---

Statement

With the Section 1 definitions ($m_n^r(a)$ the smallest sum of $r$ consecutive intervals cut by $a_1,\ldots,a_n$ on the circle of circumference $1$, $\lambda_r(a)=\liminf_n nm_n^r(a)$), for every sequence $a$ and every integer $r\ge1$:

$$\lambda_r(a)\ \le\ \frac{r}{r+1}\Big/\log\Bigl(1+\frac1r\Bigr)\ <\ r . \tag{4.3}$$

Taking the supremum over sequences, $\lambda_r\le\frac{r}{r+1}/\log(1+1/r)$, the bound the site quotes. For $r=1$ it gives $\lambda_1\le1/\log4$, attained by the Section 2 sequence. Footnote 2 (p. 15) credits van Aardenne-Ehrenfest with an independent discovery of the Section 4 proof.

Source. N. G. de Bruijn and P. Erdős, Sequences of points on a circle, Proc. 52 (1949), 14--17; Section 4 on printed pp. 15--16, the display (4.3) on p. 16 (PDF pp. 3--4 of the TU/e portal PDF), read on the page images. The edition read is identified in the source digest.

Read depth. Claims checked: the display and its hypotheses were read clause by clause on the page image; the proof was read for its structure and not checked.

Proof pointer

Section 4 proves the $r=1$ case and says the general case follows "similarly". For $r=1$: suppose $km_k^1(a)>\varrho$ for all $k$ with $n<k\le2n$, display (4.1). Label $a_1,\ldots,a_{2n}$ in circular order as $a_{k_1},\ldots,a_{k_{2n}}$ and set $k_i^*=\max(k_i,k_{i+1},n+1)$; the arc from $a_{k_i}$ to $a_{k_{i+1}}$ has both ends among $a_1,\ldots,a_{k_i^*}$ and no point of $a_1,\ldots,a_{2n}$ inside it, so it is an interval of stage $k_i^*$ and is longer than $\varrho/k_i^*$ (the page prints "less than", a slip: (4.1) bounds each stage-$k$ interval below by $\varrho/k$, and (4.2) sums those lower bounds), and summing over the $2n$ intervals gives $1>\varrho\sum_i1/k_i^*$ (4.2). Each $k$ in $(n+1,2n]$ occurs at most twice among the $k_i^*$, so $\sum_i1/k_i^*\ge\sum_{k=n+1}^{2n}2/k$. Hence for at least one $k$ in $(n,2n]$,

$$km_k^1(a)\ \le\ \tau_n:=\Bigl(\frac2{n+1}+\cdots+\frac2{2n}\Bigr)^{-1},$$

and $\tau_n>1/\log4$, $\tau_n\to1/\log4$, so $\lambda_1(a)\le1/\log4$. For general $r$ the same count, applied to $r$-spans and to $k$ with $rn<k\le(r+1)n$, gives a bound whose limit is $\frac{r}{r+1}/\log(1+1/r)$. Since $\log(1+1/r)>1/(r+1)$, the bound is below $r$. The argument, with the general-$r$ case written out and the printed slip recorded, is reconstructed (author-recorded, unreviewed) at its reconstruction page.

Mean-normalized form

An authored remark. Dividing by the average $r$-span $r/n$, and using $(r+1)\log(1+1/r)=1+\tfrac1{2r}-\tfrac1{6r^2}+O(r^{-3})$,

$$\frac{r}{r+1}\Big/\log\Bigl(1+\frac1r\Bigr)=r-\frac12+\frac5{12r}+O(r^{-2}),$$

so the bound reads $r-\lambda_r\ge\tfrac12-\tfrac5{12r}+O(r^{-2})$, that is, $r(1-\lambda_r/r)\ge\tfrac12+o(1)$. This is the form in which the second expression of the Section 6 conjecture is a nontrivial question. Read as the site words it, without the normalization, the second expression tends to $-\infty$: a sum of $r$ consecutive intervals is at least $r$ times the smallest interval, so $nm_n^r(a)\ge r\,nm_n^1(a)$ for every $n$, hence $\lambda_r(a)\ge r\lambda_1(a)$ and $\lambda_r\ge r\lambda_1=r/\log4>1$ for $r\ge2$ (an authored one-line remark, checked here).

Dependencies

None beyond the Section 1 definitions.

Bears on

• Problem 1221: the second of the three bounds the site quotes. The literal second expression $r(1-\lambda_r)$ of the problem tends to $-\infty$ rather than to $+\infty$ by the remark above, $\lambda_r\ge r\lambda_1=r/\log4$, not by (4.3), which bounds $\lambda_r$ only from above.