Sum of divisors of Stirling numbers of the second kind

Question

In this post we denote the Stirling number of the second kind as ${n\brace k}$, I add as reference the article Stirling numbers of the second kind from the encyclopedia Wikipedia. And we denote the sum of divisors function as $\sigma(n)=\sum_{1\leq d\mid n}d$.

I wondered if it is possible to prove some of the following conjectures that I got using a Pari/GP program inspired in an article that refers [1].

Conjecture 1. The estimate $$\frac{1}{n}\sum_{1\leq k\leq n}\frac{\sigma\left({n\brace k}\right)}{{n\brace k}}>C\cdot\log\log\log n$$ holds for all $n\geq 16$, being $C>0$ a suitable constant.

Conjecture 2. One has that $$\frac{\sum_{1\leq k\leq n}\sigma\left({n\brace k}\right)}{n^{4+\frac{n}{2}}}\to\infty$$ as $n\to\infty$.

Question. Is it possible to prove Conjecture 1 or Conjecture 2? Many thanks.

I think that these conjectures aren't more interesting than studied in [1], but I've curiosity about if it can be deduced (from the few computational evidence that I have, my experiments tell me that these conjectures can aren't sharpest). I would like to know what work can be done to elucidate the veracity of our conjectures.

References:

[1] F. Luca, Sums of divisors of binomial coefficients, Int. Math. Res. Not. IMRN (2007), Art. ID rnm 101.

I know it from the introductory section of

Florian Luca and Juan Luis Varona, Multiperfect numbers on lines of the Pascal triangle, Journal of Number Theory, Volume 129, Issue 5 (2009).

Answer 1

Conjecture 2 seems to be true. If $n \geq 2$ then

$$\frac{1}{2}(k^2+k+2)k^{n-k-1}-1 \leq \left\{{n \atop k}\right\} \leq \frac{1}{2}{n \choose k} k^{n-k} < 2^n k^{n-k}. $$ (Inequalities from Here.)

For a lower bound for your sum we have $$\sum_{1\leq k\leq n}\sigma\left({n\brace k}\right) \geq \sum_{1\leq k\leq n} \frac{1}{2}(k^2+k+2)k^{n-k-1}-1 . $$ Let's try to estimate then $$T(n) = \sum_{1\leq k\leq n} (k^2+k+2)k^{n-k-1}.$$ Note that the function $f(n,x) = x^{n-x}$ has its maximum when $x \approx \frac{n}{\log n}.$ So we'll say that $T(x)$ is at least about $$\frac{n^2}{\log^2 n} \left(\frac{n}{\log n}\right)^{(n- \frac{n}{\log n}-1)} = \frac{n}{\log n} \left(\frac{n}{\log n}\right)^{(n- \frac{n}{\log n})} . $$ One can see by comparing logs that this last grows faster than $n^{c+\frac{n}{2}}$ for any $c$. Notice that we haven't used any properties of $\sigma$ here at all, just bounds on the size of Stirling numbers.

If one does want an idea of the upper bound, we can do a little with that. $\sigma(n) = O(n \log \log n). So, $$\sigma({n\brace k}) =O\left(2^n n^{n-k} \log (n \log 2 + (n-k) \log k) \right) = O(2^n n^{n-k} \log \log n ).$$

$$\sum_{1\leq k\leq n}\sigma\left({n\brace k}\right) = \sum_{1 \leq k \leq n} O(2^n n^{n-k} \log \log n ) = O\left((2^n \log \log n)\sum_{1 \leq k \leq n} n^{n-k}\right). $$

We have that $\sum_{1 \leq k \leq n} n^{n-k} = O(n^{n-1})$, so $$\sum_{1\leq k\leq n}\sigma\left({n\brace k}\right) = O(2^n n^n \log \log n).$$ Since we've engaged in some pretty heavy overestimating here (especially in regards to the binomial coefficients) this should be able to improved a fair bit.