# Sum of divisors of Stirling numbers of the second kind

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).

• All, is there any idea to prove these conjectures? Do you think that it is possible to improve the statements of my conjectures? Sep 26, 2019 at 10:01

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.

• Many thanks for your great answer, I'm going to read it and study your nice computations. Since I was asking for some of those conjectures I should accept an answer as soon as I understand the computations and reasonings. And as soon I have enough reputation I'm going to upvote for your statement. Many thanks for your attention and patience. Apr 25, 2020 at 19:19