Independent examination of White's upper bound for Problem 1063

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Subject and independence

Reviewer: independent mathematical reviewer, disclosed model gpt-6-astra, reasoning effort max. Examination date: 2026-10-15 UTC. Frozen subject date: 2026-10-15 at 05:52:55.296610 UTC. Charge: attempt to refute the entire eventual quantitative bound and both of its stated asymptotic consequences, including every essential deduction of the supplied proof. This is a whole-proof examination, not an acceptance or grading record.

The convention is $n\ge2k$ with exactly one index $0\le j<k$ for which $n-j$ fails to divide $\binom nk$; it is never relaxed to at most one.

The assessed mathematical subject consists exactly of the following frozen fragments. The paths identify the source pages, not permission to read their other sections.

The first two fragments and their provenance belong with this report under wiki/problems/integer_sequences/E1063/claims/evidence/assets/white_review_subject/. The README is identified by its immutable external URL. Its copied text is private: no open license for republication was established, and this report does not reproduce it. The independent reasoning is retained below, with canonical home wiki/problems/integer_sequences/E1063/claims/evidence/verify/white_upper_bound/report.md. There is no computational warrant, keeper program, or executable rerun requirement. Credit for the source construction and bound remains with Patrick White and the collaborators credited by the source.

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The mathematical read set was only that manifest and its three enumerated fragments. Operating reading comprised the supplied root instructions; README.md; the introduction, Claims, Evidence, Tiers, and Project claims in docs/anatomy.md; docs/verification.md, docs/evidence.md, docs/approach_vetting.md, docs/math_authoring.md, and docs/weaving.md; campaign/AGENTS.md; campaign/guides/verification.md; and Reading in campaign/guides/research.md. The named charter, step instructions, configuration, template record, and applicable installed operating skills were also inspected. Other named operating files were checked for identity without using them as mathematical inputs. Operating examples and links were not followed into the corpus.

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Restatement

The variables $k,n,j,E,S,m$ are integers; logarithms are natural. For $k\ge2$, let

$$\mathcal A_k=\left\{n\ge2k: \#\left\{j\in\{0,\ldots,k-1\}:n-j\nmid\binom nk\right\}=1\right\}.$$

The extremal quantity is $n_k=\min\mathcal A_k$. The assertion examined is that absolute constants $C>0$ and $K$ exist such that, for every $k\ge K$,

$$n_k\le\exp\left(C\frac{k\log\log k}{\log k}\right).$$

The two further assertions are $\log n_k=o(k)$ and

$$\frac{n_k}{k\operatorname{lcm}(1,\ldots,k-1)}\longrightarrow0.$$

Exactly one failure is essential. The theorem is eventual; it asserts neither a numerical value of $K$ nor a sharp value of $n_k$. Nonemptiness and admissibility are examined below, including the definition for small $k$.

Premises and their actual basis

On 2026-10-15 UTC, reading the entire frozen statement, argument, and consequence sentences identified no consumed native L-claim or ledger row. This finding was derived from the mathematical text, independently of its negative dependency sentence. Consequently there is no local tier or staleness premise to check, no pending native grade, and no batch acceptance order. No ledger was read.

The following are the precise external interfaces allowed by the commission and used by the source. Here $\theta(t)=\sum_{p\le t}\log p$ and $\psi(t)=\sum_{p^a\le t,\,a\ge1}\log p$.

The exact source version and locators for these uses are the immutable README listed above. It supplies no separate bibliographic editions for the classical facts in the assessed fragments. Reliance is therefore explicitly on the named, permitted interfaces, not on an independently inspected literature proof or a source card's coverage label. The verdict below is relative to these interfaces as authorized; it does not certify their external proof coverage.

The pigeonhole argument, valuation comparisons, binomial estimate, finite counting, and asymptotic deductions are examined directly. Kummer's theorem is not an additional premise: the needed carry identity is derived from Legendre. Neither Erdős–Selfridge nor a Chinese remainder theorem is used. The small-$k$ parenthetical about Cambie's construction is unread support for that particular attribution. It is not needed for the eventual theorem, and the given finite construction itself supplies nonemptiness as checked below. No assertion about Cambie's proof is made.

Weakest steps and full proof coverage

1. Valuation coverage and the unique failure

Fix $1\le E<k$, set $r=k-1-E$ and $\rho=\max(E,r)$, and write $x=Sm$, $n=E+x$, and $W_j=n-j$. In the construction, the small, exceptional, and regular prime sets are subsets of the primes $p\le k$. This scope is required by equation (21), the bound $d\le\pi(k)$, and the separate treatment of $p>k$ in Section 4. An unbounded interpretation of the regular set would not support the displayed pigeonhole argument. Making this implicit scope explicit uses the scope already present in those proof sites.

To fix the notation independently, let $y=\max(E,\lfloor\sqrt k\rfloor)$ and, for $p\le k$, define

$$\begin{aligned} A_p&=\lfloor\log_p k\rfloor,\\ \sigma_p&=\sum_{a=1}^{A_p} \mathbf1_{E\bmod p^a<k\bmod p^a},\\ T_p&=\max_{\substack{0\le j<k\\j\ne E}}v_p(E-j),\qquad H_p=\max(0,T_p-\sigma_p). \end{aligned}$$

Valuations of negative nonzero integers mean valuations of their absolute values. A prime $y<p\le k$ is exceptional when $k\bmod p<2E$ and regular otherwise. The factors prescribed for $S$ are $p^{A_p+H_p}$ for small primes and $p^2$ for exceptional primes, together with $p_*^{A_{p_*}+1}$ for the selected prime $p_*$ described below.

For a fixed $p\le k$, abbreviate $A=A_p$. At level $q=p^a$, writing $n=uq+v$ and $k=hq+s$ with $0\le v,s<q$ shows that the number of multiples of $q$ in the window, denoted $c_{p,a}(n)$, is $h+\mathbf1_{v<s}$. In particular, the extra term is zero when $q\mid k$. Legendre therefore gives

$$v_p\binom nk=\sum_{a\ge1}\varepsilon_a, \qquad \varepsilon_a=\mathbf1_{n\bmod p^a<k\bmod p^a}.$$

These sums are finite: when $p^a>n\ge k$, the count and the carry are zero.

Divisibility of $W_j$ requires the valuation inequality for all primes, not for one selected prime. This is the universal interpretation of Section 1's equations (1)–(2). For a fixed prime its deficit is

$$v_p\binom nk-v_p(W_j) =\sum_{a=1}^{A}\varepsilon_a-\min(v_p(W_j),A) +\sum_{a>A}\left(c_{p,a}(n)-\mathbf1_{p^a\mid W_j}\right).$$

Every summand in the final sum is nonnegative: it counts multiples elsewhere in the window. For $p^a>k$ there is at most one multiple, including at a boundary of the window. Thus a very high power on a coordinate $j\ne E$ cannot invalidate the truncated covering lemma. The equality case $p^a=k$ belongs in the initial sum; it has no extra carry. For $p>k$, the simpler identity $v_p(k!)=0$ already gives $v_p\binom nk=\sum_i v_p(W_i)\ge v_p(W_j)$.

This also checks Section 2's obstruction at its stated scope. If $q\mid k$ and $q<k$, then $q\le k/2$ and the window has at least two multiples of $q$. At least one is not the excused coordinate. Its removal lowers that level's count below $\lfloor k/q\rfloor$, so a requirement imposed separately at each prime-power level cannot succeed. This is an obstruction to that sufficient recipe, not to the valuation sum or to every possible proof.

For the accounting identity, write $E=uq+e$ and $r=vq+f$. Since $k=E+r+1$, the difference $\lfloor k/q\rfloor-\lfloor E/q\rfloor-\lfloor r/q\rfloor$ is zero when $e+f+1<q$ and one otherwise. In the former case $e<k\bmod q$; in the latter case $e\ge k\bmod q$. Summing gives

$$A_p-\sigma_p=v_p(D),\qquad D=\frac{k!}{E!r!}=k\binom{k-1}{E}.$$

The nonzero differences $E-j$ have absolute values covering the integers from $1$ through $\rho$. Consequently $T_p=\lfloor\log_p\rho\rfloor\le A_p$, and $H_p=\max(0,T_p-\sigma_p)\le A_p-\sigma_p$. This includes $r=0$ and $\rho=1$, with $T_p=0$ in the latter case. There is no valuation of zero in $T_p$, because $j=E$ is excluded.

Now check each prime class for $j\ne E$, writing $d_j=E-j$.

These cases cover all primes. A possible concern is the removal of $p_*$ from the regular set. In fact, no regular prime can be chosen as $p_*$: if $p>y\ge E$ is regular, then $A_p=1$ and $E<2E\le k\bmod p$, so $\sigma_p=1$ and $v_p(D)=0$. Thus the selected prime is already in a covered class. Its extra anchor only strengthens a divisibility condition used above.

It remains to prove a failure rather than just an upper bound on their number. The integer $D\ge k>1$ has a prime divisor $p_*$, necessarily at most $k$ since $D$ divides $k!$. Let $A=A_{p_*}$ and $b=v_{p_*}(x)\ge A+1$, as forced by the augmentation. The first $A$ carries total $\sigma_{p_*}$, and the next $b-A$ carries are one. Write $x=p_*^b u$ with $p_*\nmid u$. Its next residue satisfies

$$n\bmod p_*^{b+1}=E+p_*^b(u\bmod p_*), \qquad k<p_*^b<E+p_*^b(u\bmod p_*)<p_*^{b+1}.$$

Residues modulo subsequent powers cannot decrease: each adds a nonnegative multiple of the previous modulus. Thus all later carries vanish, including when $m$ has made $b$ arbitrarily large. It follows that

$$v_{p_*}\binom nk=\sigma_{p_*}+b-A =b-v_{p_*}(D)<b=v_{p_*}(W_E).$$

The proof for $j\ne E$ and this strict deficit establish exactly one failure. No theorem asserting that some failure must exist is imported.

2. A common multiplier and an admissible integer

Choose $S$ to be the least positive integer satisfying the prescribed prime power divisibilities. Such an integer exists, being the least common multiple of a finite list. This makes the size accounting precise; arbitrary extra factors are not silently charged the same bound. The selection of $p_*$ and its anchor use only $k$ and $E$, not the unknown $m$.

Let $\mathcal R$ be the finite regular set, $d=|\mathcal R|$, and $Q=\lceil k/E\rceil\ge2$. Partition $[0,1)^d$ into half-open cubes of side $1/Q$. Use the $Q^d+1$ points $P_t=(\{tS/p\})_{p\in\mathcal R}$ for $t=0,\ldots,Q^d$, where braces denote fractional part. Two lie in one cube. Order their indices and subtract to obtain $1\le m\le Q^d$. In every coordinate the difference of fractional parts has absolute value strictly less than $1/Q$, and it differs from $mS/p$ by an integer. Hence there is an integer residue $z_p$ of $mS$ modulo $p$ satisfying

$$|z_p|<\frac pQ\le\frac kQ\le E.$$

For a regular prime, $k\bmod p\ge2E$, so

$$1\le E+z_p\le2E-1<k\bmod p<p.$$

Thus the signed approximation cannot wrap past zero or $p$, and it gives exactly the carry required in Step 1. The argument needs no independence or irrationality of the numbers $S/p$, no equidistribution, no coprimality condition, and no probabilistic assumption. If $d=0$, take $m=1$; the empty-dimensional assertion is vacuous and the same bound $m\le Q^0$ holds. In particular, the construction supplies an actual common positive integer, not a list of separately realizable congruences.

Admissibility survives two checks. The source's eventual check uses $S\ge\prod_{p\le y}p$ and the permitted lower bound on $\theta(y)$; with the Section 5 choice of $E$, it gives $S>2k$ for large $k$. There is also a direct check from the same anchor requirements, which addresses small $k$ without reading another construction. Every prime factor of $k$ is either small, in which case its exponent in $S$ is at least its exponent in $k$, or large, in which case it has exponent one in $k$ and is exceptional because $k\bmod p=0<2E$. Therefore $k\mid S$. The $p_*$ anchor gives $S\ge p_*^{A_{p_*}+1}>k$. Consequently

$$S\ge2k,\qquad n=E+Sm\ge2k+1.$$

This uses only the prescribed divisibilities. It introduces no new construction or external premise. It also places $n$ in the positive factorial domain before any use of the carry formula. For any $k\ge2$, the allowed choice $E=1$ in Sections 3–4 now gives an element of $\mathcal A_k$. Well-ordering gives its least element $n_k$. This all-$k$ existence check is an explicit reconstruction from the supplied parameterized construction; it is not a verification of the Cambie parenthetical or an all-$k$ extension of the quantitative estimate.

3. Uniform cost and each asymptotic consequence

Put $L=\log k$ and $F(k)=k\log L/L$. For sufficiently large $k$, the specified $E=\lceil k/L^3\rceil$ satisfies

$$\frac{k}{L^3}\le E\le\frac{2k}{L^3},\qquad \sqrt k<E<k,\qquad y=E.$$

All subsequent constants and thresholds are absolute. This is a statement for every sufficiently large integer, with no selection of a subsequence.

For the small-prime base, all prime powers with exponent at least two have prime base at most $\sqrt k<y$. Thus

$$\sum_{p\le y}A_p\log p =\theta(y)+\psi(k)-\theta(k) =O\left(\frac{k}{L^3}+\sqrt k\,L\right).$$

The higher-power estimate can also be seen directly: for each $p\le\sqrt k$, its contribution $(A_p-1)\log p$ is at most $L$; there are at most $\lfloor\sqrt k\rfloor$ such primes. No prime-distribution claim beyond the separately stated bound on $\theta(y)$ is needed for this term.

For the extra depth, the accounting identity gives $\sum_{p\le y}H_p\log p\le\log D$. The elementary estimates $E!\ge(E/e)^E$ and $\binom{k-1}{E}\le k^E/E!$ yield

$$\log D\le L+E\log(ek/E) \le L+\frac{2k}{L^3}(1+3\log L).$$

For completeness, the factorial inequality follows from $\sum_{j=1}^E\log j\ge\int_1^E\log t\,dt=E\log E-E+1$. This bound includes the endpoint $E=1$ and uses no unproved entropy estimate.

For exceptional primes, let $M=\lfloor k/y\rfloor\le L^3$. Each such prime has a unique positive integer $\ell=\lfloor k/p\rfloor\le M$ and lies in the interval $((k-2E)/\ell,k/\ell]$. An interval of that length has at most $2E/\ell+1$ integers. Counting all integers overcounts the primes and therefore remains an upper bound, regardless of overlaps. It follows that

$$\begin{aligned} \sum_{\substack{y<p\le k\\p\text{ exceptional}}}\log p &\le L\left(2E\sum_{\ell=1}^{M}\frac1\ell+M\right)\\ &\le L\bigl(2E(1+\log M)+M\bigr)\\ &=O\left(\frac{k\log L}{L^2}+L^4\right). \end{aligned}$$

The square anchors double this contribution. Adding the forced prime's anchor costs at most $(A_{p_*}+1)\log p_*\le2L$, even if none of that exponent had already been included. Overlaps between prescriptions only reduce the cost of their least common multiple. Combining these explicit estimates gives

$$\log S =O\left(\frac{k\log L}{L^2}+\sqrt k\,L+L^4\right) =o(F(k)).$$

The last equality is stronger than merely $o(k)$. Dividing its three terms by $F(k)$ gives respectively $1/L$, $L^2/(\sqrt k\log L)$, and $L^5/(k\log L)$, all tending to zero. The source's final appeal to its summary $\log S=o(k)$ is compressed: that summary alone would not imply the quantitative theorem. The explicit estimates preceding it do, as this substitution shows. No new estimate or missing essential lemma is needed.

The approximation dimension satisfies $d\le\pi(k)=O(k/L)$, while $Q\le L^3+1$ and $\log Q=O(\log L)$. Hence

$$\log m\le d\log Q=O(F(k)).$$

Since $E<k\le S/2\le Sm/2$, we have $n=E+Sm\le2Sm$ and therefore $\log n\le\log2+\log S+\log m=O(F(k))$. This supplies absolute $C,K$ for the constructed admissible $n$, and $n_k\le n$ supplies the bound in the stated extremal units. The original admissibility estimate is valid as well: $\theta(E)\ge c_0E$ and $E/\log k\to\infty$ imply $\log S>\log(2k)$ eventually.

For the first consequence, the two-sided inequality

$$0\le\frac{\log n_k}{k} \le C\frac{\log\log k}{\log k}$$

forces $\log n_k=o(k)$; it is not an inference from a finite set of $k$. For the second, prime factorization gives the exact identity $\log\operatorname{lcm}(1,\ldots,k-1)=\psi(k-1)$. The permitted prime number theorem gives $\psi(k-1)=k+o(k)$, including the shift by one. Consequently

$$\log\frac{n_k}{k\operatorname{lcm}(1,\ldots,k-1)} =\log n_k-\log k-\psi(k-1)=-k+o(k).$$

Thus the ratio tends to zero, and the source's stated exponential decay on the $k$ scale is justified. Both consequences have been checked in their own variables, rather than inferred from a general description of the construction.

Strongest attempted refutation and other attacks

The strongest attack was to make the prime-power argument lose exactness: cancellation in a nonexcused coordinate might create a second failure, or additional carries above the forced valuation might remove the claimed failure at $E$. Both would invalidate the target even if all size estimates were sound.

A concrete cancellation test within the specified construction is $k=7$, $E=2$, $S=44100=2^2\cdot3^2\cdot5^2\cdot7^2$, and $m=1$. Here $D=105$ permits $p_*=3$, all large primes are exceptional, and the small prime $2$ has $A_2=T_2=\sigma_2=2$ and $H_2=0$. For $j=6$, the difference $E-j=-4$ has valuation two but

$$n-j=44096=2^6\cdot689$$

has valuation six. A proof equating these valuations in the $H_p=0$ case would fail on this exact input. The actual truncated lemma survives: at $n=44102$, the six carry levels for $p=2$ are all present. The residues modulo $2,4$ are $0,2$, and modulo $8,16,32,64$ they are all $6$; the residue modulo $128$ is $70$, so no subsequent level contributes. This is exposed exact arithmetic, not a computational or finite-to-infinite warrant.

For the exceptional coordinate, a separate deletion calculation attacks the source's assertion about the infinite tail without assuming it. At $p_*$, we have $p_*^b>k>|E-j|$ for every $j\ne E$. Therefore

$$\sum_{j\ne E}v_{p_*}(n-j) =\sum_{j\ne E}v_{p_*}(E-j)=v_{p_*}(E!r!).$$

Subtracting the factorial denominator gives the exact identity

$$v_{p_*}(n-E)-v_{p_*}\binom nk=v_{p_*}(D)>0.$$

This confirms the strict failure through a finite product identity, as well as through the carry-tail proof in Step 1. Increasing $v_{p_*}(m)$ does not alter the deficit. These two derivations do not invoke a theorem about the existence of a nondivisor.

Other attempted failure mechanisms were checked as follows.

• Signed residues crossing the permitted interval: the strict inequality $|z_p|<E$ and the full margin $k\bmod p\ge2E$ place $E+z_p$ in $[1,2E-1]$, so neither endpoint can produce a lost carry.
• A prime omitted when $p_*$ is removed: a regular prime has $A_p-\sigma_p=0$, so it cannot be $p_*$. Primes above $k$ are covered directly by the denominator valuation.
• A nonexistent or too-small construction: the finite pigeonhole gives a positive $m$ even for $d=0$, while $k\mid S$ and $S>k$ give $n\ge2k$. No separate congruence solutions are being composed without a common witness.
• An asymptotic weakening hidden in the cost summary: $o(k)$ alone would be insufficient, since, for example, $k/\sqrt{\log k}$ is $o(k)$ but larger than $F(k)$. The explicit anchor estimates instead give $o(F(k))$.
• A stronger consequence than the proof supplies: each limit is derived separately, and the lcm limit's stronger external premise is identified.

The reconstructions above make explicit the prime universe, the least anchor choice, empty-dimensional pigeonhole, elementary accounting, all-$k$ nonemptiness from the same construction, and the substitution of explicit cost estimates. They neither modify the frozen subject nor replace an unsupported essential step with a different proof. No missing essential lemma was found in the commissioned theorem and consequences.

Checklist

Canonical failure modes

Named patterns

The neighboring disciplines are also respected: there are no statistical samples for which independence or correlation must be inferred, and every retained argument is matched to its stated scope. The finite arithmetic example is not assigned universal coverage.

Erdos-specific checklist

This is a single noncomputational examination. A computational batch's structurally different reproduction leg and a second cycle's disjoint-attack requirement are inapplicable; no extra review cycle or unseen prior attack record is claimed.

Verdict and grading

refutation-failed. The supplied argument survives this attempted refutation of the whole commissioned quantitative theorem and both asymptotic consequences, with the explicitly permitted classical interfaces at the reading depths stated above. The examination covers existence and admissibility, every prime and prime-power case, the unique failed divisor, the common multiplier, all quantitative costs, and the minimum's asymptotic consequences. No counterexample or unsupported essential step was found.

The quantitative conclusion uses the source's explicit cost estimates, not the weaker $o(k)$ summary alone. The prime universe and least anchor choice are made explicit, and the additional nonemptiness check is derived from the very same prescribed factors. These are disclosed reconstructions, not an assessment of repaired source bytes.

The warrant excludes source checker code, finite checker claims, prior reviews, Cipollini's proof, LEAP developments, Cambie's construction, and any statement outside the supplied Theorem and Sections 1–5. It does not independently certify the unread external proofs of Chebyshev estimates or the prime number theorem. It provides no numerical optimal constants, no Lean verification, no report grade, and no alteration of tier or catalog standing. A separate eligible grader must judge the report's contract and independence before any acceptance rests on it.