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The period in the literal v1 Proposition 3.2 can contain prime powers that divide no admissible denominator. Its exact-target conclusion is therefore false, even when all its displayed parameter and probability conditions hold. This is an obstruction to that auxiliary statement; the counting target one is treated by the actual-period replacement.

Source. Liu–Sawhney, arXiv:2404.07113v1, Proposition 3.2, pp. 9–10. The counterexample below is a compilation deduction, not an author erratum or a claim about the uninspected published version.

Bears on. Problem 297, by clarifying the exact Fourier input to the counting proof.

Proof

Fix any positive constant $C$ in the printed parameter assumptions. Take arbitrarily large prime integers $N$, and put

$$L=\log N,\qquad \ell=\log\log N,\qquad M=\lfloor N/10\rfloor, \quad K=\left\lfloor10^{-7}N/L\right\rfloor, \quad S=\lfloor N/L^4\rfloor.$$

For sufficiently large $N$, these obey

$$N^{.9999}\le S\le K\le M\le N/10, \qquad N/L^{10}\le K\le10^{-7}N/L,$$

and both printed upper bounds on $S$. Indeed,

$$\frac{S}{M^2/(CN)}\sim\frac{100C}{L^4}\longrightarrow0, \qquad \frac{S}{K^3/(CN^2\ell^5)} \sim\frac{C10^{21}\ell^5}{L}\longrightarrow0.$$

Let $A$ be exactly the set in the printed statement. Its restriction $\Omega(N)\le10\ell$ holds because $N$ is prime. Its other restrictions are that $n\in[M,N]$ has all prime-power divisors at most $S$ and $\widetilde\Omega(n)\le5\ell$.

The maximum-exponent exceptions are a subset of $\{n\le N:\Omega(n)>5\ell\}$, whose count is $o(N)$ by the proved reciprocal-mass bound multiplied by $N$. For $t=N/S\sim L^4$, the prime-power deletion bound removes $O(N\ell/L)=o(N)$ more integers. Every removed denominator in this interval is at least $M\sim N/10$, so its reciprocal contribution is at most $1/M$. Thus

$$R(A):=\sum_{n\in A}\frac1n=\log10+o(1).$$

Every member of $A$ has 2-adic exponent at most $\lfloor5\ell\rfloor$. A finite sum of their reciprocals has reduced denominator dividing $\operatorname{lcm}(A)$, so its 2-adic denominator exponent is also at most $\lfloor5\ell\rfloor$.

In contrast, the printed period is

Q_{\rm src}=\operatorname{lcm}\{q\le S:q\text{ is a prime power}}.

Its 2-adic exponent is $\lfloor\log_2S\rfloor>5\ell+1$ eventually. In particular $Q_{\rm src}/2$ is even. Set

$$x=Q_{\rm src}/2+1,\qquad \tau=x/Q_{\rm src},\qquad p=\tau/R(A).$$

The integer $x$ is odd and lies in $[1,Q_{\rm src}]$. Moreover, $p\to1/(2\log10)\in(0,1/2)$, so $1/\ell\le p\le1/2$ eventually. Select each member of $A$ independently with this common probability. Then $\mathbb ER(B)=pR(A)=x/Q_{\rm src}$, as required by the printed statement. But that target has the full 2-adic denominator exponent of $Q_{\rm src}$ and cannot be attained. Its probability is zero instead of at least $1/(4Q_{\rm src})$.

The counterexample remains valid if the source's $\Omega(N)$ is first changed to the intended member-wise $\Omega(n)\le10\ell$. That extra restriction removes only $o(N)$ integers by the same reciprocal bound, so $R(A)=\log10+o(1)$ and the entire argument persist.

Scope

The period defect is independent of the uppercase-variable typo. Using the actual period $Q=\operatorname{lcm}(A)$ removes this obstruction. For the counting target one, choose the integer $x=Q$; there is no 2-adic target obstruction. The counting proof also checks all other hypotheses of the sufficient restricted proposition.