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Statement

Theorem (p. 82, quoted). "There exists an absolute constant $c$ such that, if $n>ck$, and if $\frac{a_1}{b_1},\frac{a_2}{b_2},\ldots$ are the Farey fractions of order $n$, then $\frac{a_x}{b_x}$ and $\frac{a_{x+k}}{b_{x+k}}$ are similarly ordered."

The paper uses Mayer's term without defining it. In the corpus's gloss, two fractions are similarly ordered when their numerators and denominators do not move in opposite directions, $(a_y-a_x)(b_y-b_x)\ge0$; the proof's first sentence makes the negation explicit: a pair $a_x/b_x<a_y/b_y$ that fails to be similarly ordered has $a_y\ge a_x+1$ and $b_y\le b_x-1$. The paper does not name a value of $c$; its proof establishes the conclusion under $n>192k$ in Case I ($a_x/b_x<1/6$, p. 83, where the last step is printed "for $k\geqslant3$") and under $n>400k$ in Case II ($a_x/b_x\ge1/6$, p. 84). The proof does not treat $k\le2$ separately; for $k\ge3$ the printed thresholds give the theorem with $c=400$, which is the constant $1/400$ van Doorn's 2025 paper reads from it. In the notation of Problem 1005, where $f(n)$ is the largest integer such that every pair at index distance at most $f(n)$ is similarly ordered, the theorem gives $f(n)\ge n/c-1$ (every $k<n/c$ is covered). Erdős adds (p. 84): "I have not been able to find the best possible value for the constant $c$ in the above result."

Source. P. Erdős, A note on Farey series, Quart. J. Math. Oxford Ser. 14 (1943), 82--85; the Theorem on printed p. 82 (PDF p. 1 of the Rényi archive scan), the two thresholds on pp. 83--84 (PDF pp. 2--3), read on the rendered page images. The artifact is identified in the source digest.

Read depth. Claims checked: the statement, the reduction and the two thresholds were read clause by clause on the page images; the whole proof was read for structure and not checked step by step.

Proof pointer

Pp. 82--84. It suffices to show that at least $k$ Farey fractions of order $n$ lie between $a_x/b_x$ and $(a_x+1)/(b_x-1)$. Case I ($a_x/b_x<1/6$): the interval $(a_x/b_x,\ a_x/b_x+1/n)$ lies inside; with $a_x/b_x,\ldots,a_y/b_y$ its Farey fractions, consecutive differences $1/(b_jb_{j+1})<2/(n\min(b_j,b_{j+1}))$ give $\Sigma=\sum_{j=x}^{y}1/\min(b_j,b_{j+1})>1/2$ (display (1)); the part of $\Sigma$ over $j$ with $\min(b_j,b_{j+1})<8k$ is at most $64k/n<1/3$ once $n>192k$ (through $\sum1/b_{r_i}<32k/n$ for the small denominators), so the remaining part exceeds $1/6$, and since each of its terms is at most $1/(8k)$ there are more than $\tfrac43k$ of them: $y-x+1>k+1$ for $k\ge3$. Case II ($a_x/b_x\ge1/6$): the same argument on $(a_x/b_x,\ a_x/b_x+7/(6n))$, where at most one denominator $b_r\le5$ occurs and, if it does, every other $b_j>n/10>40k$ when $n>400k$; the sum over the remaining $j$ exceeds $1/20$ and each term is below $1/(40k)$, so $y-x+1>2k\ge k+1$. Not reconstructed here.

Dependencies

Elementary properties of consecutive Farey fractions ($b_j+b_{j+1}>n$; the difference $1/(b_jb_{j+1})$, hence at most $1/n$, and at most $1/(2(n-1))$ when neither denominator is $1$: the bounds the proof uses on pp. 82 and 84, printed there as "less than $\frac1n$" and "at most $1/2(n-1)$"); Mayer's observation on non-similarly-ordered pairs (p. 82). Self-contained otherwise.

Bears on

• Problem 1005: the linear lower bound $f(n)\gg n$ that the site credits to this note, with the constant $1/400$ that van Doorn's 2025 paper reads from the proof; van Doorn's paper gives the lower bound $f(n)\ge(\frac1{12}-o(1))n$ and Cipollini's 2026 preprint $f(n)\ge(\frac14-o(1))n$. The problem asks whether $f(n)=(c+o(1))n$ for a constant $c>0$; Erdős writes (p. 84) that he could not find the best possible value of the constant $c$ in his theorem.