Improved bounds for the Mayer-Erd\H{o}s phenomenon on similarly ordered Farey fractions

TL;DR

This paper improves bounds on the Mayer-Erd"H{o}s phenomenon for Farey fractions, showing a larger range where the product of differences is non-negative, and identifying cases where it becomes negative.

Contribution

It establishes sharper bounds for the Mayer-Erd"H{o}s phenomenon on Farey fractions, extending previous results by Erd"H{o}s.

Findings

Proves the inequality $(a_l - a_k)(b_l - b_k) \\ge 0$ for l - k \\le (1/12 - o(1))n.

Shows existence of pairs with negative product for l < k + n/4 + 5.

Sharpens the understanding of the distribution of Farey fractions and their ordering properties.

Abstract

Let $\frac{a_1}{b_1}, \frac{a_2}{b_2}, \ldots$ be the Farey fractions of order $n$. We then prove that the inequality $(a_l - a_k)(b_l - b_k) \ge 0$ holds for all $k$ and $l > k$ with $l-k \le \left(\frac{1}{12} - o(1) \right)n$, sharpening an old result by Erd\H{o}s. On the other hand, we will show that for all $n \ge 4$ there are $k, l$ with $k < l < k + \frac{n}{4} + 5$ for which the product $(a_l - a_k)(b_l - b_k)$ is negative.

Equations

$$\frac{m}{2m+1},\frac{m+1}{2m+3},\dots ,\frac{2m-1}{4m-1},\frac{1}{2},\frac{2m}{4m-1}.$$

$$\frac{1}{2},\frac{2m+1}{4m+1},\frac{2m}{4m-1},\dots ,\frac{m+1}{2m+1},\frac{2m+1}{4m}.$$

$$\frac{15}{19},\frac{19}{24},\frac{23}{29},\frac{27}{34},\frac{31}{39},\frac{4}{5},\frac{29}{36},\frac{25}{31},\frac{21}{26},\frac{17}{21}.$$

$$\frac{p+ca}{q+cb},\frac{p+(c+1)a}{q+(c+1)b},\dots ,\frac{p+da}{q+db},\frac{a}{b},\frac{r+d^{\prime }a}{s+d^{\prime }b},\frac{r+(d^{\prime }-1)a}{s+(d^{\prime }-1)b},\dots ,\frac{r+c^{\prime }a}{s+c^{\prime }b}.$$

$$b_l$$

$$\ge n+1-\left(s+\frac{n-2s-b}{2}+b\right)$$

$$=\frac{n-b+2}{2}$$

$$>2b.$$

$$a_l$$

$$>\frac{2b{c}^{\prime }a}{2c^{\prime }b}$$

$$=a.$$

$$bX-aY$$

$$=1+1.$$

$$\min (d-c,d^{\prime }-c^{\prime })$$

$$\ge \min \left(\frac{n-q}{b}-\frac{n-2q-b-1}{2b},\frac{n-s}{b}-\frac{n-2s-b-1}{2b}\right)-2$$

$$=\frac{n+b+1}{2b}-2.$$

$$N\left(\alpha -\frac{1}{n}\right)\le A_n(\alpha )\le N\left(\alpha +\frac{1}{n}\right).$$

$$\sum _{j=1}^t\frac{1}{b_{i_j}}<\frac{x}{6}+\frac{1}{n^{1/3}}.$$

$$\frac{x}{n}+\frac{6}{n^{4/3}}$$

$$=\frac{6}{n^{4/3}}+\sum _{j=1}^{t-1}\left(\frac{a_{i_{j+1}}}{b_{i_{j+1}}}-\frac{a_{i_j}}{b_{i_j}}\right)$$

$$\ge \frac{6}{n^{4/3}}+\sum _{j=1}^{t-1}\frac{1}{b_{i_j}{b}_{i_{j+1}}}$$

$$\ge \frac{6}{n}\left(\max \left(\sum _{j=1}^{t-1}\frac{1}{b_{i_j}},\sum _{j=2}^t\frac{1}{b_{i_j}}\right)+\frac{1}{n^{1/3}}\right)$$

$$>\frac{6}{n}\sum _{j=1}^t\frac{1}{b_{i_j}},$$

$$\frac{x}{n}$$

$$=\sum _{i=k}^{l-1}\left(\frac{a_{i+1}}{b_{i+1}}-\frac{a_i}{b_i}\right)$$

$$=\sum _{i=k}^{l-1}\frac{1}{b_i{b}_{i+1}}$$

$$=\sum _{i\in S_1}\frac{1}{b_i{b}_{i+1}}+\sum _{i\in S_2}\frac{1}{b_i{b}_{i+1}}$$

$$\sum _{i\in S_2}\frac{1}{b_i{b}_{i+1}}<\frac{36(l-k)}{5n^2},$$

$$l-k>\frac{5n^2}{36}\sum _{i\in S_2}\frac{1}{b_i{b}_{i+1}}.$$

$$\sum _{i\in S_1}\frac{1}{b_i{b}_{i+1}}$$

$$\le \frac{12}{5n}\sum _{j=1}^t\frac{1}{b_{i_j}}$$

$$<\frac{12}{5n}\left(\frac{x}{6}+\frac{1}{n^{1/3}}\right)$$

$$=\frac{2x}{5n}-\frac{12}{5n^{4/3}}.$$

$$l-k$$

$$=\frac{5n^2}{36}\left(\frac{x}{n}-\sum _{i\in S_1}\frac{1}{b_i{b}_{i+1}}\right)$$

$$>\frac{5n^2}{36}\left(\frac{x}{n}-\frac{2x}{5n}-\frac{12}{5n^{4/3}}\right)$$

$$=\frac{nx}{12}-\frac{n^{2/3}}{3}$$

$$>\frac{n}{12}\left(1-\frac{4}{n^{1/3}}\right).\text{qed}$$

Full text

Improved bounds for the Mayer-Erdős phenomenon on similarly ordered Farey fractions

Wouter van Doorn

Abstract

Let $\frac{a_{1}}{b_{1}},\frac{a_{2}}{b_{2}},\ldots$ be the Farey fractions of order $n$. We then prove that the inequality $(a_{l}-a_{k})(b_{l}-b_{k})\geq 0$ holds for all $k$ and $l>k$ with $l-k\leq\left(\frac{1}{12}-o(1)\right)n$, sharpening an old result by Erdős. On the other hand, we will show that for all $n\geq 4$ there are $k,l$ with $k<l<k+\frac{n}{4}+5$ for which the product $(a_{l}-a_{k})(b_{l}-b_{k})$ is negative.

1 Introduction

If two fractions $\frac{a}{b}$ and $\frac{a^{\prime}}{b^{\prime}}$ are such that the product $(a^{\prime}-a)(b^{\prime}-b)$ is non-negative, then we say that $\frac{a}{b}$ and $\frac{a^{\prime}}{b^{\prime}}$ are similarly ordered. For example, $\frac{2}{5}$ and $\frac{3}{7}$ are similarly ordered, while $\frac{2}{5}$ and $\frac{3}{4}$ are not. With this definition in mind, let $\frac{a_{1}}{b_{1}},\frac{a_{2}}{b_{2}},\ldots$ be the Farey sequence of order $n\geq 4$ and let $f(n)$ be the largest integer such that $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are similarly ordered for all $k$ and $l$ with $|l-k|\leq f(n)$. The condition $n\geq 4$ here ensures that the Farey sequence of order $n$ actually contains fractions (e.g. $\frac{1}{4}$ and $\frac{2}{3}$) which are not similarly ordered, so that $f(n)$ is unambiguously defined.

In [1] Mayer proved $f(n)\geq 3$ for all $n\geq 5$, which he subsequently improved in [2] to $f(n)\rightarrow\infty$ if $n\rightarrow\infty$. This was further improved by Erdős in [3], where he showed $f(n)>cn$ for some suitable constant $c$. Moreover, his proof showed that one can take $c=\frac{1}{400}$. A generalization to arbitrary linear forms was then obtained by Zaharescu in [4] (with a constant $c=\frac{1}{480}$), after which Meng and Zaharescu generalized it even further in [5], to arbitrary linear forms in multiple variables.

Concerning the original problem, Erdős remarked in [3] that he was not able to find the optimal value of $c$. And as far as we are aware, in the better part of a century since, no improvements have occurred in the literature. In this paper we take another look at Erdős's proof, try to optimize its arguments, and find a better lower bound.

We start off by looking at upper bounds, however. We will prove that $f(n)\leq\frac{n}{4}+O(1)$ holds for all $n\geq 4$, and conjecture that this is optimal.

2 Upper bounds

Recall that $\frac{a_{1}}{b_{1}},\frac{a_{2}}{b_{2}},\ldots$ is the Farey sequence of order $n$, and, in order to upper bound $f(n)$, we aim to find $k$ and $l>k$ with $(a_{l}-a_{k})(b_{l}-b_{k})<0$ and $l-k$ as small as possible. We claim that such $k$ and $l$ exist with $l-k<\frac{n}{4}+5$.

Theorem 1**.**

For all $n\geq 4$ we have $f(n)\leq\left\lfloor\frac{n}{4}\right\rfloor+d$, with $d=1,2,2,4$, depending on whether $n\equiv 0,1,2,3\pmod{4}$.

To prove this theorem, we will use the following well-known property of consecutive Farey fractions.

Lemma 1**.**

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two reduced fractions with $0\leq\frac{a}{b}<\frac{c}{d}\leq 1$. Then they are consecutive fractions in the Farey sequence of order $n$ if, and only if, $bc-ad=1$ and $\max(b,d)\leq n<b+d$.

Proof of Theorem 1.

If $n=4m$ for some $m\in\mathbb{N}$, consider the fraction $\frac{a_{k}}{b_{k}}=\frac{2m-1}{4m}$. One can then check by Lemma 1 that the Farey sequence continues as follows:

[TABLE]

With $\frac{a_{l}}{b_{l}}$ equal to this final fraction, we notice that $\frac{a_{k}}{b_{k}}=\frac{2m-1}{4m}$ and $\frac{a_{l}}{b_{l}}=\frac{2m}{4m-1}$ are not similarly ordered. Since $l=k+m+2$, this shows $f(n)\leq m+1$.

If $n=4m+1$ or $n=4m+2$, consider $\frac{a_{k}}{b_{k}}=\frac{2m}{4m+1}$ instead. These are then the next Farey fractions:

[TABLE]

With $\frac{a_{l}}{b_{l}}=\frac{2m+1}{4m}$ we have $l=k+m+3$ and $(a_{l}-a_{k})(b_{l}-b_{k})<0$, so that $f(n)\leq m+2$.

Finally, for $n=4m+3$ we also take $\frac{a_{k}}{b_{k}}=\frac{2m}{4m+1}$ and $\frac{a_{l}}{b_{l}}=\frac{2m+1}{4m}$. In this case however, the two fractions $\frac{2m+1}{4m+3}$ and $\frac{2m+2}{4m+3}$ are contained in the sequence we just mentioned as well (right before and right after $\frac{1}{2}$ respectively). We therefore have $l=k+m+5$, implying $f(n)\leq m+4$. ∎

Based on computer calculations we tentatively believe Theorem 1 to be optimal for large enough $n$.

Conjecture**.**

For all $n\geq 4$ we have $f(n)>\frac{n}{4}$. More precisely, for all $n\geq 92$ we have the equality $f(n)=\left\lfloor\frac{n}{4}\right\rfloor+d$, with $d$ as in Theorem 1.

We have checked this conjecture for all $n\leq 5000$ and have not been able to find any counterexamples. In fact, the only positive integers $n$ with $4\leq n<92$ for which $f(n)$ is strictly smaller than the upper bound from Theorem 1 are $n=7,9,11,15,19,23,25,27,31,35,39,49,51,63,91$.

It is possible to strengthen the above conjecture in the following way: given any integer $d$, it seems plausible that for large enough $n$ one can actually classify all pairs of Farey fractions $(\frac{a_{k}}{b_{k}}$, $\frac{a_{l}}{b_{l}})$ with $l-k=\left\lfloor\frac{n}{4}\right\rfloor+d$ that are not similarly ordered. In particular, for every $d$ there should be an $e$ such that for all $n$ there are at most $e$ such pairs of fractions, with $e=0$ for $d\leq 0$ in particular. We leave the exact formulation (and proof) of such a stronger conjecture to the interested reader.

3 Lower bounds

To improve upon the lower bound $f(n)>\frac{n}{400}$ that was proven in [3], we will first show that, given any fraction with small denominator, there is a small interval around it that only contains similarly ordered Farey fractions. To give an idea of what such an interval looks like, let us consider the fraction $\frac{4}{5}$. These are then the Farey fractions of order $40$ around this fraction:

[TABLE]

One can notice that, to the left of $\frac{4}{5}$, both the numerators and the denominators form an increasing arithmetic progression (with common difference $4$ and $5$ respectively), whereas to the right of $\frac{4}{5}$ the numerators and denominators form decreasing arithmetic progressions. Such a result turns out to be true in general, which we will apply in the proof of our next lemma.

Lemma 2**.**

Let $\frac{a_{k}}{b_{k}}$, $\frac{a}{b}$ and $\frac{a_{l}}{b_{l}}$ be fractions in the Farey sequence of order $n$ with $\frac{a_{k}}{b_{k}}\leq\frac{a}{b}\leq\frac{a_{l}}{b_{l}}$. Then $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are similarly ordered if $l-k\leq\frac{n+b+1}{2b}$.

Proof.

If $b=1$ the result is trivial as it forces either $\frac{a_{k}}{b_{k}}=\frac{0}{1}$ or $\frac{a_{l}}{b_{l}}=\frac{1}{1}$ in which case $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are certainly similarly ordered, so without loss of generality we may assume $b\geq 2$. Moreover, if $\frac{n+b+1}{2b}<3$, then the lemma follows from Mayer's result in [1], so we may further assume $n\geq 5b-1$. Now, in the Farey sequence of order $b$, let $\frac{p}{q}$ and $\frac{r}{s}$ be the two fractions immediately to the left and right of $\frac{a}{b}$ respectively, and note that both $q$ and $s$ are smaller than $b$. Then, analogously to what we saw earlier in the case $\frac{a}{b}=\frac{4}{5}$, it follows from Lemma 1 that the segment of the Farey sequence of order $n$ around $\frac{a}{b}$ is as follows:

[TABLE]

Here, $c=\left\lfloor\frac{n-2q-b}{2b}\right\rfloor+1$, $c^{\prime}=\left\lfloor\frac{n-2s-b}{2b}\right\rfloor+1$, $d=\left\lfloor\frac{n-q}{b}\right\rfloor$, and $d^{\prime}=\left\lfloor\frac{n-s}{b}\right\rfloor$. The values of $c$ and $c^{\prime}$ ensure that any sum of two consecutive denominators is larger than $n$ (which is required by Lemma 1), while $d$ and $d^{\prime}$ are the largest values for which all denominators are smaller than or equal to $n$.

In order to prove Lemma 2, we now have three different cases to consider: either $\frac{a_{k}}{b_{k}}=\frac{a}{b}$, or $\frac{a_{l}}{b_{l}}=\frac{a}{b}$, or $\frac{a_{k}}{b_{k}}<\frac{a}{b}<\frac{a_{l}}{b_{l}}$. As for the first case, it is clear that $\frac{a_{k}}{b_{k}}=\frac{a}{b}$ and $\frac{a_{l}}{b_{l}}$ are similarly ordered if $\frac{a_{l}}{b_{l}}$ is one of the elements in the segment, as both $a_{l}>a$ and $b_{l}>b$. Moreover, if $\frac{a_{l}}{b_{l}}$ is the smallest Farey fraction larger than $\frac{r+c^{\prime}a}{s+c^{\prime}b}$, then we claim $b_{l}>2b$. Indeed, applying Lemma 1 and $n\geq 5b-1$,

[TABLE]

By the inequalities $s+c^{\prime}b<(c^{\prime}+1)b\leq 2c^{\prime}b$ and the fact that $\frac{r+c^{\prime}a}{s+c^{\prime}b}$ and $\frac{a_{l}}{b_{l}}$ are consecutive Farey fractions, we (once again by Lemma 1) then get

[TABLE]

Since both $a_{l}>a$ and $b_{l}>b$, we deduce that, even when $\frac{a_{l}}{b_{l}}>\frac{a}{b}$ is the smallest Farey fraction outside of the segment, $\frac{a}{b}$ and $\frac{a_{l}}{b_{l}}$ are still similarly ordered. We therefore conclude that $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are similarly ordered in this case if $l-k\leq d^{\prime}-c^{\prime}+2$ holds, so in particular whenever $l-k\leq\min(d-c,d^{\prime}-c^{\prime})+2$.

Analogously, if $\frac{a_{l}}{b_{l}}=\frac{a}{b}$, then $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are similarly ordered as well, as long as $l-k\leq\min(d-c,d^{\prime}-c^{\prime})+2$.

As for the third and final case, assume that $\frac{a_{k}}{b_{k}}=\frac{p+ea}{q+eb}$ and $\frac{a_{l}}{b_{l}}=\frac{r+e^{\prime}a}{s+e^{\prime}b}$ are two fractions contained in the segment, with $\frac{a_{k}}{b_{k}}<\frac{a}{b}<\frac{a_{l}}{b_{l}}$, $c\leq e\leq d$ and $c^{\prime}\leq e^{\prime}\leq d^{\prime}$. We then aim to prove that they are similarly ordered too. Define $X:=a_{l}-a_{k}=r+e^{\prime}a-p-ea$ and $Y:=b_{l}-b_{k}=s+e^{\prime}b-q-eb$. We then get

[TABLE]

Here, the second equality follows from the fact that $\frac{p}{q},\frac{a}{b}$ and $\frac{r}{s}$ were consecutive fractions in the Farey sequence of order $b$. Since $bX-aY=2$ with $a\geq 1$ and $b\geq 2$, this implies that $X$ and $Y$ cannot have opposite signs, which is what we wanted to show. So in this third case we conclude that $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are similarly ordered whenever $l-k\leq\min(d-c,d^{\prime}-c^{\prime})+2$ as well.

It therefore remains to calculate this latter quantity. By applying the aforementioned values of $c,c^{\prime},d,d^{\prime}$ we obtain

[TABLE]

We conclude that if $l-k\leq\frac{n+b+1}{2b}$, then $l-k\leq\min(d-c,d^{\prime}-c^{\prime})+2$, which in all three cases was sufficient to deduce that $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are similarly ordered. ∎

Note that, in light of the proof of Theorem 1, Lemma 2 is essentially optimal for $b=2$. Now, before we continue with the statement and proof of our main lower bound, we need two more preliminary lemmas, where we define $N$ to be the number of Farey fractions of order $n$.

Lemma 3**.**

For all positive integers $n$ we have $N>\frac{n^{2}}{4}$.

Proof (sketch)..

With a computer one can check the inequality for all $n<56$, so assume $n\geq 56$. With $\varphi(n)$ Euler's totient function, we have $N=1+\sum_{i\leq n}\varphi(i)$. By applying Möbius inversion to the identity $n=\sum_{d|n}\varphi(d)$ and rewriting the sum $\sum_{i\leq n}\varphi(i)$, we obtain $N=1+\frac{1}{2}\sum_{i\leq n}\mu(i)\left\lfloor\frac{n}{i}\right\rfloor\left(\left\lfloor\frac{n}{i}\right\rfloor+1\right)$. Since $\left\lfloor\frac{n}{i}\right\rfloor\left(\left\lfloor\frac{n}{i}\right\rfloor+1\right)>\frac{n^{2}}{i^{2}}-\frac{n}{i}$, $\sum_{i\geq 1}\frac{\mu(i)}{i^{2}}=\frac{6}{\pi^{2}}$ and $\sum_{i\leq n}\frac{1}{i}<\log(n)+1$, with some algebra one can deduce $N>\frac{3n^{2}}{\pi^{2}}-\frac{n}{2}\left(\log(n)+2\right)$ for all $n\geq 1$. Since the latter is larger than $\frac{n^{2}}{4}$ for $n\geq 56$, this finishes the proof. ∎

We will furthermore make use of the following tight result that was obtained by Dress in [6].

Lemma 4**.**

For $\alpha\in[0,1]$, let $A_{n}(\alpha)$ be the number of Farey fractions of order $n$ in the interval $(0,\alpha)$. For all $\alpha\in[0,1]$ and all $n\in\mathbb{N}$ we then have the bounds

[TABLE]

We are now ready to prove our main lower bound.

Theorem 2**.**

If $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}>\frac{a_{k}}{b_{k}}$ are two fractions in the Farey sequence of order $n$ with $l-k\leq\frac{n}{12}\left(1-\frac{4}{n^{1/3}}\right)$, then $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are similarly ordered.

Proof.

Taking the contrapositive, let us assume that $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are not similarly ordered. We then see $\frac{a_{l}}{b_{l}}\geq\frac{a_{k}+1}{b_{k}-1}>\frac{a_{k}+1}{b_{k}}\geq\frac{a_{k}}{b_{k}}+\frac{1}{n}$, so write $\frac{a_{l}}{b_{l}}-\frac{a_{k}}{b_{k}}=\frac{x}{n}$ for some $x>1$. We now aim to show $l-k>\frac{n}{12}\left(1-\frac{4}{n^{1/3}}\right)$, and by Lemma 2 we may assume $b_{i}>6$ for all $i$ with $k\leq i\leq l$. We may further assume $n\geq 4^{3}=64$, as otherwise our upper bound is negative and the statement is trivially true.

Let $S_{1}$ be the set of indices $i$ with $k\leq i\leq l-1$ and $\min(b_{1},b_{i+1})\leq\frac{n}{6}$, and let $S_{2}$ be those $i$ with $\min(b_{1},b_{i+1})>\frac{n}{6}$. Furthermore, let $i_{1},i_{2},\ldots,i_{t}$ be the actual indices for which $b_{i_{j}}\leq\frac{n}{6}$. With these definitions in mind, we can show that we may assume that at least one of $b_{i_{1}},b_{i_{t}}$ is larger than $n^{1/3}$.

Lemma 5**.**

If $n\geq 64$, $t\geq 2$, and $\max(b_{i_{1}},b_{i_{t}})\leq n^{1/3}$, then $l-k>\frac{n}{2}$.

Proof.

If $\max(b_{i_{1}},b_{i_{t}})\leq n^{1/3}$, then $\frac{a_{l}}{b_{l}}-\frac{a_{k}}{b_{k}}\geq\frac{a_{i_{t}}}{b_{i_{t}}}-\frac{a_{i_{1}}}{b_{i_{1}}}\geq\frac{1}{b_{i_{1}}b_{i_{t}}}\geq\frac{1}{n^{2/3}}$. Applying Lemma 4 with $\alpha=\frac{a_{k}}{b_{k}}$ and $\alpha=\frac{a_{k}}{b_{k}}+\frac{1}{n^{2/3}}$, and we obtain that there are at least $N\left(\frac{1}{n^{2/3}}-\frac{2}{n}\right)=\frac{N(n^{1/3}-2)}{n}$ Farey fractions in between $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$. Since $\frac{N(n^{1/3}-2)}{n}>\frac{n(n^{1/3}-2)}{4}$ by Lemma 3 and the latter is at least $\frac{n}{2}$ for $n\geq 64$, the proof is finished. ∎

With the help of Lemma 5 we can bound the sum of the reciprocals of the $b_{i_{j}}$.

Lemma 6**.**

We have the upper bound

[TABLE]

Proof.

If $t=1$, then we are done by the assumption $b_{i_{1}}>6$. If $t>1$, then

[TABLE]

where the final inequality uses Lemma 5. Multiplying both sides by $\frac{n}{6}$ gives the desired result. ∎

In the spirit of Erdős [3], we will now write $\frac{x}{n}$ as the sum of two sums.

[TABLE]

Applying $b_{i}+b_{i+1}>n$ for all $i$, we see that for the second sum (where $\min(b_{i},b_{i+1})>\frac{n}{6}$) we have $b_{i}b_{i+1}>\frac{n}{6}\frac{5n}{6}=\frac{5n^{2}}{36}$. This gives

[TABLE]

or

[TABLE]

As for the first sum we have $b_{i}b_{i+1}>\min(b_{i},b_{i+1})\frac{5n}{6}$, while every element in $S_{1}$ occurs at most twice as an $i$ with $\min(b_{i},b_{i+1})\leq\frac{n}{6}$. By furthermore applying Lemma 6 we then get

[TABLE]

We can now finish our proof as follows:

[TABLE]

4 A few final remarks

The proof of Theorem 2 more generally shows the following result on the local density of Farey fractions.

Theorem 3**.**

Let $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ be two Farey fractions of order $n$ with $\frac{a_{l}}{b_{l}}-\frac{a_{k}}{b_{k}}=\frac{x}{n}$ for some $x>0$. Then either there exists a Farey fraction $\frac{a}{b}$ with $b<\frac{6}{x}$ and $\frac{a_{k}}{b_{k}}\leq\frac{a}{b}\leq\frac{a_{l}}{b_{l}}$, or $l-k>nx\left(\frac{1}{12}-o(1)\right)$.

However, one can check that a direct application of Lemma 4 already improves upon this more general theorem for $x>2.76$, so its value seems to stem mostly from small values of $x$.

And on that note, for $\frac{a_{k}}{b_{k}}\geq\frac{1}{2}-o(1)$ we have $x\geq\frac{3}{2}-o(1)$ if $\frac{a_{k}}{b_{k}}$ and $\frac{a_{l}}{b_{l}}$ are not similarly ordered. In this case we get the improved lower bound $l-k>n\left(\frac{1}{8}-o(1)\right)$ which in turn is at most a factor $2$ off from optimal, by the proof of Theorem 1.