Is there a connection between the average 'compositeness' of a rational number and $\phi$ (golden ratio)?

Question

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?

Answer 1

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 $$ = \frac6{\pi^2} \prod_p \bigg( 1 + \frac1{p^2-p-1} \bigg) \approx \frac6{\pi^2} \cdot 2.67411 \approx 1.62567. $$