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Reviewed against this repository as it stood on 2026-09-10T07:43:28Z, the state this record's filing of 2026-09-10T09:12:21Z built on. As they stood then, library/discrepancy/kesten_1966_bounded_remainder/theorem_4.md and library/discrepancy/kesten_1966_bounded_remainder/_index.md carry the two baseline preimages the reviewer read (unchanged since the baseline state of 2026-09-10T06:09:00Z), and the PDF is the canonical source; the rule pages are cited as reconciled to that state in the source-reading record. The reviewed subject itself, the reconstruction patch applied to those two pages, was never committed at those paths and is retained byte-exact as reviewed_theorem_4.md.txt and reviewed_source_index.md.txt.

This is the full substantive rendition of the independent whole-claim reviewer's historical report, with the documentary corrections mapped in the source-reading record. First-person mathematical statements remain attributed to that reviewer. The retained original subject, not this page's later prose, is the review target. All line citations to the theorem or source digest below refer to those historical snapshots.

Verdict: refutation-failed for anchored irrational necessity only. The distinct grader passed the report contract and independence, subject to the retained documentary corrections. This discharges literature-compilation proof-coverage review for that subdirection; it gives no Bohl, rational, sufficiency, E0998-status, native claim tier or whole-Theorem-4 coverage. The transformation review of this native filing was completed and accepted before it was filed.

0. Subject, independence, artifact identity, exposure

Reviewer: independent whole-claim reviewer, fresh context. The first-person derivations below remain attributed to that reviewer. The distinct grader's report-contract and independence grades are retained separately.

The reviewer disclosed no prior exposure to the reconstruction or its author. Allowed material, recorded here from the supplied assignment wrapper, was the two frozen payload pages, their full patch, the two baseline preimage copies, the canonical PDF page images, and the three rules pages docs/verification.md, docs/evidence.md, docs/anatomy.md. This explicit permitted-material list is a filing addition; it was absent from the original report body, as the distinct grader found.

The frozen subject evidence/assets/reviewed_theorem_4.md.txt carried the reconstruction's own standing text at lines 4, 11–15, 59, 482–499 and 501–506 (the review_status: unreviewed scalar, the "author-recorded, not independently accepted" and "review remains due" sentences, and the link to the Ostrowski translated-interval review with its "acceptance unchanged" sentence), and evidence/assets/reviewed_source_index.md.txt carried it at lines 30–47; a separately spawned materiality grader (model Claude Fable 5.1) ruled this exposure immaterial on 2026-09-18 by the content test, because the exposed text states only that the proof was unreviewed, the acceptance text concerns excluded scopes the reviewer did not read, and the verdict rests on the reviewer's own rederivations and attacks.

Actually read: both frozen pages, the full patch, both preimage copies, all three named rules pages, and PDF sheets 1–4 and 7–11. The reviewer did not read the author's freeze/handoff, reading receipt or renderings, sibling verdicts, material from other repositories, plans, triage, research pages, conversation history or URLs. No repository program, Lean, or mathematical code was executed; the reviewer's derivations and W2 cross-check were by hand.

Historical subject identities and native snapshot links are recorded in the source-reading record. The reviewer verified both payloads, both preimage digests, the full two-path patch and the PDF. The patch changed only the source digest and theorem_4.md; the preimage identities are recorded in the source-reading record rather than treated as a substitute mathematical premise.

PDF sheet → printed page map (derived and verified by me from the running heads, not taken from the page): sheet $k$ carries printed pages $190+2k$ (left leaf) and $191+2k$ (right leaf). Sheet 1 = 192 | 193; sheet 2 = 194 | 195; sheet 3 = 196 | 197; sheet 4 = 198 | 199; sheet 7 = 204 | 205; sheet 8 = 206 | 207; sheet 9 = 208 | 209; sheet 10 = 210 | 211; sheet 11 = 212 | back matter. The page's own claim that the article starts on the right-hand leaf of sheet 1 at printed p. 193 is correct.

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1. Restatement, with every quantifier

Let $\xi$ be irrational with $0<\xi<1$. Let $b$ be fixed with $0<b<1$. For each integer $M\ge 1$ put

$$N(M)=\#\{k\in\mathbb Z: 1\le k\le M,\ \{k\xi\}<b\},\qquad R(M)=N(M)-Mb,$$

where $\{x\}=x-\lfloor x\rfloor\in[0,1)$. The interval is $[0,b)$: the left endpoint $0$ is included, the right endpoint $b$ is excluded, and the counting index starts at $k=1$, so $k=0$ (which would contribute the point $0$) is not counted.

Claim. If there exists $C<\infty$ with $|R(M)|\le C$ for every integer $M\ge 1$, then there exists $j\in\mathbb Z$ with $b=\{j\xi\}$.

The quantifier order is: $\forall \xi$ irrational in $(0,1)$, $\forall b\in(0,1)$, [$\exists C\ \forall M\ge1: |R(M)|\le C$] $\Rightarrow$ [$\exists j\in\mathbb Z: b=\{j\xi\}$]. The hypothesis is boundedness over all positive $M$, not eventual boundedness (the page separately, and correctly, observes on line 43 that the two are equivalent). The conclusion asserts membership of $b$ in the full two-sided orbit $\{\{j\xi\}: j\in\mathbb Z\}$, not the forward orbit; inside the contradiction's eventual terminal regime, the invariant produces $j\le 0$ (see §7, W1). This is a conditional endgame observation, not a one-sided strengthening of the conclusion.

This is verbatim what theorem_4.md lines 49–57 state, and it is exactly the claim in the brief.

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2. Scope confirmation — is this a proper subdirection of Theorem 4?

Yes, confirmed. Kesten's Theorem 4 (printed p. 193) is: for $0\le a<b\le1$, $b-a<1$, and fixed $\xi\in[0,1]$, $R(M,\xi,a,b)=N(M,\xi,a,b)-M(b-a)$ is bounded in $M$ iff $b-a=\{j\xi\}$ for some integer $j$. The reconstruction proves only:

• necessity (sufficiency is Hecke/Ostrowski, cited by Kesten at (4.2), p. 204, and explicitly excluded on lines 13–15, 60–62);
• $a=0$ (Kesten reduces general $a$ to $a=0$ by Bohl [1], p. 226, on printed p. 205; the reconstruction excludes that reduction, lines 13, 60–61);
• $\xi$ irrational in $(0,1)$ (Kesten's footnote 1 on p. 193 and his remark on p. 204 dispose of rational $\xi$; excluded, lines 14, 61).

The proof body never touches $a\ne0$, never touches rational $\xi$, and never invokes sufficiency. The pages do not claim more than this direction. Lines 11–15, 49–63, 482–499 of theorem_4.md and lines 41–47 of _index.md all restrict scope, and line 498–499 states outright that "this partial direction must not be described as full local proof coverage of Theorem 4." The historical frontmatter carried review_status: unreviewed. That newly introduced scalar is removed in the filed page; scoped standing is recorded in prose.

The source argument was read across pp. 204–212. The reconstruction does not reproduce every Section 4 step: besides excluding the Bohl and rational interfaces, it bypasses (4.27)–(4.31) and cases (i)/(ii)/(iii) through the direct nonterminal long-cell exclusion checked in §5.5. This is full review of the selected local proof, not reconstruction of every source assertion.

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3. Definitions, parity and endpoint conventions — checked against pp. 193–194

PDF p. 193 (verbatim content). "$N(M,\xi,a,b)$ the number of integers $k$, $1\le k\le M$, for which $a\le\{k\xi\}<b$. ($\{c\}$ denotes the fractional part of $c$.)" Then (1.1) $R=N-M(b-a)$, and Theorem 4 as above. Footnote 1: rational $\xi$'s "constitute a trivial case for theorem 4." Footnote 2: $a_0(\xi)$ is dropped since $a_0=0$ throughout. Continued fraction written $[a_1,a_2,\dots]=1/(a_1+1/(a_2\dots))$ for irrational $\xi\in(0,1)$.

Checks.

• Endpoint: page line 24 writes $\#\{m:1\le m\le M,\ a\le\{m\xi\}<b\}$ — identical to the source; $[a,b)$, left closed, right open. ✔
• Index base: $k$ (page: $m$) runs from $1$; $k=0$ excluded. ✔ This matters: including $k=0$ would shift $R$ by $+1$ and break the exact identity (L) even though it would not change boundedness. The page is consistent with the source everywhere.
• $M=0$: the page adopts $N(0)=R(0)=0$ (line 353) purely as a telescoping convention. Kesten's (4.2) is stated "for all $M\ge0$", so this agrees. The claim under review quantifies over $M\ge1$; the convention is used only at the empty tail $S_u=0$ of (M). ✔
• $n=0$ (CF index): the page sets $p_{-1}=1,p_0=0,q_{-1}=0,q_0=1$ and restricts the geometry to $n\ge2$. With $p_n=a_np_{n-1}+p_{n-2}$, $q_n=a_nq_{n-1}+q_{n-2}$ this reproduces Kesten's (1.3)–(1.4) exactly ($p_1=1,q_1=a_1$), and Kesten defines $q_{-1}=0$ on p. 195 and again on p. 196. ✔
• Fractional part: braces $=$ fractional part in $[0,1)$, stated on line 57, matching p. 193. ✔
• $T_i$ and $A_n$: page sets $T_1=1/\xi$, $a_i=\lfloor T_i\rfloor$, $T_{i+1}=1/(T_i-a_i)$, $A_n=T_{n+1}$. Kesten's (1.5) is $a'_{n+1}=a_{n+1}+1/a'_{n+2}$; $T_{n+1}=a_{n+1}+1/T_{n+2}$ satisfies the same recursion with the same seed, so $A_n=a'_{n+1}$. Kesten's (1.6) is $q'_{n+1}=a'_{n+1}q_n+q_{n-1}$, so $Q_n=A_nq_n+q_{n-1}=q'_{n+1}$. The page's identification on line 111 is correct. ✔

Continued-fraction identities (page lines 89–139), rederived by me.

• $p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1}$: base $n=0$ gives $0\cdot0-1\cdot1=-1$; step $p_{n+1}q_n-p_nq_{n+1}=(a_{n+1}p_n+p_{n-1})q_n-p_n(a_{n+1}q_n+q_{n-1})=-(p_nq_{n-1}-p_{n-1}q_n)$. ✔ Hence $\gcd(p_n,q_n)=1$. ✔
• $\xi=(p_nT_{n+1}+p_{n-1})/(q_nT_{n+1}+q_{n-1})$: base $n=0$ is $1/T_1=\xi$; substituting $T_{n+1}=a_{n+1}+1/T_{n+2}$ turns numerator into $(p_{n+1}T_{n+2}+p_n)/T_{n+2}$ and denominator likewise. ✔
• $\varepsilon_n=q_n\xi-p_n=(q_np_{n-1}-p_nq_{n-1})/Q_n=(-1)^n/Q_n=\sigma_n\delta_n$. ✔ This is Kesten's (2.1), $\xi=p_l/q_l+(-1)^l/(q_lq'_{l+1})$. ✔
• (A) $Q_{n+1}=A_{n+1}Q_n$: $A_{n+1}q_{n+1}+q_n=A_{n+1}(a_{n+1}q_n+q_{n-1})+q_n=A_{n+1}\big((a_{n+1}+1/A_{n+1})q_n+q_{n-1}\big)=A_{n+1}Q_n$. This is exactly Kesten's (1.6) tail $q'_{n+1}=q'_{n+2}/a'_{n+2}$. ✔ Then $\delta_{n+1}=\delta_n/A_{n+1}$ ✔ and $\delta_{n-1}=A_n\delta_n=(a_{n+1}+1/A_{n+1})\delta_n=a_{n+1}\delta_n+\delta_{n+1}$ ✔.
• (B) for $n\ge2$: $q_{n-1}<q_n$ (since $q_n\ge q_{n-1}+q_{n-2}$, and $q_1<q_2$ directly) ✔; $q_{n+2}=a_{n+2}q_{n+1}+q_n\ge q_{n+1}+q_n\ge2q_n$ ✔; $A_n\in(a,a+1)$ strictly gives $Q_n>aq_n+q_{n-1}=q_{n+1}$ and $Q_n<(a+1)q_n+q_n=(a+2)q_n$ ✔.
• (C) $a_{j+1}\delta_j=\delta_{j-1}-\delta_{j+1}$, so $\sum_{j=s}^{u}=\delta_{s-1}+\delta_s-\delta_u-\delta_{u+1}\to\delta_{s-1}+\delta_s$, and this $\to0$. ✔

The page's claim (lines 138–139, 71–72) that no external continued-fraction theorem is left unproved is accurate. The two basic identities are proved by a compressed but correct induction sketch; I completed both inductions above and they hold.

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4. The consumed Theorem 1 geometry — checked against pp. 196–199

What the source proves (pp. 196–199). Theorem 1: each $(r/q_m,(r+1)/q_m)$ contains exactly one $\{k\xi\}$ with $1\le k\le q_m$, called $P_r$ (for odd $m$, $P_r$ sits in $(\tfrac{q_m-r-1}{q_m},\tfrac{q_m-r}{q_m})$ and $J_r=(P_{r+1},P_r)$); exactly $q_m-q_{m-1}$ intervals $J_r$ are short of length $a'_{m+1}/q'_{m+1}$ and exactly $q_{m-1}$ are long of length $(a'_{m+1}+1)/q'_{m+1}$; long occurs precisely when $P_r=\{\lambda_r\xi\}$ with $1\le\lambda_r\le q_{m-1}$; the next $(c_m-1)q_m$ points cut each $J_r$ into $c_m$ pieces at $P_r+(-1)^ms/q'_{m+1}$, with $c_m-1$ pieces of length $1/q'_{m+1}$ and a last piece $J'_r$ of length $(a'_{m+1}-c_m+1)/q'_{m+1}$ (short) or $(a'_{m+1}-c_m+2)/q'_{m+1}$ (long). Proof: (2.1)–(2.10), even $m$ only — "the case where $m$ is odd being entirely analogous."

What the reconstruction uses, and my check of each item. The reconstruction re-proves, rather than cites, and the re-proofs are correct:

• Oriented coordinate $y_n(k)=\{\sigma_nk\xi\}$, $\sigma_n=(-1)^n$. For $k\ge1$ and $\xi$ irrational, $k\xi\notin\mathbb Z$, so $y_n$ is the identity ($n$ even) or the circle reflection $x\mapsto1-x$ ($n$ odd). Both are circle isometries. ✔
• (D) $\lambda_r$ = unique element of $[1,q]$ with $\sigma p_n\lambda_r\equiv r\ (\mathrm{mod}\ q)$, and $Y_r=y_n(\lambda_r)=r/q+\lambda_r/(qQ)\in(r/q,(r+1)/q)$. Even $n$: $k\xi=kp_n/q_n+k/(q_nQ_n)$ and $0<k/(q_nQ_n)<1/q_n$ for $1\le k\le q_n<Q_n$, so $\{k\xi\}=\varrho_k/q_n+k/(q_nQ_n)$ — Kesten's (2.2)–(2.5) exactly. Odd $n$: $\xi=p_n/q_n-1/(q_nQ_n)$, so $-k\xi=-kp_n/q_n+k/(q_nQ_n)$ and the same computation runs with $\varrho'_k\equiv-kp_n$. ✔ In particular $\lambda_0=q_n$ and $Y_0=\delta_n$, consistent with $q_n\xi=p_n+\sigma_n\delta_n$. ✔
• (E) $\lambda_{r+1}=\lambda_r-v$ if $\lambda_r>v$, else $\lambda_r+q-v$, with $\lambda_q=\lambda_0$. From the determinant, $\sigma p_nv\equiv(-1)^n(-1)^{n-1}=-1$, so $\lambda_{r+1}\equiv\lambda_r-v$; representatives in $[1,q]$ force the case split. This is Kesten's (2.8) verbatim, and the cyclic wrap matches his footnote 7. ✔
• (F) short $=A\delta$ when $\lambda_r>v$ (since $Y_{r+1}-Y_r=1/q-v/(qQ)=(Q-v)/(qQ)=A/Q$), long $=(A+1)\delta$ when $\lambda_r\le v$ (since $1/q+(q-v)/(qQ)=(A+1)/Q$). Matches Kesten's (2.9) $a'_{m+1}/q'_{m+1}$ and $(a'_{m+1}+1)/q'_{m+1}$, and the long/short criterion $\lambda_r\le q_{m-1}$. ✔ The zero-crossing arc is handled by the lift $Y_q=Y_0+1$, which reproduces the same formula. ✔ All lengths $<2/q$: $(A+1)/Q<2/q\iff(A-1)q+2v>0$. ✔
• Odd parity is genuinely supplied, not assumed. Under $y_n$, the odd-$n$ point $1-Y_r$ lies in $(\tfrac{q-r-1}{q},\tfrac{q-r}{q})$ — exactly Kesten's odd-$m$ placement — and $[Y_r,Y_{r+1}]$ reflects onto his $J_r=(P_{r+1},P_r)$. The correspondence is exact; the page does not spell it out, but it does not need to, because it re-derives rather than cites.
• (G) For $1\le k\le q_{n+1}$, $k$ is uniquely $\lambda_r+sq$ with $0\le s\le m_r$, $m_r=a-1$ (short) / $a$ (long), and $y_n(\lambda_r+sq)=Y_r+s\delta<(r+1)/q$ because $\lambda_r+sq\le q_{n+1}<Q$. This is Kesten's (2.10). I checked the $s$-range: $\lambda_r>v\Rightarrow sq<aq\Rightarrow s\le a-1$; $\lambda_r\le v\Rightarrow s\le a$ with $s=a$ attained. Count check $\sum_r(m_r+1)=(q-v)a+v(a+1)=aq+v=q_{n+1}$. ✔
• (H) last piece $=L_r-m_r\delta_n=(A_n-a_{n+1}+1)\delta_n=(1/A_{n+1}+1)\delta_n=\delta_n+\delta_{n+1}$, in both the short and the long case. ✔ Matches Kesten's $J'_r$ lengths at $c_m=m_r+1$.
• Level $n+1$. Short length $A_{n+1}\delta_{n+1}=\delta_n$, long length $(A_{n+1}+1)\delta_{n+1}=\delta_n+\delta_{n+1}$. So the $\delta_n$-pieces are the short level-$(n+1)$ arcs and the last piece is long. I verified this against the label criterion too: a regular piece's $y_{n+1}$-initial point has label $\lambda_r+(s+1)q_n>q_n$ (short), the last piece's has label $\lambda_{r+1}\le q_n$ (long). Counts agree: $q_{n+1}-q_n$ short, $q_n$ long. ✔

Verdict on §4: the reconstruction consumes only the $N\le q_{n+1}$ specialization of Theorem 1, states it correctly, and proves it. It does not consume Corollary 1, the three-distance statement, or the general-$N$ / Ostrowski-expansion form of Theorem 1 — and it correctly says so.

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5. Section 4 as reconstructed — every essential deduction, checked against pp. 204–212

Source landmarks, read in full: Section 4 opens on p. 204; Bohl reduction and (4.3)–(4.7) on p. 205; (4.8)–(4.12) on p. 206; (4.13)–(4.18) and the $\varepsilon_n$ stability clause on p. 207; (4.19)–(4.23) on pp. 208–209; (4.24)–(4.31) on pp. 209–210; the case analysis (i)/(ii)/(iii) and the terminal telescoping on pp. 211–212.

5.1 Setup (page lines 235–258). $z_n=b$ ($n$ even), $1-b$ ($n$ odd) — this is exactly the image of $b$ under $y_n$, and $z_{n+1}=1-z_n$. ✔ The contradiction hypothesis ($b\ne\{j\xi\}$ for all $j\in\mathbb Z$) is imposed at the start; Kesten imposes it later (p. 206, after (4.10)), which is harmless. Choosing $n$ with $2/q_n<\min(b,1-b)$ makes the arc containing $z_n$ miss $0$, for that $n$ and all larger ($q_n$ nondecreasing $\to\infty$); hence $r_n\le q_n-2$ and $h_n=z_n-Y_{r_n}$ is a plain real difference. ✔ Strictness $0<h_n<L_n$ holds because $Y_{r_n},Y_{r_n+1}$ are orbit points and $z_n$ is not — and $z_n=Y_r\iff b=\{\lambda_r\xi\}$ in both parities. ✔ $d_n:=\max\{d: 0\le d\le m_{r_n},\ d\delta_n<h_n\}$ is well defined ($d=0$ always qualifies). ✔ This is Kesten's (4.7), including his "reversal of the inequality" for odd $n$: his odd-$n$ condition $\{(\lambda+dq)\xi\}\ge b$ becomes $y_n(\lambda+dq)\le1-b=z_n$. ✔

5.2 (I) and the transitions (J), (K). Nonterminal ($d_n<m_{r_n}$) gives $d_n\delta_n<h_n<(d_n+1)\delta_n$, strict because $Y_{r_n}+(d_n+1)\delta_n$ is the orbit point of index $\lambda_n+(d_n+1)q_n\le q_{n+1}$. ✔

(J) The arc $[Y_{r_n}+d_n\delta_n,\,Y_{r_n}+(d_n+1)\delta_n]$ is a regular piece, hence short at level $n+1$, and since $y_{n+1}=1-y_n$ its $y_{n+1}$-initial endpoint is its right $y_n$-endpoint, of index $\lambda_n+(d_n+1)q_n$. Then $h_{n+1}=(1-z_n)-\big(1-Y_{r_n}-(d_n+1)\delta_n\big)=(d_n+1)\delta_n-h_n\in(0,\delta_n)$. ✔ This is Kesten's (4.23).

(K) Terminal ($d_n=m_{r_n}$): the arc is the last piece, long at level $n+1$, $y_{n+1}$-initial endpoint $Y_{r_n+1}$ with label $\lambda_{r_n+1}$, so by (E) $\lambda_{n+1}=\lambda_n-q_{n-1}$ (old short) or $\lambda_n+q_n-q_{n-1}$ (old long), and $h_{n+1}=L_n-h_n\in(0,\delta_n+\delta_{n+1})$. ✔ This is Kesten's (4.26a)–(4.26c), including his remark that $\lambda(n+1)\le q_n$ forces $J(n+1)$ long — "will be crucial for our argument."

5.3 (L), the counting identity. For nonterminal $d_n$, $t=d_n+1\in[1,a_{n+1}]$, $M_n=tq_n<q_{n+1}$. Every cell $(r/q_n,(r+1)/q_n)$ holds exactly $t$ of the first $M_n$ points, because $\lfloor(tq_n-\lambda_r)/q_n\rfloor=t-1$ for $\lambda_r\in[1,q_n]$, and all of them satisfy $\lambda_r+sq_n\le tq_n<Q_n$ so (G) applies. In cell $r_n$ the largest is $Y_{r_n}+d_n\delta_n<z_n$ by (I); and $z_n<Y_{r_n}+t\delta_n=r_n/q_n+(\lambda_n+tq_n)/(q_nQ_n)<(r_n+1)/q_n$, using $\lambda_n+tq_n\le q_{n+1}<Q_n$. So $z_n$ lies strictly inside cell $r_n$, earlier cells contribute all $t$, later cells contribute none. Count $=t(r_n+1)$; with $q_nz_n=r_n+\lambda_n/Q_n+q_nh_n$,

$$D_n(M_n)=t(r_n+1)-tq_nz_n=t\Big(1-\frac{\lambda_n}{Q_n}-q_nh_n\Big).$$

✔ This reproduces Kesten's (4.13)–(4.16). The parity bridge: for even $n$, $D_n(M)=N(M,\xi,0,b)-Mb=R(M)$. For odd $n$, $\{-k\xi\}<1-b\iff\{k\xi\}>b\iff\lnot(\{k\xi\}<b)$ — the second $\iff$ needs exactly $\{k\xi\}\ne b$, which is the contradiction hypothesis — so the count is $M-N(M,\xi,0,b)$ and $D_n(M)=-R(M)$. Hence $D_n=\sigma_nR$ in both parities. ✔ The page's lines 314–323 state precisely this and it is correct.

5.4 (M), block accumulation. Three sub-steps, all correct:

1. Finite-prefix stability. $\eta(M):=\min_{1\le k\le M}\min\big(d(\{k\xi\},0),d(\{k\xi\},b)\big)>0$ (finite min of positive numbers; positive because $\{k\xi\}\ne0$ by irrationality and $\ne b$ by hypothesis). A move of circle distance $<\eta(M)$ cannot cross $0$ or $b$, so cannot change membership in $[0,b)$. ✔ Kesten asserts this in one clause on p. 207 ("there exists an $\varepsilon_n>0$ such that…"); the reconstruction supplies the proof.
2. Tail bound. $S_i\xi=\sum_{j>i}c_{n_j}(p_{n_j}+\varepsilon_{n_j})$, so $\|S_i\xi\|\le\sum_{j>i}c_{n_j}\delta_{n_j}\le\sum_{s\ge n_{i+1}}a_{s+1}\delta_s=\delta_{n_{i+1}-1}+\delta_{n_{i+1}}$ by (C). The single pairwise separation condition therefore controls the whole tail — I checked this, it is not an oversight. ✔ Kesten's cruder $\le4/q_{n+s}\le2^{2-(n+s)/2}$ (p. 207) does the same job.
3. Reverse-order telescoping. With $T_i=\sum_{j\ge i}M_{n_j}$, $T_{u+1}=0$: $N(T_i)-N(T_{i+1})=N(M_{n_i})$, so $N(T_1)=\sum N(M_{n_i})$ and, subtracting $b\,T_1$, $R(T_1)=\sum_{i=1}^uR(M_{n_i})$. Common parity $\Rightarrow$ all summands share sign $\sigma$ and $|R(M_{n_i})|\ge c$ $\Rightarrow$ $|R(T_1)|\ge uc\to\infty$. ✔ This is Kesten's (4.19)–(4.20) exactly, including the same block ordering (block $n_1$ placed last, preceded only by higher-index blocks). The page's cross-reference "(4.19)–(4.20)" is accurate.

5.5 Digit exclusions. With $a=a_{n+1}$, $B=a_{n+2}$, $t=d_n+1$:

• (N) $D_n(M_n)>\frac tQ(Q-\lambda_n-tq)=\frac tQ\big((A-d_n-2)q+v\big)$, from $h_n<t\delta_n$ and $\lambda_n\le q$. This is Kesten's (4.16) term for term. ✔
• Case $d_n\le a-3$ (so $a\ge3$, $1\le t\le a-2$): $D_n>\frac{t(a-1-t)}{a+2}\ge\frac{a-2}{a+2}\ge\frac15$. The extremal step — minimizing the concave $t\mapsto t(a-1-t)$ over the integer interval $[1,a-2]$, both endpoints giving $a-2$ — is correct. ✔
• Case $d_n=a-2\ge0$, $B\le6$: $A-a=1/A_{n+1}>1/7$ since $A_{n+1}<B+1\le7$; then $D_n>\frac{a-1}{7(a+2)}\ge\frac1{28}$ for $a\ge2$. ✔ Together these give (O), matching Kesten's (4.21a)/(4.21b), and Kesten's single constant $1/28$ covers both.
• (P) From (J), $h_n=t\delta_n-h_{n+1}<t\delta_n-d_{n+1}\delta_{n+1}$, and $Q_n\delta_{n+1}=1/A_{n+1}$, so $q_nQ_nh_n<tq-\frac{d_{n+1}q}{A_{n+1}}$ and $D_n>\frac tQ\big(Q-\lambda_n-tq+\frac{d_{n+1}q}{A_{n+1}}\big)$. ✔
• Case $d_n=a-2$, $B\ge7$: (O) at index $n+1$ gives $d_{n+1}\ge B-2$; with $t=a-1$ the parenthesis is $q+\frac q{A_{n+1}}+v-\lambda_n+\frac{d_{n+1}q}{A_{n+1}}\ge v+\frac{(B-1)q}{A_{n+1}}\ge\frac{(B-2)q}{A_{n+1}}$, giving $D_n>\frac{a-1}{a+2}\cdot\frac{B-2}{B+1}\ge\frac14\cdot\frac58=\frac5{32}$. ✔ Kesten obtains the identical $\tfrac14\cdot\tfrac58$ on p. 209. (At $a=2$, $\frac{a-1}{a+2}=\frac14$, so the $\tfrac14$ bound is sharp there. This corrects the original report's parenthetical claim of $\tfrac13$ and looseness; the $\tfrac5{32}$ argument is unchanged.) Result: eventually $d_n\ge a_{n+1}-1$ at every index, both parities (page line 431–432 makes the both-parities point that Kesten compresses into "The same conclusion is valid if (4.18) holds for infinitely many odd $n$").
• Case $d_n=a-1$, long (the only nonterminal case left, since a short cell has $m_r=a-1$): $\lambda_n\le v$ and $d_{n+1}\ge B-1$, so with $t=a$ the parenthesis is $v-\lambda_n+\frac{(1+d_{n+1})q}{A_{n+1}}\ge\frac{Bq}{A_{n+1}}$ and $D_n(aq)>\frac a{a+2}\cdot\frac B{B+1}\ge\frac16$. ✔ Kesten's p. 211 gives $\le-\frac{a_{n+2}}{a_{n+2}+2}\cdot\frac{a_{n+3}}{a_{n+3}+1}\le-\frac16$ — same structure, same constant, opposite sign from his parity convention. The page's cross-reference "(4.24)–(4.31)" is accurate.

Case exhaustion, checked by me. Nonterminal $d_n$ satisfies $0\le d_n\le m_{r_n}-1\le a-1$. The three exclusion classes partition $\{d_n\le a-3\}\cup\{d_n=a-2\}\cup\{d_n=a-1\}$, and $d_n=a-1$ nonterminal forces a long cell. Nothing escapes. Hence eventually every $d_n$ is terminal; by (K) the next cell is long; a long terminal cell has $d_n=m_{r_n}=a_{n+1}$, which is (Q). ✔ The reconstruction reaches (Q) by a shorter route than Kesten's cases (i)/(ii)/(iii) on p. 211 — it excludes "$d_n=a_{n+1}-1$ with $J(n)$ long" directly at every large $n$, whereas Kesten excludes "terminal at $n$ and $d_{n+1}=a_{n+2}-1$". I checked that (P) needs only nonterminality at $n$ and the definition of $d_{n+1}$ — not the long/short status at level $n+1$ — so the shortcut is valid and subsumes Kesten's case split.

5.6 The endgame. In the (Q) regime the long case of (K) applies at every step, so $\lambda_{n+1}-q_n=\lambda_n-q_{n-1}$: the integer $j:=\lambda_n-q_{n-1}$ is constant for all large $n$. The initial point of the arc containing $b$ is the genuine orbit point $P_n=\{\lambda_n\xi\}$ (the coordinate $y_n$ is an isometry, so $d(P_n,b)=h_n<L_n<2/q_n\to0$). Since $\lambda_n\xi=j\xi+p_{n-1}+\varepsilon_{n-1}$ and $\varepsilon_{n-1}\to0$, $d(P_n,\,j\xi\bmod1)\to0$; by the triangle inequality $d(b,\,j\xi\bmod1)=0$, so $b-j\xi\in\mathbb Z$, and $0<b<1$ forces $j\ne0$ and $b=\{j\xi\}$ — contradicting the hypothesis. ✔

Cross-check against pp. 211–212: Kesten's final integer is $\lambda(n_3)-q_{n_3-1}$ — identical to the reconstruction's $j$. His route is a telescoped limit $b=\lim P(n)=\{\lambda(n_3)\xi\}-\{q_{n_3-1}\xi\}+\tfrac12(1+(-1)^{n_3})$. I verified this converges (the $-\tfrac12(-1)^n$ term exactly compensates $\{q_{n-1}\xi\}$ oscillating between $\delta_{n-1}$ and $1-\delta_{n-1}$) and that his last equality holds in both parities after a representative check he does not state. The reconstruction's invariant route avoids that check entirely.

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6. Steps the reconstruction supplies where the PDF elides — and my judgment on each

• #: S1

  • Elided in source: Odd-$m$ / odd-$n$ cases (p. 197 "entirely analogous"; p. 205 "reverse most of the inequalities")
  • Supplied by reconstruction: Signed coordinate $y_n=\{\sigma_nk\xi\}$, $z_n\in\{b,1-b\}$, complement identity
  • Correct?: Yes — reflection is an isometry; complement identity proved and needs exactly $\{k\xi\}\ne b$

• #: S2

  • Elided in source: Existence of the stability threshold $\varepsilon_n$ (p. 207, one clause)
  • Supplied by reconstruction: $\eta(M)=\min$ of distances to $0$ and $b$
  • Correct?: Yes

• #: S3

  • Elided in source: Tail bound via $4/q_{n+s}$ (p. 207)
  • Supplied by reconstruction: Exact telescoping (C) $\sum_{j\ge s}a_{j+1}\delta_j=\delta_{s-1}+\delta_s$
  • Correct?: Yes, and sharper

• #: S4

  • Elided in source: Transition rules derived only in the two even-$n$ situations (4.23), (4.26)
  • Supplied by reconstruction: (J)/(K) as integer-label rules in both parities
  • Correct?: Yes

• #: S5

  • Elided in source: Final limit with $\pm\tfrac12(-1)^n$ and an unstated representative check (p. 212)
  • Supplied by reconstruction: Constant invariant $j=\lambda_n-q_{n-1}$, then circle limit
  • Correct?: Yes, and cleaner

• #: S6

  • Elided in source: "It is easy to conclude from this" (4.17), p. 207
  • Supplied by reconstruction: Split into $1/5$ and $1/28$ with an explicit concave-quadratic minimization
  • Correct?: Yes

• #: S7

  • Elided in source: Kesten's cases (i)/(ii)/(iii), p. 211
  • Supplied by reconstruction: Direct exclusion of "$d_n=a-1$, long" via (P) with $t=a$
  • Correct?: Yes, and it subsumes them

• #: S8

  • Elided in source: Theorem 1 stated via the Ostrowski expansion (1.7)
  • Supplied by reconstruction: Re-derivation of only the $N\le q_{n+1}$ geometry, no expansion theorem
  • Correct?: Yes

• #: S9

  • Elided in source: Level-$(n+1)$ short/long identification implicit in $J'_r,J''_r$
  • Supplied by reconstruction: Explicit, with the label criterion and counts
  • Correct?: Yes

• #: S10

  • Elided in source: $R$ vs. the reflected count
  • Supplied by reconstruction: $D_n=\sigma_nR$ proved
  • Correct?: Yes

No supplied step is incorrect. S1, S4, S5 and S7 are the substantive ones; S3 and S6 are improvements; S2, S8, S9, S10 are gap-filling.

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7. Three weakest steps, independently selected and rederived

W1 — (K), the terminal transition and orientation reversal (page lines 274–285). This carries the whole endgame: it alone produces the invariant $j$. It is the most likely home for a parity or orientation slip.

Rederivation. Terminal means $m_{r_n}\delta_n<h_n<L_{r_n}$, so $z_n$ lies in the arc $\big(Y_{r_n}+m_{r_n}\delta_n,\ Y_{r_n+1}\big)$, which is a level-$(n+1)$ arc of length $\delta_n+\delta_{n+1}$ (long). Because $y_{n+1}=1-y_n$ pointwise on the circle, $y_{n+1}$-order is reverse $y_n$-order, so the arc's $y_{n+1}$-initial endpoint is its $y_n$-right endpoint $Y_{r_n+1}=y_n(\lambda_{r_n+1})$; hence $\lambda_{n+1}=\lambda_{r_n+1}$, which by (E) is $\lambda_n-q_{n-1}$ (short) or $\lambda_n+q_n-q_{n-1}$ (long). And $h_{n+1}=(1-z_n)-(1-Y_{r_n+1})=L_n-h_n$, in $(0,\delta_n+\delta_{n+1})$. Consistency check: $\lambda_{r_n+1}\in[1,q_n]$, which is exactly the level-$(n+1)$ "long" criterion — matching Kesten's (4.26c) and his remark that this inequality "will be crucial." Composition: inside the contradiction's eventual (Q) regime only the long branch fires, giving $\lambda_{n+1}-q_n=\lambda_n-q_{n-1}$, so $j$ is constant; conditionally in this regime, $j\le0$ (long $\Rightarrow\lambda_n\le q_{n-1}$) and $j\ne0$ (else $b\in\mathbb Z$). Sound; agrees with Kesten's final integer $\lambda(n_3)-q_{n_3-1}$.

W2 — (L), the counting identity with both parities (page lines 290–323). This converts geometry into discrepancy, and it is where an off-by-one in the index base or an endpoint convention would hide.

Rederivation: as in §5.3. Independent numerical stress test, by hand, no code. The displayed decimal corrections below are adopted from the distinct grader's disclosed floating-point cross-check; both transition and direct $h_3$ are corrected. Neither computation is a premise or mathematical evidence for the theorem. Take $\xi=[0;\overline3]=(\sqrt{13}-3)/2=0.3027756\ldots$, so $a_i\equiv3$, $A_n\equiv1/\xi=3.3027756$, $q_{-1..4}=0,1,3,10,33,109$, $p_{0..3}=0,1,3,10$. Let $b=0.35$. Even level $n=2$: $Q_2=36.027756$, $\delta_2=0.0277563$; $\lambda_r\equiv7r\ (10)$ gives $\lambda_{0..9}=10,7,4,1,8,5,2,9,6,3$, satisfying (E) at every step including the wrap $\lambda_9=3\le v=3\Rightarrow\lambda_{10}=10=\lambda_0$. Long cells are $r=3,6,9$ — exactly $q_1=3$ of them. $Y_3=0.3027756=\{\xi\}$, $Y_4=0.4222051=\{8\xi\}$, $L_3=0.1194295=(A+1)\delta_2$. ✔ So $r_2=3$, $\lambda_2=1$, long, $h_2=0.0472244$, $d_2=1$ (nonterminal), $t=2$, $M_2=20$. (L) predicts $N(20)=t(r_2+1)=8$ and $D_2=2(1-1/36.027756-10h_2)=0.99999\ldots$ Direct enumeration of $\{k\xi\}<0.35$ for $k\le20$ gives $k=1,4,7,10,11,14,17,20$ — 8 points, $R(20)=8-7=1$. ✔ Odd level $n=3$ (the parity that tests the complement identity): (J) predicts $\lambda_3=1+2\cdot10=21$ and $h_3=2\delta_2-h_2=0.0082884$; directly $y_3(21)=1-\{21\xi\}=0.6417116$ and $0.65-0.6417116=0.0082884$. ✔ $Q_3=118.99159$, $\delta_3=0.0084040>h_3$, so $d_3=0$, $t=1$, $M_3=33$. (L) gives $D_3=1-21/118.99159-33h_3=0.5500$, i.e. $\sigma_3R(33)=+0.55$, i.e. $N(33)=11.55-0.55=11$. Direct enumeration gives 11. ✔ Also $d_3=0\le a-3=0$, so step 1 predicts $D_3>1/5$ — and $0.55>0.2$. ✔ The identity, the sign convention, the odd-parity complement and the transition all reproduce exact integers.

W3 — (M), block accumulation (page lines 327–367). "Separated discrepancy blocks add" is the classic place where an argument silently assumes what it needs.

Rederivation: as in §5.4. The three risk points are (i) whether $\eta$ is positive — yes, and it needs the contradiction hypothesis, which is in force; (ii) whether one pairwise separation condition controls the entire preceding tail — yes, because the bound is taken over all $s\ge n_{i+1}$, not only over the chosen indices; (iii) whether the blocks are laid out so that the sum of the preceding blocks is the high-index tail — yes, the sum is built in reverse index order, $T_i=M_{n_i}+T_{i+1}$, exactly as in Kesten's (4.20). Same-parity selection makes the summands co-signed, so cancellation is impossible and $|R(T_1)|\ge uc$. Sound.

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8. Strongest attempted refutation

I did not attempt to refute the statement (it is a standard bounded-remainder-set characterization and Kesten's own proof stands); I attacked the chain.

Primary attack — find a nonterminal configuration that escapes all four exclusion classes, or a circular dependency among them. Nonterminal forces $0\le d_n\le m_{r_n}-1\le a_{n+1}-1$, so the value-partition $\{d_n\le a-3\}$, $\{d_n=a-2\}$, $\{d_n=a-1\}$ is exhaustive, and $d_n=a-1$ nonterminal forces $m_{r_n}=a$, i.e. a long cell. I then checked the dependency order: class 1 and class 2 are unconditional and give (O) for $n\ge N_0$; class 3 invokes (O) at index $n+1$ (legitimate, $n+1>n\ge N_0$) and gives $d_n\ge a_{n+1}-1$ for $n\ge N_1$; class 4 invokes that at index $n+1$ (legitimate) and gives terminality for $n\ge N_2$. No stage uses its own conclusion; the attack failed. I also checked the degenerate small-partial-quotient regimes: $a_{n+1}=1$ (classes 1–3 vacuous, class 4 gives $\ge\frac13\cdot\frac12=\frac16$) and $a_{n+1}=2$ (class 1 vacuous). All constants stay absolute.

Secondary attack — break (L) by putting $z_n$ on a grid point or in the wrong cell. The arc $[Y_{r_n},Y_{r_n+1}]$ straddles the grid line $(r_n+1)/q_n$, so "cell $r_n$" is not automatic. The reconstruction closes this by proving $z_n<Y_{r_n}+t\delta_n<(r_n+1)/q_n$ in the nonterminal case, which pins $z_n$ strictly inside cell $r_n$ and simultaneously rules out $z_n$ being a grid point there. Failed. I also checked that $b$ rational (e.g. $b=1/2$) creates no exception.

Tertiary attack — parity leakage in the accumulation lemma. If the four exclusion classes only bounded one parity, boundedness would not follow. But the lemma's contrapositive is stated for the whole class (pigeonhole then extracts a parity), and the page says so explicitly at line 431–432. Failed.

Quaternary attack — numerical. The hand computation in W2 at $\xi=[0;\overline3]$, $b=0.35$, levels $n=2$ (even) and $n=3$ (odd), reproduced $N(20)=8$ and $N(33)=11$ exactly against the formulas, including the sign flip and the (J) transition. Failed.

No attack succeeded.

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9. Premise interfaces and reading depth

Local claim premises consumed: none. The reconstruction consumes no native L-claim, no other corpus page, and no evidence artifact. The links it carries (problems/irrationality/E0998, …/ostrowski_1927…/equation_3, …/evidence/verify/translated_interval_review) all sit in surrounding prose about other scopes; the anchored proof is independent of every one of them. I did not read any of them.

External source premises (Kesten 1966, Acta Arithmetica 12, 193–212; local PDF kesten_1966_bounded_remainder.pdf):

• Item: Definitions of $N,R$; braces convention; index base $1\le k\le M$

  • Locator: p. 193, (1.1)
  • Hypotheses / specialization: none
  • Interface: fixes the object under review
  • Reading depth: proof-irrelevant; read in full, verbatim

• Item: Theorem 4 statement

  • Locator: p. 193, (1.2)
  • Hypotheses / specialization: $0\le a<b\le1$, $b-a<1$, fixed $\xi$
  • Interface: the ambient theorem; only the $a=0$ irrational necessity half is proved
  • Reading depth: claims checked verbatim

• Item: Rational $\xi$ trivial

  • Locator: p. 193 fn. 1; p. 204
  • Hypotheses / specialization: —
  • Interface: excluded from scope
  • Reading depth: statement read; not consumed

• Item: CF recurrences (1.3)–(1.6)

  • Locator: p. 194
  • Hypotheses / specialization: irrational $\xi\in(0,1)$
  • Interface: notation only; re-derived locally
  • Reading depth: claims checked + independently proof verified by me

• Item: Ostrowski expansion (1.7)

  • Locator: p. 194
  • Hypotheses / specialization: $0\le c_i\le a_{i+1}$ etc.
  • Interface: not consumed (page correctly says so)
  • Reading depth: statement read only

• Item: Theorem 1 (long/short lengths, refinement)

  • Locator: pp. 196–197, proof (2.1)–(2.10) pp. 197–199
  • Hypotheses / specialization: $\xi$ irrational, $N$ with $m(N,\xi)=n$; only $N\le q_{n+1}$ used
  • Interface: supplies (D)–(H); re-proved locally, both parities
  • Reading depth: proof verified for (2.1)–(2.10); odd case is "analogous" in the source and is supplied by the page

• Item: Corollary 1 / three-distance

  • Locator: p. 199
  • Hypotheses / specialization: —
  • Interface: not consumed
  • Reading depth: statement read only

• Item: Theorems 2, 3 (Farey, metric)

  • Locator: pp. 195–196
  • Hypotheses / specialization: —
  • Interface: not consumed
  • Reading depth: statement read only

• Item: Bohl reduction ([1], p. 226)

  • Locator: p. 205
  • Hypotheses / specialization: reduces general $a$ to $a=0$
  • Interface: not consumed; explicitly excluded
  • Reading depth: Kesten's paraphrase read; Bohl's paper unread

• Item: Hecke [6] / Ostrowski [10] sufficiency (4.2)

  • Locator: p. 204
  • Hypotheses / specialization: $b-a=\{k\xi\}$
  • Interface: not consumed
  • Reading depth: statement read only

• Item: Section 4 necessity argument (4.3)–(4.31) + endgame

  • Locator: pp. 205–212
  • Hypotheses / specialization: $\xi$ irrational, $a=0$, $0<b<1$, $b\notin\{ \{j\xi\}\}$
  • Interface: the reconstructed subject
  • Reading depth: proof verified against the complete selected reconstruction; (4.27)–(4.31) and cases (i)/(ii)/(iii) were read but are bypassed by its direct shortcut

Explicit assumptions inside the proof: (i) $\xi$ irrational, $0<\xi<1$; (ii) $0<b<1$ fixed; (iii) $|R(M)|\le C$ for all $M\ge1$; (iv) the contradiction hypothesis $b\ne\{j\xi\}$ for every $j\in\mathbb Z$; (v) $n\ge$ a threshold depending on $\xi$ and $b$ only. Nothing else.

Dependency on excluded material: none. I confirmed independently that Kesten's own $a=0$ argument does not invoke Bohl (Bohl is used solely for the $a\ne0$ reduction), does not invoke sufficiency, and does not invoke Section 3.

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10. Audit checklist — explicit verdict on all ten items

1. Quantifiers and scope. PASS. "Bounded for all $M\ge1$" vs. eventual: correctly identified as equivalent, and the proof uses only all-$M$ boundedness (through infinitely many block sums). Every "for sufficiently large $n$" threshold is finite and depends only on $\xi,b$. $j$ ranges over all of $\mathbb Z$ in the conclusion; the contradiction's eventual terminal regime delivers $j\le0$, $j\ne0$. This does not restrict the theorem's two-sided orbit conclusion. Boundary cases $b\to0,1$ are excluded by $0<b<1$ and enter only through $\min(b,1-b)>0$. The full interval and the empty interval are handled in the surrounding prose (lines 41) and are outside the anchored claim.
2. Circularity. PASS. The exclusion chain has a strict order: classes 1–2 unconditional $\Rightarrow$ (O); class 3 uses (O) at $n+1$; class 4 uses class 3's conclusion at $n+1$; (Q) uses class 4. No stage presupposes its own conclusion. No induction assumes the target.
3. Model and convention changes. PASS — this is the item most at risk here. The proof substitutes the twisted coordinate $y_n(k)=\{\sigma_nk\xi\}$ and the twisted target $z_n\in\{b,1-b\}$ for the actual objects. The transfer is proved, not asserted: $y_n$ is the identity or the circle reflection $x\mapsto1-x$ (an isometry), and the resulting counting transfer $D_n=\sigma_nR$ is derived, with the one place it needs the standing hypothesis ($\{k\xi\}\ne b$) made explicit. I checked the transfer against Kesten's odd-$m$ statement on p. 196 and they coincide exactly.
4. Finite and statistical overreach. PASS (no instance). No finite case, sample, or heuristic average is used as a universal proof anywhere in the reconstruction. My own numerical spot-check in W2 is a reviewer's cross-check, not part of the argument, and I do not treat it as evidence for the theorem.
5. Uniformity. PASS. The four exclusion constants $\tfrac15,\tfrac1{28},\tfrac5{32},\tfrac16$ are absolute — independent of $n$, $\xi$, $b$ — which is exactly what the accumulation lemma requires (a single $c>0$ across infinitely many $n$). The interchange of limit and sum in (C) is an absolutely convergent telescoping with an explicit closed form. The $\eta(M)$ threshold depends on $M$, and the construction correctly chooses $n_{i+1}$ after $M_{n_i}$ is fixed.
6. Extremal conclusions. PASS. One genuine extremal step: $\min_{1\le t\le a-2}t(a-1-t)=a-2$, by concavity with both endpoint values equal to $a-2$. I verified it. The other minima ($\frac{a-2}{a+2}\ge\frac15$ at $a=3$; $\frac{a-1}{7(a+2)}\ge\frac1{28}$ at $a=2$; $\frac a{a+2}\cdot\frac B{B+1}\ge\frac16$ at $a=B=1$) are monotone-in-parameter checks, all correct. Existence and boundedness of every infimum used are clear.
7. Consequences and composition. PASS. I checked each "hence" separately: (A)→(B)→(C); (D)+(E)→(F); (F)+(G)→(H)→level-$(n+1)$ classification; (I)→(J)/(K); (G)+(I)→(L); (L)+$\eta$+(C)→(M); (L)+(I)→(N); (L)+(J)→(P); (N)/(P)+(M)→(O)→terminality→(Q); (K)+(Q)→invariant→conclusion. Every consumed clause is supplied at its actual strength; in particular (P)'s use with $t=a$ needs only nonterminality at $n$ plus the definition of $d_{n+1}$, which is available.
8. Computation. INAPPLICABLE, with reason. The reconstruction contains no code, no numerical enumeration, no certified enclosure, and no claim resting on computation; it is a pure hand argument. There are therefore no exact inputs, ranges, coverage claims, or failure exits to audit. (My own arithmetic in W2 is a reviewer cross-check performed by hand; it is not part of the subject and I claim no tier from it.)
9. Reproduction. INAPPLICABLE, with reason. No rerun commands, no retained inputs, no coverage claims, and no cached success are asserted by the pages, because there is no computational leg. The only reproducibility obligation actually present — that the source resolve from an ordinary clone — is met: the PDF is the tracked LFS artifact kesten_1966_bounded_remainder.pdf, whose digest the source card then recorded and which I verified byte-for-byte; the file is identified by its path and its Git LFS pointer.
10. Source and verdict fidelity. PASS, with two minor characterization notes. All page/label/locator references I could check are correct: Theorem 4 on p. 193; definitions pp. 193–194; Theorem 1 geometry pp. 196–199; Section 4 pp. 204–212; Bohl on p. 205; accumulation = (4.19)–(4.20); terminal/transition = (4.24)–(4.31); endgame = pp. 211–212; $A_n=a'_{n+1}$, $Q_n=q'_{n+1}$ per (1.5)–(1.6); "Labels (A)–(Q) belong to this reconstruction, not to the source." The two notes are in §11.

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11. Convention drift, and source-side defects the reconstruction navigates

Between the two pages and the PDF — no substantive drift. Every definitional element matches: the half-open $[a,b)$, the index base $k\ge1$, $R=N-M(b-a)$, the theorem's hypotheses $0\le a<b\le1$ and $b-a<1$, the rational footnote, the CF conventions with $a_0$ dropped, and the $A_n/Q_n\leftrightarrow a'_{n+1}/q'_{n+1}$ dictionary. Minor, non-substantive items:

• theorem_4.md line 29 says bounded "as $M$ ranges over the positive integers"; Kesten's (4.2) says "all $M\ge0$". Immaterial since $R(0)=0$, and the page states that convention at line 353.
• The page reuses the letter $m$ as a counting index (line 24) and $m_r$ as a cell multiplicity (line 198–204), while Kesten uses $m=m(N,\xi)$ for the Ostrowski top index. No content collision — the source's $m$ never appears on the page — but a symbol-by-symbol reader should be warned.
• The page uses $\varepsilon_n=q_n\xi-p_n$; Kesten uses $\varepsilon_n$ on p. 207 for the stability threshold (which the page calls $\eta$). Again a pure letter collision, unflagged.
• For odd $n$ the page's $\lambda_r$ is defined by its own congruence $\sigma p_n\lambda_r\equiv r$, not literally by Kesten's (2.6). I verified the two coincide under his odd-$m$ reindexing ($1-Y_r$ lands in $(\frac{q-r-1}{q},\frac{q-r}{q})$, his odd-$m$ $P_r$), so this is unstated, not wrong.

Source-side defects I found in Kesten, which the reconstruction either repairs or is unaffected by (recorded because checklist item 10 requires fidelity of characterizations in both directions):

• p. 204, last line: "boundedness of $R(M,\xi,a,b)$ implies that $b=\{k\xi\}$ for some $\xi$" — should read $b-a=\{k\xi\}$ for some $k$. Typo; irrelevant to the reconstruction.
• p. 207: the chain $\{\sum e_jq_j\xi\}\le\sum|e_j|\{q_j\xi\}\le\sum a_{j+1}/q_{j+1}$ is false as printed ($\{q_j\xi\}=1-\delta_j$ for odd $j$); the braces must be read as $\|\cdot\|$, the distance to the nearest integer defined by Kesten in footnote 4 on p. 196 (the distinct grader's additional source observation). The reconstruction repairs this by using the signed $\varepsilon_j$ and (C), and its remark at line 367 ("without … confusing a fractional part with a small signed error") is a fair and correct observation about the source.
• p. 196 vs. (4.4) on p. 205: the odd-case $J_r$ is written once open, $(P_{r+1},P_r)$, and once half-open, $(P_{r_n+1},P_{r_n}]$. The reconstruction sidesteps this by working with strict inequalities under the standing hypothesis.
• One characterization nit against the reconstruction. Line 479–480 says its invariant "avoids taking an unjustified ordinary real limit through a fractional-part discontinuity." Kesten's limit on p. 212 is in fact justified: his $+\tfrac12(-1)^{n_3}-\tfrac12(-1)^{n}$ terms exactly compensate the oscillation of $\{q_{n-1}\xi\}$ between $\delta_{n-1}$ and $1-\delta_{n-1}$, and I verified the combination converges. The residual gap in the source is narrower — the final equality $\{\lambda\xi\}-\{q_{s-1}\xi\}+\tfrac12(1+(-1)^s)=\{(\lambda-q_{s-1})\xi\}$ needs a representative check Kesten does not state (it holds in both parities; I checked). So "unjustified" overstates the source's defect by a small margin. This is a prose calibration issue, not a mathematical defect, and it does not touch the reconstruction's own correctness. Line 490–491 already correctly declines to call any of this an author-issued erratum.

Do the pages claim more than this direction? No. See §2. The accepted coverage remains only the anchored irrational necessity direction.

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12. Verdict

refutation-failed.

Every essential deduction of the anchored irrational necessity reconstruction on theorem_4.md (retained as reviewed_theorem_4.md.txt) is correct and is supported by printed pp. 193–194, 196–199 and 204–212 of the canonical PDF. The reconstruction reproduces Kesten's own bounds term for term where they overlap ((4.16)$\equiv$(N), the $\tfrac14\cdot\tfrac58$ of p. 209, the $\tfrac16$ of p. 211, the final integer $\lambda(n_3)-q_{n_3-1}$), supplies ten steps the source elides — chiefly the odd-parity case, the stability threshold, and the terminal-transition/endgame invariant — and every supplied step is correct. The four attacks I mounted (case-exhaustion escape, cell-containment/grid-point, parity leakage, numerical) all failed. No defect found.

Limitations of this verdict.

• It covers only the claim restated in §1: anchored ($a=0$), irrational $\xi\in(0,1)$, $0<b<1$, interval $[0,b)$, necessity. It does not cover the Bohl arbitrary-translate reduction, rational $\xi$, the sufficiency direction, Theorem 1 beyond the consumed $N\le q_{n+1}$ geometry, Corollary 1, Theorems 2–3, or any formalization. It borrows no earlier acceptance verdict.
• The external premise Bohl [1], p. 226 is unread by me and is not consumed; likewise Hecke [6] and Ostrowski [9], [10]. Their content is outside this scope.
• Printed pp. 200–203 (Section 3, the metric result) were not read. I verified from Section 4's own internal citations that it references only Bohl, Theorem 1, (1.6), (2.5), (2.6), (2.10), and the external sufficiency papers — nothing from Section 3 — so this is a justified non-read, not a coverage gap.
• Preimage identity was established from the two supplied byte copies matching the pinned baseline digests, not from a substitute source or argument.
• Two basic continued-fraction identities are proved on the page by a compressed induction sketch. I completed both inductions and they hold, but a reader wanting a fully written proof will find the page terse there.
• This report is a reviewer's record only. Under docs/verification.md, a distinct grader must assess the report contract and independence. That distinct passing grade is now retained separately. I assert no tier; this library source review does not create a native claim tier.

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13. Source coverage — PDF pages read and what each supplied

• PDF sheet: 1

  • Printed pp.: 192 | 193
  • Read?: Yes
  • What it supplied: 192 = tail of the preceding Kesten–Sós article (not consumed). 193: definition of $N(M,\xi,a,b)$ with $a\le\{k\xi\}<b$ and $1\le k\le M$; (1.1) $R=N-M(b-a)$; Theorem 4 verbatim with (1.2); footnote 1 (rational $\xi$ trivial); footnote 2 ($a_0$ dropped); CF notation $[a_1,a_2,\dots]$

• PDF sheet: 2

  • Printed pp.: 194 | 195
  • Read?: Yes
  • What it supplied: 194: (1.3)–(1.4) recurrences; (1.5) $a'_{n+1}$; (1.6) $q'_{n+1}=a'_{n+1}q_n+q_{n-1}=q'_{n+2}/a'_{n+2}$; (1.7) Ostrowski expansion (read, not consumed); the "$N$ intervals, identify 0 and 1" convention. 195: Theorem 2 (Farey), Theorem 3 (metric) — statements only, not consumed; $q_{-1}=0$

• PDF sheet: 3

  • Printed pp.: 196 | 197
  • Read?: Yes
  • What it supplied: 196: end of Theorem 3's distribution formula; Section 2 preamble ($q_{-1}=0$); Theorem 1 statement, both parities, short/long lengths $a'_{m+1}/q'_{m+1}$ and $(a'_{m+1}+1)/q'_{m+1}$, long$\iff\lambda_r\le q_{m-1}$. 197: refinement clause ($c_m$ points, $J'_r$, $J''_r$, long-first ordering); proof begins, (2.1)

• PDF sheet: 4

  • Printed pp.: 198 | 199
  • Read?: Yes
  • What it supplied: 198: (2.2)–(2.8) — the residue map $\varrho_k$, one point per cell, $\lambda_r$ via (2.5)/(2.6), $\lambda_{r+1}-\lambda_r\equiv-q_{m-1}$, and the case split (2.8). 199: (2.9) the two interval lengths; (2.10) $\{(\lambda_r+sq_m)\xi\}=P_r+s/q'_{m+1}$ valid for $\lambda_r+sq_m\le q'_{m+1}$; long-first subdivision; Corollary 1 (read, not consumed)

• PDF sheet: 5–6

  • Printed pp.: 200–203
  • Read?: No — deliberate
  • What it supplied: Section 3 (metric result, Friedman–Niven / Farey techniques). Verified from Section 4's internal citations that nothing there is consumed

• PDF sheet: 7

  • Printed pp.: 204 | 205
  • Read?: Yes
  • What it supplied: 204: Section 4 opens; (4.1)/(4.2) sufficiency attributed to Hecke and Ostrowski ("we only have to prove that (4.1) is a necessary condition"); rational-$\xi$ remark. 205: the Bohl reduction ([1] p. 226) to $a=0$, $0<b<1$ — the exact boundary of the reviewed scope; (4.3)–(4.7): $P_r^{(n)},\lambda_r^{(n)},J^{(n)}_{r_n}\ni b$, permissible $d$ ranges ($a_{n+1}$ long / $a_{n+1}-1$ short), and the definition of $d_n$ with its odd-$n$ inequality reversal

• PDF sheet: 8

  • Printed pp.: 206 | 207
  • Read?: Yes
  • What it supplied: 206: $0\notin J(n)$ for large $n$; (4.8)–(4.9) with strictness once $b\ne\{k\xi\}$; (4.10); (4.11) $M_n=(d_n+1)q_n$; (4.12a–c) the interval/counting decomposition. 207: (4.13) $N(M_n)= (d_n+1)(r_n+1)$; (4.14)–(4.15); (4.16) — the bound the page reproduces as (N); (4.17)/(4.18) with the constant $1/28$; the one-clause stability threshold $\varepsilon_n$ and the $\le4/q_{n+s}$ tail bound (with the $\{\cdot\}$-vs-$\|\cdot\|$ slip)

• PDF sheet: 9

  • Printed pp.: 208 | 209
  • Read?: Yes
  • What it supplied: 208: (4.19)–(4.20) — the block-addition and reverse-order telescoping the page reproduces as (M); $R(M,b)\ge t/28$; (4.21a)/(4.21b) = the page's (O). 209: (4.22)–(4.23) the nonterminal transition $\lambda(n+1)=\lambda(n)+(d_n+1)q_n$ = the page's (J); the $d_{n+1}\ge a_{n+2}-2$ sharpening giving $\tfrac14\cdot\tfrac58$; (4.24a)/(4.24b) the terminal cases; (4.25)

• PDF sheet: 10

  • Printed pp.: 210 | 211
  • Read?: Yes
  • What it supplied: 210: (4.26a–c) the terminal transition $\lambda(n+1)=\lambda^{(n)}_{r_n+1}\le q_n$ and "$J(n+1)$ is long" = the page's (K); (4.27)–(4.31) the $d_{n+1}=a_{n+2}-1$ exclusion with $M_{n+1}=a_{n+2}q_{n+1}$. 211: the $\le-\tfrac16$ bound; the (i)/(ii)/(iii) case analysis; conclusion $d_n=a_{n+1}$ for $n\ge n_3$; $P(n+1)=P(n)+(-1)^n(a'_{n+1}+1)/q'_{n+1}$

• PDF sheet: 11

  • Printed pp.: 212 | back matter
  • Read?: Yes
  • What it supplied: 212: the telescoped limit and the final conclusion $b=\{(\lambda(n_3)-q_{n_3-1})\xi\}$ — the same integer the page's invariant produces; reference list (Bohl [1], Hecke [6], Ostrowski [9],[10], Sós [11],[12], Surányi [13]); "Reçu par la Rédaction le 25.3.1966"

Totals: 9 of 11 sheets read (printed pp. 192–199, 204–212); 2 sheets (printed pp. 200–203) deliberately not read and justified above. Reading was by direct page-image inspection of the canonical hash-verified PDF, not text extraction.