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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 9 "Digit-budget propagation (DB)" with displays (9.1) and (DB), physical pp. 10--11, in the seventeen-page PDF held by its library source card, Yu and Chen (2026). The phase-mesh step ("maximum circular gap less than $3/q$") and the overlap of the windows are stated without proof in the source; both are written out below.

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

Definitions

The normalized pair, $a_i,b_i$, $(u_i,v_i)$, $K_n$, $P_n$ are as on the normalization page; $R_n$, $c_0$ and $a$ are as on the finite-event decay page. Let $\theta=\alpha/\beta\in(1,2)$ and $\{x\}=x-\lfloor x\rfloor$. A good rational is a reduced $p/q$ with $q\ge1$ and $|\theta-p/q|<1/q^2$. At a prefix depth $n\ge0$ and for a good rational with $q\ge2$ put

$$\lambda=2^n\beta,\qquad k=\lceil\log_2(8q)\rceil,\qquad K=2^k\ge8q,\qquad E_{n,k}=\sum_{i=n}^{n+k-1}\bigl(\{2^i\alpha\}+\{2^i\beta\}\bigr).$$

Statement

Window lemma. Let $n\ge0$ and let $p/q$ be a good rational with $q\ge2$. If $P_n$ contains all integers of an interval $[a,b]$ with

$$b-a\ \ge\ \frac{3\lambda}q+E_{n,k},$$

then every sufficiently large integer lies in $\bigcup_tP_t$ (the normalized sequence is complete).

(9.1). $0\le E_{n,k}<2(K_{n+k}-K_n)+2$.

(DB). If the normalized sequence is incomplete, then for every $n\ge1$ and every good rational with $q\ge2^n\beta$,

$$K_{n+k}\ \ge\ K_n+\frac{c_0}2e^{aK_n}-3.$$

Proof

Step 1: the fractional-part budget (9.1)

Let $r_i=\{2^i\alpha\}$ and $s_i=\{2^i\beta\}$. From $2r_i=\{2^{i+1}\alpha\}+\lfloor2r_i\rfloor$ and $u_i=\lfloor2r_i\rfloor$ (normalization page, item 4), $r_{i+1}=2r_i-u_i$. Summing $2r_i-r_{i+1}=u_i$ over $n\le i<n+k$ gives

$$\sum_{i=n}^{n+k-1}r_i=\sum_{i=n}^{n+k-1}u_i-r_n+r_{n+k},$$

and the same for $\{2^i\beta\}$ with $v_i$. Adding, and using $-r_n-s_n\le0$, $r_{n+k}+s_{n+k}<2$ and $\sum_{i=n}^{n+k-1}(u_i+v_i)\le2(K_{n+k}-K_n)$ (each nonzero conversion at an index in $[n,n+k)$ is an event at a position in $(n,n+k]$ and contributes at most $2$), we get $0\le E_{n,k}<2(K_{n+k}-K_n)+2$.

Step 2: ideal and actual suffix sums

A selection of the weights of indices $n,\ldots,n+k-1$ is a pair of digit strings $(\xi_i),(\eta_i)\in\{0,1\}^k$; put $x=\sum_i\xi_i2^{i-n}$ and $y=\sum_i\eta_i2^{i-n}$, both in $[0,K-1]$, and every pair $0\le x,y<K$ arises exactly once. The ideal sum is $\sum_i(\xi_i2^i\alpha+\eta_i2^i\beta)=\lambda(\theta x+y)$, and the actual sum $v=\sum_i(\xi_ia_i+\eta_ib_i)$ satisfies

$$\lambda(\theta x+y)-E_{n,k}\le v\le\lambda(\theta x+y),$$

since $a_i=2^i\alpha-\{2^i\alpha\}$ and the selected fractional parts total at most $E_{n,k}$.

Step 3: the phase mesh

For $0\le j<q$, $|j\theta-jp/q|=j|\theta-p/q|\le(q-1)/q^2<1/q$, and the residues of $jp/q$ modulo $1$ are exactly the $q$ points $0,1/q,\ldots,(q-1)/q$ because $\gcd(p,q)=1$. So every point of the circle $\mathbb R/\mathbb Z$ is within $1/(2q)$ of some $jp/q$ and within $1/(2q)+1/q$ of some phase $j\theta$; hence every arc of length $3/q$ contains a phase, and the points of the set

$$\Lambda=\{j\theta+y:0\le j<q,\ y\in\mathbb Z\}\subset\mathbb R$$

have consecutive differences less than $3/q$.

Put $t_0=\lceil(q-1)\theta\rceil$ and fix $0\le\ell\le K-q$. For every real $\xi'\in[t_0,K-1]$ there is a point $\mu\in\Lambda$ with $\xi'\le\mu<\xi'+3/q$: the point $K-1\in\Lambda$ ($j=0$, $y=K-1$) is $\ge\xi'$, so the least point $\mu$ of $\Lambda$ with $\mu\ge\xi'$ exists and satisfies $\mu\le K-1$; if $\mu>\xi'$ its predecessor in $\Lambda$ is less than $\xi'$ and within $3/q$ of $\mu$. Writing $\mu=j\theta+y$, we have $y=\mu-j\theta\ge t_0-(q-1)\theta\ge0$ and $y\le\mu\le K-1$. Hence with $x=\ell+j\in[0,K-1]$ and this $y$, for every $\xi\in[\ell\theta+t_0,\ell\theta+K-1]$ there are $0\le x,y<K$ with

$$\xi\le\theta x+y<\xi+\frac3q .$$

The windows $[\ell\theta+t_0,\ell\theta+K-1]$ for consecutive $\ell$ overlap, because their length $K-1-t_0>8q-1-(2q-1)=6q$ exceeds the shift $\theta<2$, using $t_0\le(q-1)\theta+1<2q-1$. Their union over $0\le\ell\le K-q$ is $[s,t]$ with

$$s=t_0,\qquad t=\theta(K-q)+K-1,\qquad t-s>(K-q)+K-1-(2q-1)=2K-3q\ge K+1,$$

the last step because $K\ge8q$ and $q\ge2$. So for every $\xi\in[s,t]$ there are $0\le x,y<K$ with $\xi\le\theta x+y<\xi+3/q$.

Step 4: the window lemma

Let $[a,b]$ be as in the statement, $W=b-a\ge3\lambda/q+E_{n,k}$. Put $A=\lambda s+b-E_{n,k}$ and $B=\lambda t+b-E_{n,k}$, and let $z$ be an integer with $\lceil A\rceil\le z\le\lfloor B\rfloor$. Then $\xi=(z-b+E_{n,k})/\lambda\in[s,t]$. Take $x,y$ from Step 3 and let $v$ be the actual suffix sum of the corresponding selection. By Step 2,

$$v\ \ge\ \lambda\xi-E_{n,k}=z-b,\qquad v\ <\ \lambda\Bigl(\xi+\frac3q\Bigr)=z-b+E_{n,k}+\frac{3\lambda}q\ \le\ z-a.$$

So $z-v$ is an integer in $[a,b]$, hence in $P_n$, and $z=(z-v)+v$ is a subset sum using indices below $n$ for $z-v$ and indices in $[n,n+k)$ for $v$: $z\in P_{n+k}$. Thus $P_{n+k}$ contains every integer of $[\lceil A\rceil,\lfloor B\rfloor]$, an interval of width

$$\lfloor B\rfloor-\lceil A\rceil\ \ge\ B-A-2=\lambda(t-s)-2 \ >\ \lambda(K+1)-2\ \ge\ \lambda K,$$

since $\lambda=2^n\beta\ge\beta\ge N\ge2$. This width exceeds $b_{n+k}=\lfloor\lambda K\rfloor$, the smallest weight not used in $P_{n+k}$. The consequence on the mesh lemma page, applied with gap $1$ and the weights $b_{n+k}<a_{n+k}<b_{n+k+1}<\cdots$ (each at most twice its predecessor, normalization page item 4), shows that for every $t\ge n+k$ the set $P_t$ contains an integer interval with left endpoint $\lceil A\rceil$ whose width grows without bound. Hence every integer $\ge\lceil A\rceil$ lies in some $P_t$.

Step 5: the digit budget (DB)

Suppose the sequence is incomplete. By the window lemma, no integer interval of width at least $3\lambda/q+E_{n,k}$ lies in $P_n$, that is, $R_n<3\lambda/q+E_{n,k}$. If $q\ge2^n\beta=\lambda$, then $3\lambda/q\le3$ and with (9.1)

$$R_n<3+2(K_{n+k}-K_n)+2,\qquad\text{so}\qquad R_n\le2(K_{n+k}-K_n)+4$$

by integrality. For $n\ge1$, (FE-R) gives $c_0e^{aK_n}\le R_n+2$, hence

$$c_0e^{aK_n}\le2(K_{n+k}-K_n)+6,\qquad K_{n+k}\ge K_n+\frac{c_0}2e^{aK_n}-3.$$

Scope. The window lemma holds for every normalized pair and every good rational with $q\ge2$; (DB) needs incompleteness, $n\ge1$ and a good denominator at least $2^n\beta$. Irrationality enters only through the existence of good rationals with large denominators, used on the windows page.