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Source. Y. Yu and K. Chen, Erdős Problem 354(i): Strong Completeness of Two Dyadic Floor Sequences, manuscript of 13 September 2026, Lemma 2.1 with its identity (2.1), physical p. 3, in the seventeen-page PDF held by its library source card, Yu and Chen (2026). The source gives the lemma three sentences; the proof below writes them out.

Standing. This is an author-recorded reconstruction. It is not an independent review, changes no status and assigns no tier.

Definitions

Let $d\ge1$ and let $X\subseteq\mathbb Z/d\mathbb Z$ be nonempty. A missing run of $X$ is a set $\{s,s+1,\ldots,s+r-1\}$ of $r\ge1$ consecutive residues, taken modulo $d$, none of which lies in $X$; it is maximal when $s-1\in X$ and $s+r\in X$. The quantity $h(X)$ is the largest length of a missing run of $X$, and $h(X)=0$ when $X=\mathbb Z/d\mathbb Z$. Because $X$ is nonempty, every missing run has length at most $d-1$, and every missing run is contained in a unique maximal one. For an integer $c$, $X+c=\{x+c:x\in X\}$.

Statement

Lemma 2.1. For every nonempty $X\subseteq\mathbb Z/d\mathbb Z$,

$$h\bigl(X\cup(X+1)\bigr)=\max\bigl(0,\,h(X)-1\bigr).$$

Proof

If $X$ is the whole circle, both sides are $0$. Assume $X$ is not full, so $h(X)\ge1$ and $X$ has at least one maximal missing run.

A residue $y$ is missing from $X\cup(X+1)$ exactly when $y\notin X$ and $y-1\notin X$: the complement of the union is $X^c\cap(X^c+1)$.

Let $\{s,\ldots,s+r-1\}$ be a maximal missing run of $X$, so that $s-1\in X$ and $s+r\in X$. A residue $y$ of this run is missing from the union exactly when $y-1\notin X$ as well, which fails for $y=s$ (since $s-1\in X$) and holds for $y=s+1,\ldots,s+r-1$ (since those $y-1$ lie in the run). Hence the run contributes the missing set $\{s+1,\ldots,s+r-1\}$, of length $r-1$, and contributes nothing when $r=1$.

Every residue missing from the union is missing from $X$, so it lies in one maximal missing run of $X$. Therefore the set of residues missing from the union is the disjoint union of the shortened sets $\{s+1,\ldots,s+r-1\}$ over the maximal runs of $X$. Two shortened sets never join into a longer run: between the last residue $s+r-1$ of one shortened set and the first residue $s'+1$ of the next lies the residue $s+r\in X$, which is not missing from the union. So the maximal missing runs of the union are exactly the shortened sets of length $r-1\ge1$, and their largest length is $h(X)-1$ when $h(X)\ge2$; when $h(X)=1$ every shortened set is empty and the union is full. This is the identity.