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Statement

With the Section 1 definitions ($M_n^r(a)$ and $m_n^r(a)$ the largest and smallest sums of $r$ consecutive intervals cut by $a_1,\ldots,a_n$ on the circle of circumference $1$, $\mu_r(a)=\limsup_nM_n^r(a)/m_n^r(a)$ and $\mu_r$ its infimum over sequences), for every sequence $a$:

(5.1), p. 16. For all integers $r\ge1$ and $n\ge1$,

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

(5.7), p. 17. For every integer $r\ge1$,

$$\mu_r\ \ge\ 1+\frac1r .$$

For $r=1$ this is $\mu_1\ge2$, attained by the Section 2 sequence. The site quotes (5.7); the p. 14 introduction calls it all the authors can prove about $\mu_r$.

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 TU/e portal PDF), read on the page images. The edition read is identified in the source digest.

Read depth. Claims checked: both displays and their hypotheses were read clause by clause on the page images; the proofs were read for their structure and not checked.

Proof pointer

For (5.1) with $r>1$: let $I_{k_0}$ be the interval of the $n$-th stage into which $a_{n+1}$ falls, and take the $2r-1$ consecutive intervals $I_{k_{-r+1}},\ldots,I_{k_0},\ldots,I_{k_{r-1}}$ around it (5.2), with lengths $\beta_i$ (footnote 3: when $2r-1>n$ these are not all distinct); $a_{n+1}$ splits $I_{k_0}$ into parts $\gamma_1,\gamma_2$. Write $M=M_n^r(a)$, $m=m_{n+1}^r(a)$ and $M_1\le M$ for the largest $r$-span inside (5.2). Some $\beta_j$ with $j\ne0$ is at least $(M_1-\beta_0)/(r-1)$, and the $r$-span of the new stage that avoids $I_{k_j}$ but contains both parts of $I_{k_0}$ gives $m\le M_1-\beta_j\le\frac{r-2}{r-1}M_1+\frac{\beta_0}{r-1}$ (5.3); the two $r$-spans ending at $a_{n+1}$ from either side give $m\le M_1-\tfrac12\beta_0$ (5.4). If $\beta_0\le2M_1/(r+1)$, (5.3) gives $m\le\frac r{r+1}M_1\le\frac r{r+1}M$; if $\beta_0\ge2M_1/(r+1)$, (5.4) gives the same. For $r=1$, $m\le\min(\gamma_1,\gamma_2)\le\beta_0/2\le M/2$.

For (5.7): suppose $M_k^r(a)/m_k^r(a)<(1+1/r)/(1+1/k)^2$ for all $k$ with $nr\le k\le n(r+1)$ (5.5). Then (5.1) gives $m_{k+1}^r/m_k^r<k^2/(k+1)^2$ on that range, so $m_{rn+n}^r/m_{rn}^r<r^2/(r+1)^2$ (5.6). But $m_{rn}^r\le1/n$ trivially, while (5.5) at $k=rn+n$ with $M_{rn+n}^r\ge r/(rn+n-1)$ gives $m_{rn+n}^r>\frac r{1+r}\cdot\frac r{rn+n-1}\ge\frac{r^2}{(r+1)^2}\cdot\frac1n$, contradicting (5.6). So (5.5) fails for some $k$ in every such range, and $\mu_r(a)\ge1+1/r$ for every $a$. The argument is reconstructed (author-recorded, unreviewed), with the block counts made explicit, at its reconstruction page; the denominator $rn+n-1$ above is the note's printed chain, and the mean identity gives $M_{rn+n}^r\ge r/(rn+n)$, which suffices.

Dependencies

The trivial inequalities $nM_n^r(a)\ge r\ge nm_n^r(a)$ of Section 1.

Bears on

• Problem 1221: the third of the three bounds the site quotes, so $r(\mu_r-1)\ge1$ for every $r$; the third part of the problem asks whether this expression tends to infinity (the p. 14 introduction conjectures only that it is unbounded). The fixed-$r$ improvement $\mu_r\ge1+r/(r^2-1)$ for $r\ge2$, over sequences of distinct points, is Theorem 1.1 of Korsky's 2026 note (unrefereed), and the claimed growth $\mu_r-1\ge\log r/(100r)$ for large $r$, over sequences of distinct points, is Theorem 1.1 of Korsky's 2026 preprint (claimed, unreviewed).