Let $σ$ be the sum-of-divisors function. Let $σ(n)/n$ be the abundancy index of $n$. Consider the density map $$f(x) = \lim_{N \to \infty} f_N(x) \ \ \text{ with } \ \ f_N(x) = \frac{1}{N} \#\{ 1 \le n \le N \ | \ \frac{\sigma(n)}{n} < x \}. $$ In this paper, Deléglise mentioned that Davenport proved that $f$ is continuous, and proved that $0.752 < f(2) < 0.7526$ (bounds improved by Kobayashi in his PhD thesis).

Let $\alpha = f^{-1}(1/2)$ be the *median abundancy index*, i.e. the number $\alpha$ such that the integers of abundancy index greater than $\alpha$ have natural density exactly $1/2$.

$$\begin{array}{c|c} N & f_N^{-1}(1/2) \newline \hline 1 &1.00000000000000 \newline \hline 10 &1.50000000000000 \newline \hline 10^2 &1.54838709677419 \newline \hline 10^3 &1.51485148514851 \newline \hline 10^4 &1.52707249923524 \newline \hline 10^5 &1.52501827363944 \newline \hline 10^6 &1.52384533012867 \newline \hline 10^7 &1.52381552194973 \newline \hline 10^8 &1.52381084043829 \end{array}$$

The above table *suggests* that $\alpha \simeq 1.52381$.

**Question 1**: What is known about the median abundancy index? Is it even mentioned somewhere? It is true that $|\alpha-1.52381|<10^{-5}$?

Let $(b_n)_{n \ge 1}$ be the sequence of integers such that for all $k<b_n$ then $$|\sigma(k)/k - \alpha| > |\sigma(b_n)/b_n - \alpha|.$$ This is the lexicographically first sequence of integers whose adundancy index *strictly* converge to the median adundancy index. Let us call this sequence the *buddhist sequence* in reference to the Middle Way in buddhism philosophy. Assuming that $|\alpha-1.52381|<10^{-5}$, here are the first terms of this sequence together with the distance of their adundancy index from $1.52381$: $$ \begin{array}{c|c}
n & b_n & |\sigma(b_n)/b_n -1.52381| \newline \hline
1 & 1 & 0.52381000000000 \newline \hline
2 & 2 & 0.02381000000000\newline \hline
3 & 21& 0.00000047619048\newline \hline
4? & 22099389? & 0.0000002693327?
\end{array} $$

Observe that $b_3=21$, $\sigma(21)/21 = 32/21$ and $|32/21-1.52381|<10^{-6}$, which is statistically unexpectable, as shown if we consider the variation $(b'_n)$ taking $22$ as initial term:

$$ \begin{array}{c|c} n & b'_n & |\sigma(b'_n)/b'_n -1.52381| \newline \hline 1&22& 0.112553636363636 \newline \hline 3&26& 0.0915746153846153 \newline \hline 4&27& 0.0423285185185187 \newline \hline 5&46& 0.0414073913043478 \newline \hline 6&58& 0.0279141379310344 \newline \hline 7&62& 0.0245770967741934 \newline \hline 8&74& 0.0167305405405405 \newline \hline 9&82& 0.0127753658536585 \newline \hline 10&86& 0.0110737209302325 \newline \hline 11&94& 0.00810489361702116 \newline \hline 12&106& 0.00449188679245283 \newline \hline 13&118& 0.00161372881355915 \newline \hline 14&122& 0.000780163934426037 \newline \hline 15&3249& 0.000659067405355485 \newline \hline 16&14337& 0.000478759154634911 \end{array} $$

So there is a very good chance that $\alpha = 32/21$. If so the buddhist sequence ends with its third term and $b_3=21$ should be called the *Buddha number*. If not, then we know that the set of abundancy indices is dense, so that the buddhist sequence must have a next term $b_4$, but $\sigma(b_3)/b_3$ is already too close to $\alpha$ compared to its above conjectured approximation, so we cannot conjecture the next term. A possible candidate for $b_4$ is mentioned in above table.

**Question 2**: Does the buddhist sequence end with its third term? If not what are the next terms?

Below are some additional computations with 10 samples of 100001 random integers between $10^{20}$ and $10^{21}$ suggesting that $\alpha = 32/21$ should be correct (sage lists are numbered from 0).

```
sage: import random
sage: for t in range(10):
....: L=[]
....: for i in range(100001):
....: b=random.randint(10**20,10**21)
....: q=sum(divisors(b))/b
....: L.append(q)
....: L.sort()
....: print((32/21-L[50000]).n())
-2.01727393333164e-8
0.00244355476044226
0.00201824866273585
-0.00130445314014877
-0.000322772616778371
0.00102756546533326
-6.74774915307343e-10
-1.48849650772673e-19
-0.0000572173485145812
-6.52303473965081e-20
```

*Observation*: One sample provides a median close to 32/21 with 20 digits, one with 19 digits, one with 10, one with 8, one with 5, one with 4 and four with 3.

How to explain such statistical irregularities?

A number with abundancy index greater (resp. less) than $2$ is called an abundant (resp. deficient) number, because the sum of its proper divisors (or aliquot sum) exceeds (resp. subceeds) itself. In the same flavour, a number with abundancy index greater (resp. less) than the median abundancy index $\alpha$ could be called an advantaged (resp. disadvantaged) number.

There is a Collatz-like problem (called Calatan-Dickson conjecture) related to the aliquot sum $s$ asking whether all the aliquot sequences $(s^{\circ r}(n))_{r \ge 0}$ are bounded. A value of $\alpha-1 \simeq 0.52381$ suggests *heuristically* a positive answer to this problem because $\alpha-1$ is the median for $s(n)/n$, although there are serious counter-example candidates like $n=276$ as $s^{\circ 100}(276)>10^{19}$. There are five such candidates less than $1000$ called the Lehmer Five (see this webpage dedicated to recent advances on the aliquot sequence).

The following picture displays $f_N$ for $N=10^7$ (which should be a good approximation of $f$, according to above table).

Observe that the function $f$ seems to make a *jump* around $\alpha$, whereas it is continuous, so it should be non-differentiable there; moreover the phenomenon happens around many other points (with a Cantor set or fractal flavour), which leads to:

**Question 3**: Is $f$ a Weierstrass function? What is the meaning of these jumps?

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