Overlapping parent preparations

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A finite preparation lemma: proved

Let $\mathcal P$ be the odd primes at most $z$, and put

$$\lambda=\sum_{p\in\mathcal P}\frac1p,\qquad k=\lfloor\lambda\rfloor,\qquad r=\lfloor2\lambda-2\sqrt\lambda\rfloor.$$

As $z\to\infty$, there is a finite singleton sieve whose distinct moduli are squarefree products of $k$ primes of $\mathcal P$, with full survivor set $S$, and a measurable set $G\subseteq S$ such that

$$\mu(G)\ge\tfrac12-o(1). \tag{1}$$

Each $x\in G$ belongs to a parent cylinder $C(x)$ whose modulus is a product of $r$ primes of $\mathcal P$, and

$$\frac{\mu(S\cap C(x))}{\mu(C(x))}\le e^{-3\lambda/2}. \tag{2}$$

The parents are auxiliary cylinders, not installed congruences. They can overlap and can have the same modulus with different residues. The actual preparatory sieve uses at most one residue per modulus. Thus (1)--(2) establish a concrete heavily overlapping preparation, outside the disjoint-parent bounds in prepared-block entropy. They do not construct the later singleton congruences needed for a counterexample.

Random singleton preparation

Define

$$R(x)=\#\{p\in\mathcal P:x\bmod p\in\{0,1\}\},\qquad T=\{|R-2\lambda|\le2\sqrt\lambda\}.$$

Haar independence gives $\mathbb ER=2\lambda$ and $\operatorname{Var}R\le2\lambda$, so $\mu(T)\ge1/2$. Put

$$b=\lceil2\lambda+4\sqrt\lambda\rceil,\qquad B=\binom bk.$$

For each $k$-element subset $J\subseteq\mathcal P$, independently install its product modulus with probability $2^k/B$, and, if installed, choose its CRT digits independently and uniformly in $\{0,1\}$ at the primes of $J$. Since $b\ge2k$, $\binom bk\ge2^k$, so this is a valid probability.

For a fixed $x$, precisely $\binom{R(x)}k$ subsets could hit it. Each hits with probability $1/B$, independently over subsets. Therefore the exact probability of surviving the random preparation is

$$f(R(x))=\left(1-\frac1B\right)^{\binom{R(x)}k}. \tag{3}$$

On $T$,

$$1-f(R(x)) \le\frac{\binom{R(x)}k}{B} \le(3/4)^{2\sqrt\lambda-1}=o(1) \tag{4}$$

for sufficiently large $\lambda$. For the last bound, increase the upper binomial argument from $R(x)$ to $b$ one step at a time: each reverse ratio is at most $3/4$, and there are at least $2\sqrt\lambda-1$ steps.

Conditional survival in every parent

A parent $C$ fixes $r$ success coordinates and their $0/1$ values. Under Haar law conditioned on $C$,

$$R=r+V,\qquad \mathbb EV=2\lambda-O(\log\lambda), \tag{5}$$

uniformly over the parent. Indeed the removed reciprocal sum is at most $H_{r+2}$, even if the chosen primes are the smallest available.

The elementary exponential-moment bound for this sum of independent Bernoulli variables gives

$$\Pr(V<12\sqrt\lambda) \le\exp\{-2\lambda+O(\sqrt\lambda\log\lambda)\}. \tag{6}$$

For example use $\Pr(V<t)\le u^{-t}\mathbb E u^V \le u^{-t}\exp(-(1-u)\mathbb EV)$, with $u=\lambda^{-1/2}$ and $t=\lceil12\sqrt\lambda\rceil$. Above that cutoff, $r+V$ exceeds $b$ by at least $5\sqrt\lambda$, for large $\lambda$. In the first that many binomial-ratio steps each ratio is at least $4/3$. Thus

$$\frac{\binom{r+V}k}{B}\ge(4/3)^{5\sqrt\lambda}, \qquad f(r+V)\le\exp\{-(4/3)^{5\sqrt\lambda}\}.$$

Integrating (3) over the parent, and using (6), proves

$$\mathbb E_{\rm sieve} \frac{\mu(S\cap C)}{\mu(C)} \le\eta_\lambda :=\exp\{-2\lambda+O(\sqrt\lambda\log\lambda)\}, \tag{7}$$

uniformly over all parents.

For each $x\in T$, choose $C(x)$ deterministically before sampling the sieve, for example by taking its $r$ smallest success primes. Markov's inequality and (7) give

$$\mathbb E_{\rm sieve}\mu\left\{x\in T: \frac{\mu(S\cap C(x))}{\mu(C(x))}>e^{-3\lambda/2}\right\} \le\eta_\lambda e^{3\lambda/2}=o(1). \tag{8}$$

Equation (4) also gives $\mathbb E_{\rm sieve}\mu(T\setminus S)=o(1)$. Choose a realization with the sum of these two defects at most its expectation. The remaining part of $T$ is the claimed $G$. In particular, no union bound over the very numerous parents is used.

Activated arithmetic version

Fix integers $X\ge2Q$ and $Q\ge z^{r+1}$ before sampling the sieve. The realization can be chosen to satisfy (1)--(2) and additionally

$$\frac{\sum_{n\in G\cap[Q,X]}1/n}{\sum_{Q\le n\le X}1/n} \ge\tfrac12-o(1). \tag{9}$$

To verify this, set $s=|\mathcal P|$. Direct CRT counting of the one-prime and two-prime conditions gives, for every $Y\ge1$,

$$\sum_{n\le Y}(R(n)-2\lambda)^2 =Y\sum_{p\in\mathcal P}(2/p-4/p^2)+O(s^2). \tag{10}$$

The mean-count error contributes $O(\lambda s)$, which is included in $O(s^2)$. Partial summation on $[Q,X]$ and Chebyshev put at least $1/2-o(1)$ of its normalized harmonic measure on $T$: the normalized error is $O(s^2/(\lambda Q\sum_{Q\le n\le X}1/n))=o(1)$.

The estimates (4) and (7)--(8) hold after replacing Haar sampling of $x$ by any fixed probability measure, because the conditional-parent expectation is uniform and $C(x)$ is fixed in advance. Apply them simultaneously to Haar measure and to this normalized harmonic measure, then choose a realization whose sum of all four bad masses is at most its expectation. This proves (9).

All installed moduli are at most $z^k<Q$, and all parent moduli are at most $z^r<Q$, for large $z$. Hence every preparatory class is active at all integers counted in (9). The installed distinct moduli can be ordered strictly increasingly. The realization may depend on $Q,X$; no simultaneous-in-$X$ assertion is made.

A quantified obstruction to the missing refinement

Any appended modulus refining a parent has at least $r$ distinct prime factors from $\mathcal P$. Let

$$D_r=\{q:\omega_{\mathcal P}(q)\ge r\},\qquad \beta_r=\Pr\left(\sum_{p\in\mathcal P} {\rm Bernoulli}(1/p)\ge r\right).$$

The full multiples union generated by $D_r$ has measure $\beta_r$. Chernoff's inequality at the parameter $\log2$ gives

$$\beta_r\le2^{-r}\prod_{p\in\mathcal P}(1+1/p) \le\exp\{-(2\log2-1)\lambda+O(\sqrt\lambda)\}. \tag{11}$$

For arbitrary singleton appended residues, distinctness and the exact multiples harmonic inequality imply

$$\frac{H_{\text{hits with positive quotient }t\le K}(X)}{H_X} \le H_{\lfloor K\rfloor} \frac{\sum_{q\le X,\ q\in D_r}1/q}{H_X} \le H_{\lfloor K\rfloor}\beta_r. \tag{12}$$

Consequently a fixed positive harmonic deletion cannot be obtained from bounded quotients, or even from quotient ranges with

$$\log K\le \exp\{(2\log2-1-\varepsilon)\lambda\}$$

for a fixed $\varepsilon>0$ and large $\lambda$.

The open step is to choose distinct appended moduli whose many active progression points remain in the protected arithmetic survivors, while their total additional conditional Haar footprint tends to zero. The preparation lemma proves neither that step nor a logarithmic density counterexample. It shows why restricting to disjoint parents does not exhaust possible preparations.

Product truncation lowers the preparatory activation scale

The preparation can be strengthened: all installed moduli may be at most $z$, all parents have modulus below $z^2$, and (9) already holds on $[Q,X]$ for $Q=\lceil z^2\rceil$ and any chosen $X\ge2Q$. The realization may depend on $X$. Replace the parameters by

$$k=\lfloor\lambda\rfloor,\quad y=z^{1/k},\quad r=\lfloor2\lambda-6\sqrt\lambda\rfloor, \quad b=\lceil2\lambda+4\sqrt\lambda\rceil,\quad B=\binom bk. \tag{13}$$

Take $z$ large enough that $k,r>0$. Independently install each $k$-element prime subset $J$ only when $\prod_{p\in J}p\le z$, with the same probability $2^k/B$ and uniform $0/1$ CRT digits. An eligible subset hits a compatible point with probability $1/B$.

Write $R_I(x)$ for its number of success primes in $I$, and set

$$T_0=\{|R_{[3,z]}-2\lambda|\le2\sqrt\lambda,\ R_{(y,z]}\le\sqrt\lambda\}. \tag{14}$$

The elementary prime-counting upper bound and partial summation give

$$\sum_{y<p\le z}\frac1p \ll1+\log\frac{\log z}{\log y} =O(\log(2k))=O(\log\lambda). \tag{15}$$

Haar Chebyshev for the first condition and Markov for the second give $\mu(T_0)\ge\tfrac12-o(1)$.

The same holds for normalized harmonic measure $\nu$ on $[Q,X]$. With $s=\#\mathcal P$, the CRT identity (10) and partial summation give

$$\mathbb E_\nu(R_{[3,z]}-2\lambda)^2 \le2\lambda+O\!\left(\frac{s^2}{QH_{[Q,X]}}\right) =2\lambda+O(1),$$

where $H_{[Q,X]}=\sum_{Q\le n\le X}1/n\gg1$. Similarly,

$$\mathbb E_\nu R_{(y,z]} =2\sum_{y<p\le z}\frac1p+ O\!\left(\frac{s}{QH_{[Q,X]}}\right) =O(\log\lambda).$$

Thus $\nu(T_0)\ge\tfrac12-o(1)$.

Every $x\in T_0$ has at least $2\lambda-3\sqrt\lambda$ success primes at most $y$. Choose $r$ of them deterministically as its parent $C(x)$, before sampling the sieve. Its modulus satisfies

$$d\le y^r=z^{r/k}<z^2. \tag{16}$$

Product truncation only decreases the old deletion bound:

$$\Pr(x\notin S)\le\frac{\binom{R_{[3,z]}(x)}k}{B} \le(3/4)^{2\sqrt\lambda-O(1)}=o(1) \qquad(x\in T_0). \tag{17}$$

Conditioned on any such low-prime parent,

$$R_{[3,y]}=r+V,\qquad \mathbb EV=2\lambda-O(\log\lambda). \tag{18}$$

Here (15) accounts for the omitted large primes, while the reciprocal sum of the $r$ fixed primes is $O(\log(r+2))$. The exponential-moment argument in (6), now with cutoff $15\sqrt\lambda$, gives

$$\Pr(V<15\sqrt\lambda) \le\exp\{-2\lambda+O(\sqrt\lambda\log\lambda)\}. \tag{19}$$

Above that cutoff, $r+V$ exceeds $b$ by at least $5\sqrt\lambda-O(1)$. Every $k$-subset of these low successes is eligible, since its product is at most $y^k=z$. Independence and the binomial-ratio estimate therefore give

$$\mathbb E_{\rm sieve}\mathbf1_S(x) \le\left(1-\frac1B\right)^{\binom{r+V}k} \le\exp\{-(4/3)^{5\sqrt\lambda-O(1)}\}. \tag{20}$$

Integrating over the parent and using (19) proves the uniform bound

$$\mathbb E_{\rm sieve}\frac{\mu(S\cap C)}{\mu(C)} \le\exp\{-2\lambda+O(\sqrt\lambda\log\lambda)\}. \tag{21}$$

Apply Markov at threshold $e^{-3\lambda/2}$ in (21), and integrate over the predetermined parents $C(x)$, separately against Haar measure and $\nu$. Together with (17), the sum of all four expected bad masses is $o(1)$. Choose a realization with that sum at most its expectation and retain

$$G=\{x\in T_0\cap S: \mu(S\cap C(x))/\mu(C(x))\le e^{-3\lambda/2}\}.$$

Then

$$\mu(G)\ge\tfrac12-o(1),\qquad \nu(G)\ge\tfrac12-o(1). \tag{22}$$

This proves the strengthened preparation.

Every installed modulus is a distinct squarefree $k$-fold prime product at most $z$, so all preparatory congruences are active on $[Q,X]$ and can be ordered strictly increasingly. The parents remain auxiliary, potentially overlapping cylinders, not repeated installed congruences. A distinct appended refinement giving a fixed positive normalized harmonic anomaly has still not been constructed.

The later bounded-overlap refinement theorem rules out this missing step at every cutoff if the appended refinements have subexponential overlap within each parent and exponentially small incremental footprint. Its low-success-count comparison with the actual preparatory survivors requires no overlap assumption. The unresolved case is the conversion of a raw sum of conditional cylinder masses to their union measure when in-parent overlap is unrestricted.

Lower-rank parents: another proved preparation

The choice of parent rank near $2\lambda$ is not forced. There is a weaker conditional-survival estimate with parents of rank $k=\lfloor\lambda\rfloor$, which avoids the exponentially sparse modulus-support condition of the rank-$2\lambda$ version. It is still only a preparation, not a counterexample.

Keep the product-truncated random sieve and the parameters $k,y,b,B$ from (13), but choose a parent using exactly $k$ success primes at most $y$, rather than $r$ of them. Every point of $T_0$ has enough such primes, and this parent has modulus at most $y^k=z$. Uniformly over all these parents,

$$\mathbb E_{\rm sieve}\frac{\mu(S\cap C)}{\mu(C)} \le\exp\{-(1-\log2)\lambda+O(\sqrt\lambda)\}. \tag{23}$$

Indeed, conditioned on the parent, the number of low-prime successes is $k+V$, with $\mathbb EV=2\lambda-O(\log\lambda)$. The Bernoulli exponential-moment bound gives, for a fixed sufficiently large absolute $A$,

$$\Pr(V\le\lambda+A\sqrt\lambda) \le2^{\lambda+A\sqrt\lambda}\mathbb E2^{-V} \le\exp\{-(1-\log2)\lambda+O(\sqrt\lambda)\}.$$

Above this cutoff, $k+V$ exceeds the binomial threshold $b$ by at least $5\sqrt\lambda$, after increasing $A$. All $k$-subsets of these low-prime successes have product at most $z$, so the same binomial-ratio calculation as (20) makes the conditional random-survival probability at most $\exp\{-(4/3)^{5\sqrt\lambda-O(1)}\}$. Integrating proves (23).

For each point of $T_0$, fix its parent before sampling the sieve by taking its $k$ smallest success primes at most $y$, with their actual 0/1 digits. This is a finite-coordinate measurable rule. Markov's inequality at threshold $\exp(-(1-\log2)\lambda/2)$, together with (17), gives a realization with a set $G\subseteq S\cap T_0$ satisfying

$$\mu(G)\ge\tfrac12-o(1),\qquad \nu(G)\ge\tfrac12-o(1),\qquad \frac{\mu(S\cap C(x))}{\mu(C(x))} \le e^{-(1-\log2)\lambda/2}\quad(x\in G), \tag{24}$$

where $\nu$ is the chosen harmonic measure on $[z^2,X]$ from the product-truncation argument. Its uniform expectation proof allows Haar and $\nu$ to be imposed simultaneously exactly as before. All actual preparatory moduli remain distinct and at most $z$; the parents are auxiliary and may overlap.

The rank threshold $k\approx\lambda$ no longer supplies an exponentially small Chernoff bound for the reciprocal support of possible refined moduli. Also a typical point can lie in roughly $\binom{2\lambda}{\lambda}$ parents, rather than the subexponential number in the rank-$2\lambda$ construction. Thus the new weighted and overlap obstructions do not apply with their vanishing errors unchanged. No distinct refinements with a fixed harmonic anomaly and small incremental union footprint have been obtained from (24). The difficulty is controlling the union over a full logarithmic epoch, not merely finding one parent of small conditional survivor measure.

Status of mixed-root and coding attempts

The last construction attempts used the lower-rank parents above, and also intermediate parent ranks between $\lambda$ and $2\lambda$. No assignment of distinct appended moduli was found that both removes a fixed positive normalized harmonic mass over an epoch and has vanishing incremental Haar union measure. The proved single-parent rarity estimates do not supply that union estimate. This is a missing step, not an impossibility theorem for these lower-rank constructions.

For mixed 0/1 residues, the exact identity available at any hit is

$$a_q^2\equiv a_q\pmod q,\quad n\equiv a_q\pmod q \quad\Longrightarrow\quad q\mid n(n-1).$$

This permits attempts to encode choices among divisors of $n(n-1)$. But separating codes for divisibility-comparable moduli does not exclude a different, incomparable divisor from covering the same integer. Neither a bound for these cross-scale collisions nor a private canonical witness construction was proved. Also, the size of $n(n-1)/q$ does not by itself control its number of old prime factors: selecting idempotent roots can bias that cofactor. No typical-prime-factor heuristic for it is used in (23)--(24) or in any claimed theorem.

Limits of endpoint accounting and asymmetric colors

The subsequent construction attempt did not produce the missing overlapping refinements. Two precise restrictions on its proposed accounting can be retained.

First, if one selected integer $n_q\ge q$ is assigned to each distinct modulus, and the selected endpoints have harmonic mass at least $cH_X$, then

$$\sum_{q\le X}\frac1q \ge\sum_{n_q\le X}\frac1{n_q}\ge cH_X. \tag{25}$$

Thus an upper certificate using only $\mu(S\cap C_q)\le\eta/q$ and then summing these bounds cannot tend to zero unless $\eta H_X\to0$. This is a limitation of that certificate, not a lower bound for the actual incremental union. The lower-rank preparation's available exponent in (24) does not meet this requirement over an epoch starting at $z^2$. Smaller actual cell masses or coordinated overlap could still change the situation.

Second, consider a finite pool of odd primes, with allowed success digits $E_p=\{0\}$ or $\{0,1\}$. Suppose every preparatory modulus is squarefree and its prescribed digits belong to these sets. Let a parent $C$ fix success digits at a subset $J$ of the pool. If $S\cap C\ne\varnothing$, then

$$\mu(S\mid C)\ge \prod_{p\notin J}(1-|E_p|/p). \tag{26}$$

Indeed, points in $C$ with nonsuccess digits at every pool prime outside $J$ avoid all preparatory classes using such a prime. Any remaining preparatory class hitting such a point would use only primes in $J$, and would then contain the whole parent, contrary to its nonempty intersection with $S$. Independence proves (26).

Writing $\lambda=\sum_p1/p$ and $\lambda_1=\sum_{p:E_p=\{0,1\}}1/p$, the convergent quadratic error in the logarithm of this product gives the weaker uniform bound

$$\mu(S\mid C)\ge\exp(-\lambda-\lambda_1-O(1)).$$

This does not control unions of overlapping parents. The attempted asymmetric-color variation supplied no rigorous improvement of their total incremental footprint; heuristic exponent balances are not used as an impossibility theorem here.

Simply enlarging polynomial epochs also did not evade the existing weighted bound: for the rank $2\lambda-o(\lambda)$ parents, it already extends to $\log X\le\exp((1+2\log2-\varepsilon)\lambda)$. No new counterexample mechanism or general exclusion was proved.