that $\sigma^k(n)$ is an odd perfect square for all $k\in\Bbb{N}$?
This seems extremely unlikely, but I do not have a proof. The examples $$\sigma(1^2)=1^2\qquad\text{ and }\qquad \sigma(9^2)=11^2,$$ show that it is possible for the divisor sum of an odd perfect square to be an odd perfect square again. Perhaps one can show that the number of prime factors is decreasing in any such sequence $(\sigma^k(n))_{k\in\Bbb{N}}$? Or perhaps there is a useful result on the related aliquot sequences, which have been studied in more detail?
We have $$\dfrac{(2c + 2)p^k - cD(p^k)}{(c + 1)p^k + D(p^k)} < I(p^k) < \dfrac{(2d + 4)p^k - dD(p^k)}{(d + 2)p^k + D(p^k)}$$ and $$\dfrac{(2e + 2)m^2 - eD(m^2)}{(e + 1)m^2 + D(m^2)} < I(m^2) < \dfrac{(2f + 4)m^2 - fD(m^2)}{(f + 2)m^2 + D(m^2)}$$ where $c > 0,e > 0,d > -1$ and $f > -1$.
Therefore, we get $$\dfrac{(2c + 2)p^k - cD(p^k)}{(c + 1)p^k + D(p^k)}\cdot\dfrac{(2e + 2)m^2 - eD(m^2)}{(e + 1)m^2 + D(m^2)}\lt I(p^k)I(m^2)=2\lt \dfrac{(2d + 4)p^k - dD(p^k)}{(d + 2)p^k + D(p^k)}\cdot\dfrac{(2f + 4)m^2 - fD(m^2)}{(f + 2)m^2 + D(m^2)}$$
From the inequality on the left, we obtain $$\bigg((2c + 2)p^k - cD(p^k)\bigg)\bigg((2e + 2)m^2 - eD(m^2)\bigg)\lt 2\bigg((c + 1)p^k + D(p^k)\bigg)\bigg((e + 1)m^2 + D(m^2)\bigg)$$ i.e. $$\bigg(2p^k +c\sigma(p^k)\bigg)\bigg(2m^2 +e\sigma(m^2)\bigg)\lt 2\bigg((c+3)p^k -\sigma(p^k)\bigg)\bigg((e + 3)m^2-\sigma(m^2)\bigg)$$ i.e. $$\bigg(2(c+3)(e+3)-4+2(2-ec)\bigg)p^km^2=\bigg(2(c+3)(e+3)-4\bigg)p^km^2+\bigg(2-ec\bigg)\sigma(p^k)\sigma(m^2)\gt \bigg(2c+6+2e\bigg)p^k\sigma(m^2)+\bigg(2e+6+2c\bigg)m^2\sigma(p^k)$$ Dividing the both sides by $(2c+2e+6)p^km^2$ gives $$I(m^2)+I(p^k)\lt 3$$
From the inequality on the right $$2\lt \dfrac{(2d + 4)p^k - dD(p^k)}{(d + 2)p^k + D(p^k)}\cdot\dfrac{(2f + 4)m^2 - fD(m^2)}{(f + 2)m^2 + D(m^2)}$$ we obtain $$2\bigg((d + 2)p^k + D(p^k)\bigg)\bigg((f + 2)m^2 + D(m^2)\bigg)\lt \bigg((2d + 4)p^k - dD(p^k)\bigg)\bigg((2f + 4)m^2 - fD(m^2)\bigg)$$ i.e. $$2\bigg((d + 4)p^k -\sigma(p^k)\bigg)\bigg((f + 4)m^2 -\sigma(m^2)\bigg)\lt \bigg(4p^k +d\sigma(p^k)\bigg)\bigg(4m^2 +f\sigma(m^2)\bigg)$$ i.e. $$\bigg(2(d+4)(f+4)-16+2(2-df)\bigg)p^km^2\lt \bigg(2(d+4)+4f\bigg)p^k\sigma(m^2)+\bigg(2(f+4)+4d\bigg)m^2\sigma(p^k)$$ Dividing the both sides by $p^km^2$ gives $$4d+4f+10\lt (d+4+2f)I(m^2)+(f+4+2d)I(p^k)$$
Now, $d+4+2f=f+4+2d$ holds if and only if $f=d$ for which we have $$I(m^2)+I(p^k)\gt \frac{8d+10}{3d+4}$$ Let $f(d)=\dfrac{8d+10}{3d+4}$. Then, $f'(d)=\dfrac{2}{(3d+4)^2}\gt 0$, so we get $$I(m^2)+I(p^k)\gt \lim_{d\to\infty}f(d)=\color{red}{\frac 83}$$ which is larger than $\dfrac{18}{7}$, but is still smaller than $\dfrac{57}{20}$.
Best Answer
We have the following (unconditional) inequality due to Robin: $$\sigma(n)<e^\gamma n\log\log n+\frac{0.6483 n}{\log\log n}$$ for $n\geq 3$. Therefore, to have the inequality you ask it's enough to have $$e^\gamma n\log\log n+\frac{0.6483 n}{\log\log n}\leq(e^\gamma+\varepsilon)n\log\log n.$$ Elementary manipulations show that this is equivalent to $$n\geq e^{e^{\sqrt{0.6483/\varepsilon}}},$$ so we may take the right-hand side as the upper bound for $b$, provided it's at least $3$.
There are better inequalities known nowadays, for instance the following due to Axler (A new upper bound for the sum of divisors function) gives $$\sigma(n)<e^\gamma n\log\log n+\frac{0.1209 n}{(\log\log n)^2},$$ for $n>2520$, and we can similarly derive an upper bound for $b$: $$b\leq e^{e^{\sqrt[3]{0.1209/\varepsilon}}}$$ if the right-hand side is at least $2520$.
As for finding the exact value of $b$, this already shows that this is in principle algorithmically possible, since we just have to find the largest $n$ smaller than the given bound for a given $\varepsilon$.
If we admit conjectural bounds, by results of Robin under RH the inequality holds for $\varepsilon=0$ for $n>5040$, and therefore $b\leq 5040$ for all $\varepsilon$, and for all small enough values of $\varepsilon$ this is exact. Therefore under RH, $b=5040$ for small enough $\varepsilon$.