Let $n\in N$, where $n = p\_{1}^{k\_{1}}p\_{2}^{k\_{2}}...p\_{m}^{k\_{m}}$ for $p\_{i}$ prime.


Define the 'density' of $n$ as: 


$d(n) = \frac{(p\_{1}+1)^{k\_{1}}(p\_{2}+1)^{k\_{2}}...(p\_{m}+1)^{k\_{m}}}{n}$


Notice that $d(n)$ gives us a measure of the 'compositeness' of a number - relative to other numbers of a similar size. Notice also that $n\_{1} \neq n\_{2} \implies d(n\_{1}) \neq d(n\_{2})$


Now extend the definition to the rational numbers so that for $r \in Q$, where $r=a/b$, and $a=p\_{1}^{k\_{1}}p\_{2}^{k\_{2}}...p\_{m}^{k\_{m}}$, $b=q\_{1}^{l\_{1}}q\_{2}^{l\_{2}}...q\_{n}^{l\_{n}}$


Define the density of $r$ as:


$d(r) = \frac{(p\_{1}+1)^{k\_{1}}(p\_{2}+1)^{k\_{2}}...(p\_{m}+1)^{k\_{m}}}{a}.\frac{b}{(q\_{1}+1)^{l\_{1}}(q\_{2}+1)^{l\_{2}}...(q\_{n}+1)^{l\_{n}}}$


Now order the rational numbers in an $n$ x $n$ grid (the same grid used to prove the countability of the rationals). Denote the average density of the first $n$ rationals in the grid by $D\_{n}$


Does $\lim\_{n\to\infty} D\_{n}$ exist, and if so what is it?


I have computed this for $n > 500,000$ and it turns out to be 1.61806, which is remarkably close to $\phi$. Is there a relationship between the density definition given above and the golden ratio?



Is there a connection between the average 'compositeness' of a rational number and $\phi$ (golden ratio)?
Probably not. I can tell you what the limiting value is when taking averages over $m\times m$ grids themselves, rather than diagonal-counting-sequences; but I suspect the averages are the same.


The average of $d(r)$ over the $m\times m$ grid is simply
$$
\frac1{m^2} \sum\_{a=1}^m \sum\_{b=1}^m d\big( \tfrac ab\big) = \bigg( \frac1m \sum\_{a=1}^m d(a) \bigg) \bigg( \frac1m \sum\_{b=1}^m \frac1{d(b)} \bigg).
$$
The first sum is a sum over a totally multiplicative function $d$ with the property that $d(p) = 1+\frac1p$. General results about sums of multiplicative functions that are close to $1$ on primes tell us that
$$
\frac1m \sum\_{a=1}^m d(a) \sim \prod\_p \bigg( 1 + \frac{d(p)}p + \frac{d(p^2)}{p^2} + \cdots \bigg) \bigg( 1-\frac1p \bigg) = \prod\_p \bigg( 1 - \frac{d(p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg)
$$
as $m\to\infty$ (where the products are over all primes $p$); in this case, we obtain
$$
\frac1m \sum\_{a=1}^m d(a) \sim \prod\_p \bigg( 1 - \frac{1+1/p}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) = \prod\_p \bigg( 1 + \frac1{p^2-p-1} \bigg) \approx 2.67411.
$$
Since $1/d$ is also totally multiplicative, the same procedure gives
\begin{align\*}
\frac1m \sum\_{b=1}^m \frac1{d(b)} &\sim \prod\_p \bigg( 1 - \frac{1/d(p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) \\
&= \prod\_p \bigg( 1 - \frac{1/(1+1/p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) = \prod\_p \bigg( 1 - \frac1{p^2} \bigg) = \frac6{\pi^2}.
\end{align\*}
Therefore the average in equation is
$$
= \frac6{\pi^2} \prod\_p \bigg( 1 + \frac1{p^2-p-1} \bigg) \approx \frac6{\pi^2} \cdot 2.67411 \approx 1.62567.
$$


