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Source. N. G. de Bruijn and P. Erdős, Sequences of points on a circle, Proc. 52 (1949), 14--17, Section 5 on printed pp. 16--17 (PDF pp. 4--5 of the retained scan, offprint pp. 5--6): displays (5.1)--(5.7) and footnote 3, read on the page images; held by its library card, de Bruijn and Erdős 1949, with the result page (5.1) and (5.7).

Standing. Author-recorded reconstruction; not an independent review; changes no status and assigns no tier. Section 5 is printed in full for general $r$; the reconstruction follows it, makes the block counts explicit, restricts (5.1) to the range where the note's intervals (5.2) are distinct, and records one printed slip.

Definitions

As on the Section 3 page: $a_1,\ldots,a_n$ cut the circle of circumference $1$ into the $n$ intervals of stage $n$; $M_n^r(a)$ and $m_n^r(a)$ are the largest and smallest sums of $r$ cyclically consecutive intervals of stage $n$, and

$$\mu_r(a)=\limsup_{n\to\infty}\frac{M_n^r(a)}{m_n^r(a)},\qquad \mu_r=\inf_a\mu_r(a).$$

The spans of stage $n$ sum to $r$, so $m_n^r(a)\le r/n\le M_n^r(a)$. When coincident points make $m_n^r(a)=0$ the ratio is read as $+\infty$; the one-step inequality is stated multiplicatively so that this case needs no separate treatment.

Statement

(5.1). For every sequence $a$, every integer $r\ge1$ and every $n\ge2r-1$,

$$M_n^r(a)\ \ge\ \Bigl(1+\frac1r\Bigr)m_{n+1}^r(a).$$

The note states (5.1) for all $n\ge1$; the case $n<2r-1$, where footnote 3 says the intervals in (5.2) are not all distinct, is not reconstructed here and is not needed below.

(5.7). For every sequence $a$ and every integer $r\ge1$, $\mu_r(a)\ge1+1/r$; hence $\mu_r\ge1+1/r$.

Proof of (5.1)

The case $r=1$. The point $a_{n+1}$ lies in an interval of stage $n$ of length $\beta_0\le M_n^1(a)$ and splits it into pieces $\gamma_1$, $\gamma_2$ with $\gamma_1+\gamma_2=\beta_0$. Both pieces are intervals of stage $n+1$, so $m_{n+1}^1(a)\le\min(\gamma_1,\gamma_2)\le\beta_0/2\le M_n^1(a)/2$.

Setting for $r\ge2$. Let $I_1,\ldots,I_n$ be the intervals of stage $n$ and let $I_{k_0}$ be the one containing $a_{n+1}$ (if $a_{n+1}$ coincides with an endpoint, take either adjacent interval; one piece then has length $0$). Write $\beta_0=|I_{k_0}|$, and let $\gamma_1,\gamma_2$ be the lengths of the two pieces into which $a_{n+1}$ cuts it, so $\gamma_1+\gamma_2=\beta_0$. Let

$$I_{k_{-r+1}},\ \ldots,\ I_{k_{-1}},\ I_{k_0},\ I_{k_1},\ \ldots,\ I_{k_{r-1}}$$

be the $2r-1$ consecutive intervals of stage $n$ centered at $I_{k_0}$, the note's (5.2), with lengths $\beta_j=|I_{k_j}|$; the hypothesis $n\ge2r-1$ makes them distinct. Put $M=M_n^r(a)$, $m=m_{n+1}^r(a)$, and let $M_1$ be the largest sum of $r$ consecutive intervals among these $2r-1$:

$$M_1=\max_{-r+1\le i\le0}\bigl(\beta_i+\beta_{i+1}+\cdots+\beta_{i+r-1}\bigr) \ \le\ M .$$

A long neighbor. Every block of $r$ consecutive intervals among the $2r-1$ has index range $\{i,\ldots,i+r-1\}$ with $-r+1\le i\le0$, which contains $0$; so every such block contains $I_{k_0}$. Take a block realizing $M_1$. Its $r-1$ intervals other than $I_{k_0}$ have total length $M_1-\beta_0$, so one of them, $I_{k_j}$ with $j\ne0$, satisfies

$$\beta_j\ \ge\ \frac{M_1-\beta_0}{r-1}.$$

Reflecting the circle exchanges $j$ with $-j$ and $\gamma_1$ with $\gamma_2$, so we may assume $1\le j\le r-1$.

Inequality (5.3). At stage $n+1$ the intervals $I_{k_{j-r+1}},\ldots,I_{k_{-1}}$ (there are $r-1-j$ of them), the two pieces of $I_{k_0}$, and $I_{k_1},\ldots,I_{k_{j-1}}$ ($j-1$ of them) are $(r-1-j)+2+(j-1)=r$ consecutive intervals. Their total length is $\beta_{j-r+1}+\cdots+\beta_{-1}+\gamma_1+\gamma_2+\beta_1+\cdots+\beta_{j-1}$, which is the block $I_{k_{j-r+1}},\ldots,I_{k_j}$ of $r$ consecutive intervals of stage $n$, of length at most $M_1$, minus $\beta_j$. Hence

$$m\ \le\ M_1-\beta_j\ \le\ M_1-\frac{M_1-\beta_0}{r-1} =\frac{r-2}{r-1}M_1+\frac{\beta_0}{r-1}.$$

Inequality (5.4). At stage $n+1$ the second piece of $I_{k_0}$ followed by $I_{k_1},\ldots,I_{k_{r-1}}$ is a block of $r$ consecutive intervals, of length $(\beta_0+\beta_1+\cdots+\beta_{r-1})-\gamma_1\le M_1-\gamma_1$; symmetrically $I_{k_{-r+1}},\ldots,I_{k_{-1}}$ followed by the first piece has length at most $M_1-\gamma_2$. So $m\le M_1-\gamma_1$ and $m\le M_1-\gamma_2$, and averaging,

$$m\ \le\ M_1-\tfrac12\beta_0 .$$

Conclusion. If $\beta_0\le2M_1/(r+1)$, then (5.3) gives

$$m\le\frac{r-2}{r-1}M_1+\frac{2M_1}{(r-1)(r+1)} =\frac{(r-2)(r+1)+2}{(r-1)(r+1)}M_1=\frac{r^2-r}{(r-1)(r+1)}M_1 =\frac r{r+1}M_1 .$$

If $\beta_0\ge2M_1/(r+1)$, then (5.4) gives $m\le M_1-M_1/(r+1)=\frac r{r+1}M_1$. In both cases $m\le\frac r{r+1}M_1\le\frac r{r+1}M$, which is (5.1).

Proof of (5.7)

Fix $r\ge1$ and an integer $n\ge2$, so that every $k\ge rn$ satisfies $k\ge2r-1$ and (5.1) applies at stage $k$. Suppose that for every $k$ with $rn\le k\le(r+1)n$,

$$\frac{M_k^r(a)}{m_k^r(a)}<\frac{1+1/r}{(1+1/k)^2},$$

the note's (5.5); in particular $m_k^r(a)>0$ on this range. For $rn\le k<(r+1)n$, (5.1) and (5.5) give

$$m_{k+1}^r(a)\le\frac{M_k^r(a)}{1+1/r}<\frac{m_k^r(a)}{(1+1/k)^2} =\frac{k^2}{(k+1)^2}m_k^r(a).$$

Multiplying these $n$ inequalities, the product telescopes:

$$\frac{m_{(r+1)n}^r(a)}{m_{rn}^r(a)}<\prod_{k=rn}^{(r+1)n-1}\frac{k^2}{(k+1)^2} =\frac{(rn)^2}{((r+1)n)^2}=\frac{r^2}{(r+1)^2},$$

the note's (5.6). The mean $r$-span at stage $rn$ is $r/(rn)=1/n$, so $m_{rn}^r(a)\le1/n$. At $k=(r+1)n$, (5.5) together with $(1+1/k)^2\ge1$ and the mean identity $M_k^r(a)\ge r/k$ gives

$$m_{(r+1)n}^r(a)>\frac r{r+1}M_{(r+1)n}^r(a)\ge\frac r{r+1}\cdot \frac r{(r+1)n}=\frac{r^2}{(r+1)^2}\cdot\frac1n\ \ge\ \frac{r^2}{(r+1)^2}m_{rn}^r(a),$$

contradicting (5.6). So for every $n\ge2$ some $k_n$ with $rn\le k_n\le(r+1)n$ violates (5.5): either $m_{k_n}^r(a)=0$, or

$$\frac{M_{k_n}^r(a)}{m_{k_n}^r(a)}\ \ge\ \frac{1+1/r}{(1+1/k_n)^2}.$$

Since $k_n\to\infty$, the ratio exceeds $1+1/r-o(1)$ along the subsequence $k_n$, and $\mu_r(a)=\limsup_kM_k^r(a)/m_k^r(a)\ge1+1/r$. Taking the infimum over $a$ gives (5.7).

Source notes

• A printed denominator. The p. 17 chain bounds $M_{rn+n}^r(a)$ below by $r/(rn+n-1)$; the mean identity at stage $rn+n$ gives $r/(rn+n)$, which is what the reconstruction uses, and it suffices because the strict inequality comes from (5.5).
• Small $n$ in (5.1). Footnote 3 says the $k_i$ in (5.2) are not all different when $2r-1>n$; the note gives no separate argument, and none is supplied here. The proof of (5.7) uses (5.1) only at stages $k\ge rn\ge2r-1$.
• Zero spans. The note's Section 1 allows coincident points, so a span can vanish; the multiplicative form of (5.1) and the convention $M/0=+\infty$ cover this. Over sequences of distinct points, the setting of the 2026 papers, all spans are positive.
• The bound is sharp for $r=1$: the Section 2 sequence has $\mu_1(a)=2$ (Section 2).

Reading addressed

The third expression of the conjecture, $r(\mu_r-1)$, needs no normalization and is the same under the literal and the mean-normalized readings; (5.7) gives $r(\mu_r-1)\ge1$ for every $r$, over all sequences, coincident points allowed. The fixed-$r$ improvement $\mu_r\ge1+r/(r^2-1)$ over sequences of distinct points is Korsky's 2026 note, and the claimed growth $\mu_r-1\ge\log r/(100r)$ is the ratio part of Korsky's 2026 preprint.