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Source. Wouter van Doorn and GPT-6 Astra Pro (the author line as printed), Practical numbers and Egyptian fractions, the definitions of $Q(k)$ and $t(k)$ and Lemma 3.3 with its proof, physical p. 4 of the seven-page PDF held by van Doorn (2026). The displays were read on the page image. Uses the Lemma 3.2 reconstruction; consumed by the Corollary 3.4 reconstruction and the Proposition 4.1 reconstruction.

Standing. Author-recorded reconstruction of a claimed result (see the Lemma 3.1 page for the note's standing); not an independent review; changes no status and assigns no tier. Imported inputs: the prime number theorem in the form $\pi(2Q)-\pi(Q)\sim Q/\log Q$ as $Q\to\infty$, of which only the lower bound $\pi(2Q)-\pi(Q)\gg Q/\log Q$ is consumed; Hölder's inequality; and the inequality $(a+b)^\alpha\le a^\alpha+b^\alpha$ for $a,b\ge0$, $0<\alpha\le1$.

Definitions

$c_0=14/\log2$, all logarithms being natural (the note fixes this at the end of its Section 1, physical p. 2), so that $14/c_0=\log2$. For integers $k\ge3$,

$$Q(k)=k^6\log k,\qquad t(k)=\Bigl\lfloor\frac{14k}{c_0\,(7\log k+3\log\log k)}\Bigr\rfloor =\Bigl\lfloor\frac{k\log2}{7\log k+3\log\log k}\Bigr\rfloor .$$

$\omega(n)$ is the number of distinct prime factors of $n$; $D(n)$, $M_d(X)$ and the representation (3.2) are as on the Lemma 3.2 page. For a finite set $J$ of primes, $a_J=\prod_{p\in J}p$. All $O$, $\ll$ and $o$ constants below are absolute.

Statement

For every sufficiently large $k$, every odd prime $p_*$, and every positive odd squarefree integer $V$ with $\omega(V)=k$ and all prime factors at most $2Q(k)$, there is an odd squarefree integer $A>1$ with

$$(A,p_*V)=1,\qquad\omega(A)=t(k),$$

all prime factors of $A$ in $(Q(k),2Q(k)]$, and every residue modulo $A$ represented as in (3.2) with $z_0,\dots,z_3\in D(V)$. The threshold for $k$ does not depend on $p_*$ or $V$.

Proof

Write $u=\log k$, $Q=Q(k)$ and $t=t(k)$; $k$ is large, so $t\ge1$ and $t\le k\log2/(7u)$.

The divisor sets. Split the prime factors of $V$ into two disjoint sets of sizes $\lfloor k/2\rfloor$ and $\lceil k/2\rceil$, and let $V_1,V_2$ be their products; then $V_1,V_2$ are coprime, odd and squarefree with $V=V_1V_2$. With $X_i=D(V_i)$ and $X=D(V)$,

$$|X_1|=2^{\lfloor k/2\rfloor},\qquad|X_2|=2^{\lceil k/2\rceil},\qquad|X|=2^k ,$$

so $|X_i|^{-1}\le\sqrt2\cdot2^{-k/2}$.

The prime pool. Let $\mathcal P$ be the set of primes in $(Q,2Q]$ that divide neither $V$ nor $p_*$. The prime number theorem gives $\pi(2Q)-\pi(Q)\sim Q/\log Q$, and $\log Q=6u+\log u\sim6u$, so, since at most $k+1$ primes are excluded,

$$|\mathcal P|\sim\frac Q{\log Q}\sim\frac{k^6}6 ,$$

uniformly in $V$ and $p_*$. Every element of $\mathcal P$ is an odd prime exceeding $Q$.

Few primes of the pool divide a difference. Every element of $X_1$, $X_2$ or $X$ lies in $[1,V]$, so a nonzero difference $\delta$ of two elements of one of these sets has $0<|\delta|<V$. If $r$ distinct primes of $\mathcal P$ divide $\delta$ then $Q^r<|\delta|<V$, so $r<\log V/\log Q$. Hence at most

$$R=\Bigl\lfloor\frac{\log V}{\log Q}\Bigr\rfloor\le\frac{k\log(2Q)}{\log Q}\le2k$$

primes of $\mathcal P$ divide $\delta$, using $\log V\le k\log(2Q)$, as $V$ has $k$ prime factors each at most $2Q$. Put $\rho=R/|\mathcal P|$; then $\rho\ll k^{-5}$, uniformly.

Random subsets. For $1\le s\le t$ let $J$ be a uniformly random $s$-element subset of $\mathcal P$. For a fixed nonzero difference $\delta$, $a_J\mid\delta$ if and only if every prime of $J$ divides $\delta$, and at most $R$ primes of $\mathcal P$ do, so

$$\Pr\bigl(a_J\mid\delta\bigr)\le\binom Rs\Big/\binom{|\mathcal P|}s =\prod_{i=0}^{s-1}\frac{R-i}{|\mathcal P|-i}\le\rho^s .$$

For $Y\in\{X_1,X_2,X\}$, $M_{a_J}(Y)$ is $|Y|^{-2}$ times the number of pairs $(x,y)\in Y^2$ with $a_J\mid x-y$; the $|Y|$ diagonal pairs always count, and each of the other pairs counts with probability at most $\rho^s$. Taking expectations,

$$\mathbb E_{|J|=s}\,M_{a_J}(Y)\le|Y|^{-1}+\rho^s\qquad(Y\in\{X_1,X_2,X\}). \tag{3.3}$$

Hölder's inequality with three exponents $3$ gives

$$\mathbb E_{|J|=s}\bigl(M_{a_J}(X_1)M_{a_J}(X_2)M_{a_J}(X)\bigr)^{1/3} \le\prod_{Y}\bigl(\mathbb E_{|J|=s}M_{a_J}(Y)\bigr)^{1/3} \ll\bigl(2^{-k/2}+\rho^s\bigr)^{2/3}\bigl(2^{-k}+\rho^s\bigr)^{1/3},$$

the last by (3.3) and $|X_i|^{-1}\le\sqrt2\cdot2^{-k/2}$.

The random modulus. Let $I$ be a uniformly random $t$-element subset of $\mathcal P$ and $A=a_I$; then $A$ is odd, squarefree, coprime to $p_*V$, with $\omega(A)=t$ and all prime factors in $(Q,2Q]$. Its divisors $d>1$ are the $a_J$ with $\varnothing\ne J\subseteq I$, and with $w=(2Q)^{2/3}$ one has $a_J^{2/3}\le w^{|J|}$. Let $S$ be the sum (3.1) of Lemma 3.2 for this $A$ and the sets $X_1,X_2,X$. A uniformly random $s$-subset of a uniformly random $t$-subset of $\mathcal P$ is a uniformly random $s$-subset of $\mathcal P$, and $I$ has $\binom ts$ subsets of size $s$, so

$$\mathbb E_I\,S\le\sum_{s=1}^{t}\binom ts w^s\, \mathbb E_{|J|=s}\bigl(M_{a_J}(X_1)M_{a_J}(X_2)M_{a_J}(X)\bigr)^{1/3} \ll\sum_{s=1}^{t}\binom ts w^s \bigl(2^{-k/2}+\rho^s\bigr)^{2/3}\bigl(2^{-k}+\rho^s\bigr)^{1/3}.$$

By $(a+b)^\alpha\le a^\alpha+b^\alpha$,

$$(2^{-k/2}+\rho^s)^{2/3}(2^{-k}+\rho^s)^{1/3} \le(2^{-k/3}+\rho^{2s/3})(2^{-k/3}+\rho^{s/3}),$$

whose expansion has the four terms $2^{-2k/3}$, $2^{-k/3}\rho^{s/3}$, $2^{-k/3}\rho^{2s/3}$ and $\rho^s$. Summing each against $\binom tsw^s$ by the binomial theorem (adding the $s=0$ term where it helps, and subtracting it in the last),

$$\mathbb E_I\,S\ll2^{-2k/3}(1+w)^t+2^{-k/3}(1+w\rho^{1/3})^t +2^{-k/3}(1+w\rho^{2/3})^t+\bigl((1+w\rho)^t-1\bigr).$$

All four terms tend to zero. Here $w=2^{2/3}k^4u^{2/3}$, so $\log(1+w)=4u+\tfrac23\log u+O(1)$, and $\rho\ll k^{-5}$ gives $\log(1+w\rho^{1/3})\le\tfrac73u+\tfrac23\log u+O(1)$ and $\log(1+w\rho^{2/3})\le\tfrac23u+\tfrac23\log u+O(1)$. Write $\tau=k\log2/(7u+3\log u)$, so $t\le\tau$.

The logarithm of the first term is at most

$$-\tfrac23k\log2+\tau\,\bigl(4u+\tfrac23\log u+O(1)\bigr) =k\log2\,\bigl(-\tfrac23+\tfrac47+o(1)\bigr) =\Bigl(-\frac4{3c_0}+o(1)\Bigr)k ,$$

because $\tau(4u+\tfrac23\log u+O(1))=\tfrac47k\log2\,(1+O(\log u/u))$ and $\tfrac2{21}\log2=\tfrac4{3c_0}$.

The logarithm of the third term is at most

$$-\tfrac13k\log2+\tau\,\bigl(\tfrac23u+\tfrac23\log u+O(1)\bigr) =k\log2\,\bigl(-\tfrac13+\tfrac2{21}+o(1)\bigr) =\Bigl(-\frac{10}{3c_0}+o(1)\Bigr)k .$$

For the second term, write $-\tfrac13k\log2=-\tau\,(\tfrac73u+\log u)$; its logarithm is at most

$$-\tau\Bigl(\tfrac73u+\log u\Bigr)+\tau\Bigl(\tfrac73u+\tfrac23\log u+O(1)\Bigr) =\tau\Bigl(-\tfrac13\log u+O(1)\Bigr),$$

which tends to $-\infty$ because $\tau\to\infty$ and $-\tfrac13\log u+O(1)\to-\infty$.

For the fourth term, $tw\rho\le\tau w\rho\ll(k/u)\,k^4u^{2/3}\,k^{-5}=u^{-1/3}$, so $(1+w\rho)^t-1\le e^{tw\rho}-1=O(u^{-1/3})=o(1)$.

All estimates depend on $k$ alone. Hence $\mathbb E_I\,S<1$ once $k$ exceeds an absolute threshold, so some $I$ has $S<1$, and Lemma 3.2 applied to $A=a_I$ with $V_1,V_2$ gives the representation (3.2) of every residue modulo $A$.

Qualifications

• The prime number theorem is used only through the lower bound $|\mathcal P|\gg Q/\log Q\asymp k^6$, hence $\rho\ll k^{-5}$, which Chebyshev-type estimates also supply; the bound $tw\rho\ll u^{-1/3}$ on the fourth term uses $\rho\ll k^{-5}$ in full, and a prime count weaker by a factor $u^{1/3}$ or more would not close it. The note cites the theorem itself.
• The note writes $|X_1|,|X_2|\asymp2^{k/2}$; the exact values are recorded above and give the same bound.