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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, Section 11 "The bounded-spacing contradiction (BG)" with Subsections 11.1--11.3 and display (11.1), physical pp. 13--14, in the seventeen-page PDF held by its library source card, Yu and Chen (2026).

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

Definitions

The normalized pair, $a_i,b_i$, $(u_i,v_i)$, the event set and $K_n$ are as on the normalization page; $\theta$, $H$, $\delta_i=qa_i-pb_i$ and the constant $L$ as on the windows page. A layer $i$ is exact (for the rational $p/q$) if $\delta_i=0$. The sequence has bounded event spacing if there are an integer $R\ge2$ and a threshold $n_0$ such that every integer $n\ge n_0$ has an event in $(n,Rn]$. For a positive integer $k$ put

$$\mathcal A=\{0\}\cup\{2^{-j}:j\in\mathbb N\},\qquad \mathcal S_k=\underbrace{\mathcal A+\cdots+\mathcal A}_{k},\qquad \mathcal C_k=\{x/y:x,y\in\mathcal S_k,\ y\ge1/2\}.$$

Statement

(BG). Let the normalized pair have irrational $\theta$. If the sequence is incomplete, it does not have bounded event spacing.

Proof

Assume incompleteness and bounded spacing with $R$ and $n_0$. All windows below come from (10.2) on the windows page, which is available under these hypotheses.

Step 1: exact layers are dense (Subsection 11.1)

Fix a window $T$, $p/q$, $H$ from (10.2). If the conversions at indices $i,\ldots,i+H-1$ are all zero, then $a_{i+H}=2^Ha_i$ and $b_{i+H}=2^Hb_i$, so $\delta_{i+H}=2^H\delta_i$. If moreover $i+H\le T$, then $|\delta_{i+H}|<2^H$ forces $|\delta_i|<1$, hence $\delta_i=0$ by integrality. Consequently every nonexact layer $i\le T-H$ has a nonzero conversion at some index in $[i,i+H-1]$, that is, an event at a position in $[i+1,i+H]\subseteq[1,T]$. A given event position is in $[i+1,i+H]$ for at most $H$ layers $i$, and there are $K_T$ event positions in $[1,T]$. Counting the last $H$ layers separately,

$$\#\{0\le i\le T:\delta_i\ne0\}\le HK_T+H=:E . \tag{11.1}$$

By (10.2), $H\le2\sqrt T$ and $K_T\le L\log T$, so $E\le2\sqrt T\,(L\log T+1)$.

Step 2: nontrivial returns cost many events (Subsection 11.2)

The set $\mathcal A$ is compact (its only accumulation point $0$ belongs to it) and consists of rationals; so $\mathcal S_k$, the image of $\mathcal A^k$ under addition, is compact and rational; and $\mathcal C_k$, the image of the compact set $\mathcal S_k\times(\mathcal S_k\cap[1/2,\infty))$ under division, is compact and rational. The irrational $\theta$ is not in the closed set $\mathcal C_k$, so some neighborhood of $\theta$ is disjoint from $\mathcal C_k$.

Let $a<b$ be exact layers for $p/q$ such that $(a,b]$ contains at least one event. Unrolling the recurrences $a_{i+1}=2a_i+u_i$ from $a$ to $b$,

$$a_b=2^{b-a}a_a+U,\qquad b_b=2^{b-a}b_a+V,\qquad U=\sum_{j=a}^{b-1}2^{b-1-j}u_j,\quad V=\sum_{j=a}^{b-1}2^{b-1-j}v_j,$$

nonnegative integers whose binary digits are the conversions. Exactness at $a$ and $b$ gives $qa_b-pb_b=0=2^{b-a}(qa_a-pb_a)$, hence $qU=pV$. Some event in $(a,b]$ means some $(u_j,v_j)\ne0$, so $(U,V)\ne(0,0)$; since $p,q>0$, $U=0$ would force $V=0$, so both are positive and $U/V=p/q$.

Now suppose $(a,b]$ contains at most $k$ events. Then $U$ and $V$ each have at most $k$ nonzero binary digits. Let $2^J$ be the largest power of two occurring in $U$ or $V$. Then $U/2^J$ and $V/2^J$ are sums of at most $k$ elements of $\{2^{-j}:j\ge0\}$, padded with zeros, so both lie in $\mathcal S_k$, and one of them is at least $1$. Since $U/V=p/q\in(1,2)$ (the window has $1<p/q<2$), $U>V$, so $U/2^J\ge1$ and $V/2^J=(q/p)(U/2^J)>1/2$. Hence $p/q=(U/2^J)/(V/2^J)\in\mathcal C_k$.

Therefore, once $p/q$ lies in the neighborhood of $\theta$ disjoint from $\mathcal C_k$, every pair of exact layers $a<b$ with an event in $(a,b]$ has more than $k$ events in $(a,b]$. This is uniform in the common multiplier $c$ with $U=cp$, $V=cq$.

Step 3: geometric capacity (Subsection 11.3)

Put $B=2R+1$ and fix a positive integer $k\ge8L\log B$. Take a window from (10.2) with $T$ large and $p/q$ so close to $\theta$ that Step 2 applies for this $k$. Write $K=K_T$ and

$$x=\max(n_0+1,E+1),\qquad r=\lfloor K/k\rfloor+1 .$$

Since $E\le2\sqrt T(L\log T+1)$, $x\le T^{3/4}$ once $T$ is large. Since $K\le L\log T$ and $k\ge8L\log B$,

$$B^r\le B\cdot B^{K/k}\le B\cdot B^{\log T/(8\log B)}=B\,T^{1/8},$$

so $2B^rx\le2BT^{7/8}\le T$ for $T$ large; enlarge $T$ accordingly.

For $i=0,\ldots,r$ the integer interval $[B^ix,2B^ix]$ lies in $[0,T]$ and contains $B^ix+1\ge x+1>E$ layers, so by (11.1) it contains an exact layer $z_i$. Then

$$z_{i+1}\ge B^{i+1}x=(2R+1)B^ix>R\cdot2B^ix\ge Rz_i ,$$

and $z_i\ge x>n_0$, so bounded spacing gives an event in $(z_i,Rz_i]\subseteq(z_i,z_{i+1}]$. Thus each of the $r$ intervals $(z_i,z_{i+1}]$, $i=0,\ldots,r-1$, is a nontrivial return between exact layers; they are pairwise disjoint and lie in $(0,T]$; by Step 2 each contains more than $k$ events. Hence $K_T\ge rk$, while $r=\lfloor K/k\rfloor+1>K/k$ gives $rk>K=K_T$. This contradiction proves (BG).

Scope. The argument uses the windows of (10.2), so it needs incompleteness and irrationality; bounded spacing enters only through the choice of $x>n_0$ and the events in $(z_i,Rz_i]$. The source supplies $R=4$ from its Section 6, reconstructed on the theorem page.