Rank and moment obstructions

---

The overlapping entropy obstruction strengthens the higher-moment failures below: overlapping first-kill potentials need not be uniformly integrable, even for zero residues. It does not refute the disjoint first-kill entropy bound.

These are bounds and counterexamples concerning intermediate estimates, not a resolution of Problem 25. The first two obstructions use zero residues; later constructions have the residues specified in their proofs.

All individual divisor-face capacities still miss the difficulty

The finite proof in bounded prime rank uses the constraints

$$\sum_{q:d\mid q,\ a_q\equiv b\ (d)}c_q\le\frac1d. \tag{1}$$

For actual normalized first-kill harmonic masses these sharpen to

$$\sum_{q:d\mid q,\ a_q\equiv b\ (d)}c_q \le\frac{H_{\lfloor X/d\rfloor}}{dH_X}. \tag{2}$$

Neither the full collection (2), nor primitivity and distinctness of the moduli together with (2), implies a rank-independent small-footprint bound.

Choose disjoint finite prime groups $P_1,\ldots,P_r$ such that

$$1<c_j:=\sum_{p\in P_j}\frac1p\le2.$$

They may all lie above any prescribed bound: take successive primes until each reciprocal sum first exceeds $1$, using divergence of the prime reciprocal sum. Let $D$ consist of all products selecting exactly one prime from each group, with residue zero. These are distinct squarefree moduli and form a primitive set. Their periodic union has density

$$\beta_r=\prod_{j=1}^r \left(1-\prod_{p\in P_j}(1-1/p)\right) \le(1-e^{-4})^r\longrightarrow0. \tag{3}$$

Indeed $\log(1-1/p)\ge-2/p$ for every prime.

Nevertheless the weights

$$\lambda_q=\frac{1/q}{\prod_jc_j}$$

have total mass $1$ and satisfy (1). If $d$ selects primes from the groups indexed by $J$, direct factorization gives

$$\sum_{q:d\mid q}\lambda_q =\frac1d\prod_{j\in J}c_j^{-1}\le\frac1d.$$

Other nonempty faces have zero mass. Put $Q=\max D$, $X=Q^3$, and $c_q=\lambda_q/2$. For each relevant $d\le Q$,

$$\frac{H_{\lfloor X/d\rfloor}}{H_X} \ge\frac{\log(X/d)}{1+\log X} \ge\frac{2\log Q}{1+3\log Q}>\frac12 \qquad(Q\ge3).$$

Thus all the stronger capacities (2) hold, while $\sum_qc_q=1/2$ and $\beta_r\to0$. These weights have not been realized as actual first-kill masses. That missing realization is precisely why this is not a counterexample to Problem 25.

The later product-capacity obstruction strengthens this failure: exact harmonic face capacities admit positive abstract mass on vanishing-footprint families even with distinct primitive moduli satisfying $\omega(q)\le3$. Large prime-power exponents replace the increasing distinct-prime rank used above. These weights, too, are not established as actual first-kill masses.

Uniform higher moments of the natural first-kill potential fail

For a finite family, write

$$c_q(X)=\frac{H_{E_q}(X)}{H_X},\qquad Z_X(\omega)=\sum_q q c_q(X)\mathbf1_{[a_q]_q}(\omega).$$

Its mean is $H_B(X)/H_X$ and its support has Haar measure $\beta$. A uniform $L^2$ bound would give a square-root small-footprint estimate by Cauchy--Schwarz. In fact there is no uniform $L^p$ bound for any fixed $p>1$.

Here is an explicit sequence disproving such bounds. For an integer $k\to\infty$, take successive primes greater than $k^3$, stopping when their reciprocal sum first reaches $k$. Denote the resulting finite pool by $P$, and put

$$L=\prod_{p\in P}p,\qquad D=\left\{\prod_{p\in S}p:S\subset P,\ |S|=k\right\}, \qquad X=L^2.$$

Use zero residues. Let $e_q=d(E_q)$ and $P_0=\prod_{p\in P}(1-1/p)$, and let $N(\omega)$ count the primes of $P$ dividing a Haar-random integer. Since

$$\sum_{p\in P}1/p=k+O(k^{-3}),\qquad \sum_{p\in P}1/p^2=O(k^{-2}),$$

we have $P_0=e^{-k}(1+O(k^{-2}))$.

The event that exactly the $k$ primes constituting $q$ divide $\omega$ has $q$ as its unique divisor from $D$. It lies in the first-kill set for $q$, regardless of the ordering. Therefore $qe_q\ge P_0$. Each $E_q$ is exactly $L$-periodic on the positive integers; activation causes no exception for zero residues. Summing over the $L$ blocks of length $L$ up to $L^2$ gives

$$H_{E_q}(L^2)\ge e_qH_L, \qquad H_{L^2}\le2H_L.$$

Consequently

$$Z_X\ge\frac{P_0}{2}\binom Nk. \tag{4}$$

The probability needed to use (4) has an elementary estimate. Put $y_p=1/(p-1)$ and $T=\sum_{p\in P}y_p=k+O(k^{-2})$. For $m=k$ or $2k$,

$$\Pr(N=m)=P_0 e_m((y_p)_{p\in P}) =(1+O(k^{-2}))e^{-k}\frac{k^m}{m!}. \tag{5}$$

Indeed $m!e_m(y)/T^m$ is the probability that $m$ independent draws with probabilities $y_p/T$ are all distinct. Its complement is at most $\binom m2\max_p(y_p/T)=O(k^{-2})$.

On $N=2k$, (4), (5), and Stirling's formula yield

$$\mathbb E Z_X^2 \ge(1+o(1))\frac{(4/e)^k}{8\pi^{3/2}k^{3/2}} \longrightarrow\infty.$$

More generally, for each fixed $p>1$,

$$\mathbb E Z_X^p\gg_p \frac{(4/e)^{(p-1)k}}{k^{(p+1)/2}}\longrightarrow\infty.$$

Thus fixed-exponent Hölder estimates cannot repair this particular first-kill potential. Likewise $\sum_q\kappa(q)c_q(X)$ is unbounded: here $\kappa(q)=(1+o(1))2^k$ uniformly, while $\sum_qc_q(X)\ge\Pr(N\ge k)/2\ge\Pr(N=k)/2$.

The exact zero-residue inequality $H_B(X)\le\beta H_X$ remains true for every family in this example. Large moments of this potential do not themselves indicate an actual harmonic-density anomaly.

A four-prime certificate against the sharper second-moment charge

Section 9 of the fourth independent record gives a small exact example of the same overlapping-potential obstruction. Put $h_q=H_{E_q}(X)$ and

$$V(\omega)=\sum_q qh_q\mathbf1_{[a_q]_q}(\omega).$$

CRT gives

$$\mathbb EV=\sum_qh_q,\qquad \mathbb EV^2=\sum_{q,r}(q,r) \mathbf1_{a_q\equiv a_r\pmod{(q,r)}}h_qh_r.$$

The proposed bound $\mathbb EV^2\le H_X\sum_qh_q$ would imply the sharp harmonic inequality by Cauchy--Schwarz on the full union. It fails already for zero residues at $2,3,5,7$, with $X=10$:

$$(h_2,h_3,h_5,h_7) =\left(\frac{137}{120},\frac49,\frac15,\frac17\right), \qquad \sum_qh_q=\frac{4861}{2520},\quad H_{10}=\frac{7381}{2520}.$$

Here the first-kill sets are respectively $\{2,4,6,8,10\},\{3,9\},\{5\},\{7\}$, so all coefficients come directly from the exact activation rule. Since the primes are distinct,

$$\mathbb EV^2 =\left(\sum_qh_q\right)^2+\sum_q(q-1)h_q^2 =\frac{862117}{151200}, \qquad \mathbb EV^2-H_{10}\sum_qh_q =\frac{329873}{6350400}>0.$$

This finite certificate was checked by rational arithmetic; the displayed sets and identities also give a direct verification. It is a failure of this charge, not of harmonic domination for zero residues. The preceding construction already proves the stronger absence of any uniform higher-moment bound for this potential.

Disjoint first-kill potentials also have no uniform higher moment

Replacing the overlapping potential $Z_X$ by a disjoint one does not fix the moment method. Let $F_q=C_q\setminus\bigcup_{r<q}C_r$ be the full first-kill regions, with densities $e_q$, and consider

$$W_X=\sum_{e_q>0}\frac{c_q(X)}{e_q}\mathbf1_{F_q}.$$

For every fixed $s>1$, $\mathbb E W_X^s$ is unbounded over finite distinct-modulus singleton systems. The following construction is explicit.

Fix an integer $M\ge2$ and set $N=M^2$. For each prime $p$ and integer $e\ge1$ with $p^e(p-1)\le N$, install all the classes

$$j p^{e-1}\pmod{j p^e},\qquad 1\le j\le p-1. \tag{6}$$

All their moduli are at most $N$ and are distinct: their largest prime factor is $p$, its exponent is $e$, and the remaining factor is $j<p$. The full survivor set is exactly $L_N\mathbb Z$, where

$$L_N=\prod_{p(p-1)\le N} p^{\lfloor\log_p(N/(p-1))\rfloor}. \tag{7}$$

To prove (7), analyze the classes by increasing $p$, and for a fixed $p$ by increasing $e$; this is only a proof ordering, not a change to the eventual increasing-modulus enumeration. Earlier primes already force divisibility by every $j<p$: if $\ell^v\mid j$, then $\ell^v(\ell-1)\le(p-1)(p-2)<p(p-1)\le N$. At the $e$th stage survivors are divisible by $p^{e-1}$. Their next nonzero $p$-adic digit $j\in\{1,\ldots,p-1\}$ places them in precisely the corresponding class (6), since they are also divisible by $j$. The zero next digit avoids all these classes. This proves (7).

Now set $Q=N+1$, translate every old residue by $Q$, and append the class $0\pmod Q$. The old full survivor set is $Q\pmod{L_N}$. All old moduli are below $Q$, so every old congruence is active at the new endpoint $Q$, which survives them all. At $X=Q$ the appended class has first-kill set $\{Q\}$, hence $c_Q=1/(QH_Q)$. If $g=\gcd(Q,L_N)$, its full first-kill density is $e_Q=g/(QL_N)$. Thus

$$\mathbb E W_Q^s \ge\frac{c_Q^s}{e_Q^{s-1}} =\frac{L_N^{s-1}}{Q g^{s-1}H_Q^s} \ge\frac{L_N^{s-1}}{Q^sH_Q^s}. \tag{8}$$

Every prime power at most $M$ divides $L_N$, so $L_N\ge\operatorname{lcm}(1,\ldots,M)\ge2^M/(M+1)$. For the last elementary inequality, each $\binom Mk$ divides the lcm: Legendre's valuation formula has at most $\lfloor\log_p M\rfloor$ nonzero summands, each at most one. The largest binomial coefficient is at least $2^M/(M+1)$. Consequently (8) tends to infinity for every $s>1$ while $Q=M^2+1$.

Again this is not a counterexample to Problem 25. It rules out uniform fixed-exponent moment control even for a potential supported on disjoint Haar first-kill regions. The substantial density cost of the old sieve has not been converted into a logarithmic-density anomaly.

A rank-sensitive additive target: now refuted

Distinctness gives a useful uniform truncation in total prime-factor rank. For a finite singleton system, let $B_{>r}$ use only moduli with $\Omega(q)>r$. Then

$$\begin{aligned} H_{B_{>r}}(X) &\le H_{\lfloor X\rfloor} \sum_{\substack{q\le X\\\Omega(q)>r}}\frac1q\\ &\le(3/2)^{-r}H_{\lfloor X\rfloor} \prod_{p\le X}(1-3/(2p))^{-1} \ll(3/2)^{-r}(\log(3X))^{5/2}. \end{aligned} \tag{9}$$

The Euler-product estimate follows from $(1-3/(2p))^{-1}(1-1/p)^{3/2}=1+O(p^{-2})$ and the already checked Mertens product bound. With $r=\lceil8\log\log(3X)\rceil$, (9) is $O(1)$.

Consequently a uniform finite estimate

$$H_B(X)\le\beta H_{\lfloor X\rfloor}+C_r, \qquad \Omega(q)\le r, \qquad C_r=\exp(o(r)), \tag{10}$$

would prove the full problem: apply it to the low-rank part, whose Haar union has density at most $\beta$, and use (9) for the rest. Its total error would be $o(\log X)$, uniformly over finite systems, permitting the usual finite-truncation squeeze. A linear bound $C_r=O(r)$ would give the stronger uniform error $O(\log\log X)$.

Estimate (10) is refuted by the energy counterexample. That construction has harmonic excess $\asymp(\log Q)^{1/21}$. Removing all moduli with $\Omega(q)>r=\lceil8\log\log(3Q)\rceil$ loses only $O(1)$ harmonic mass by (9), and can only decrease its Haar union measure. Thus any universal additive constants for rank at most $r$ must satisfy

$$C_r\ge\exp\bigl((1/168+o(1))r\bigr)$$

along the resulting sequence. In particular neither $C_r=\exp(o(r))$ nor a uniform $O(\log\log X)$ additive error is possible. The bounded-rank natural-density theorem makes no such quantitative assertion and remains valid. A uniform additive $o(\log X)$ comparison remains open.

Entropy of the disjoint potential: an unresolved stronger target

The failure of every fixed $L^p$ bound does not rule out

$$\sum_i c_i(X)\log_+\frac{c_i(X)}{e_i}\le C \tag{11}$$

with an absolute constant. No proof or counterexample to (11) was found. It would settle the full problem: for a tail with Haar mass $\eta=\sum_{i>k}e_i$ and harmonic mass $\alpha=\sum_{i>k}c_i(X)$, the log-sum inequality gives

$$\alpha\log(\alpha/\eta) \le\sum_{i>k}c_i(X)\log\frac{c_i(X)}{e_i}\le C.$$

As $\eta\to0$, this forces $\alpha\to0$ uniformly in $X$. Here zero-density finite periodic first-kill cells are empty, so they never have positive $c_i$. This is a conditional implication, not an estimate proved for arbitrary residues.

Averaging in harmonic time does not weaken the entropy target

Let $u=H_N$, and interpolate each first-kill harmonic sum linearly between successive harmonic times. Denote it by $F_i(u)$, and put

$$c_i(u)=F_i(u)/u,\qquad K(u)=\sum_i c_i(u)\log_+\frac{c_i(u)}{e_i}.$$

Terms with $e_i=0$ have $F_i=0$ and are omitted. The functions $F_i$ are nondecreasing and $\sum_i c_i(u)\le1$. Thus, for $v\ge u$, $c_i(v)\ge(u/v)c_i(u)$. Using monotonicity of $x\log_+(x/e_i)$ and

$$\lambda x\log_+\frac{\lambda x}{e_i} \ge\lambda x\log_+\frac{x}{e_i}+\lambda x\log\lambda \quad(0<\lambda\le1),$$

we obtain

$$K(v)\ge\frac uvK(u)-\frac uv\log\frac vu.$$

For every $\Lambda>1$, integration gives the exact lower bound

$$\int_u^{\Lambda u}K(v)\,dv \ge u(\log\Lambda)K(u)-\frac u2(\log\Lambda)^2.$$

Consequently an absolute averaged estimate $\int_1^T K(v)\,dv\le CT+O(1)$ would imply

$$K(u)\le\frac{C\Lambda}{\log\Lambda} +\tfrac12\log\Lambda+o(1).$$

A constant pointwise bound trivially implies an averaged one. The two targets are therefore equivalent up to constants: scale-averaging this nonnegative information does not bypass the unproved constant entropy estimate. No estimate for a different signed quantity was obtained.

Valuation-measurable systems: a proved uniform bound

Suppose each full cylinder $C_i$ is determined by prime valuations. This includes $a_i=0$ and, for even $q_i$, $a_i=q_i/2$, in any mixture. For these systems,

$$W_X\le\frac{Z_X}{H_{\lfloor X\rfloor}},\qquad Z_X=\prod_{p\le X}(1-1/p)^{-1}. \tag{12}$$

Restrict to the prefix $q_i\le X$, which does not change any earlier first-kill cell. The independent geometric valuation model on primes $p\le X$ assigns every $X$-smooth integer $n$ probability $1/(Z_Xn)$. Because $F_i$ is valuation-measurable, its probability in this model is exactly its Haar measure $e_i$. Since $E_i\subseteq F_i$,

$$H_{E_i}(X)\le Z_X\Pr(N\le X,\ N\in F_i)\le Z_Xe_i.$$

This proves (12). The Euler-product estimate $Z_X\ll\log(2X)$, checked in the Davenport--Erdős 1951 source as recorded above, makes (12) a uniform $L^\infty$ bound. In particular (11) holds in these special cases. The translated digit-sieve construction giving (8) is not valuation-measurable.

Removing $c_i$ from the logarithm destroys the target

The superficially stronger information estimate $\sum_i c_i(X)\log(1/(q_i e_i))=O(1)$ is false even for zero residues. Here is a finite construction with unbounded limiting value.

Choose primes $P$, all greater than $k^3$, whose reciprocal sum is $2k+O(k^{-3})$, and take all products of $k$ distinct primes of $P$ as zero-residue moduli. Let $P_0$ be an initial segment of $P$ with reciprocal sum $k/2+O(k^{-3})$. The least active product dividing a Haar integer is the product $q$ of its $k$ smallest dividing primes of $P$. Consequently

$$q e_q=\prod_{\substack{p\in P,\ p<\max(q)\\p\nmid q}}(1-1/p).$$

The independent prime-divisibility counts have variances at most their means. Chebyshev therefore shows that, with probability $1-O(1/k)$, fewer than $k$ primes of $P_0$ but at least $k$ primes of $P$ divide the integer. On this event, $\max(q)$ is beyond $P_0$ and

$$\log\frac1{q e_q} \ge\sum_{\substack{p\in P_0\\p\nmid q}}\frac1p \ge k/2-O(k^{-2}),$$

because at most $k$ selected primes were removed from that sum and each exceeds $k^3$. Hence

$$\sum_q e_q\log\frac1{q e_q}\ge k/2-O(1).$$

For each fixed finite system $c_q(X)\to e_q$, giving the asserted counterexample to the stronger information estimate. It does not contradict (11); in fact (12) applies to every one of these systems.

Sharp growth of the Haar-weighted information

The preceding quantity has a sharp asymptotic maximum. For a finite distinct-modulus singleton family with all $q\le Q$, put

$$I=\sum_q e_q\log\frac1{qe_q},$$

with zero summands when $e_q=0$. Then, as $Q\to\infty$,

$$\log\log Q-O((\log\log Q)^{2/3}) \le \sup I\le \log H_{\lfloor Q\rfloor}+1/e. \tag{H}$$

Thus $\sup I=(1+o(1))\log\log Q$. The lower bound already uses zero residues. This is a statement about Haar masses $e_q$, not the arithmetic masses $c_q(X)$ in the unresolved target (11).

For the upper bound let $\beta=\sum_qe_q\le1$. If $\beta>0$, the log-sum inequality and distinctness give

$$I\le\beta\log\frac{\sum_q1/q}{\beta} \le\beta\log H_{\lfloor Q\rfloor}-\beta\log\beta \le\log H_{\lfloor Q\rfloor}+1/e.$$

The case $\beta=0$ is immediate.

For the lower bound put

$$L=\log\log Q,\qquad k=\lfloor L-2L^{2/3}\rfloor,\qquad z=Q^{1/k},$$

and use every product of $k$ distinct primes from $P=\{p:k^3<p\le z\}$, with residue zero. These moduli are distinct and at most $Q$. The elementary prime reciprocal estimates already used in these notes give

$$\lambda:=\sum_{p\in P}\frac1p =L-\log k-\log(3\log k)+O(1) \ge k+k^{2/3}$$

for sufficiently large $Q$. Choose an initial prime segment $P_0\subset P$ with reciprocal sum $k-k^{2/3}+O(k^{-3})$. Under Haar measure, prime divisibility is independent and each count has variance at most its mean. Chebyshev's inequality shows that, with probability $1-O(k^{-1/3})$, fewer than $k$ primes of $P_0$ but at least $k$ primes of $P$ divide the sampled point.

Its first-kill modulus $q$ is then the product of its $k$ smallest dividing primes of $P$. Writing $P^+(q)$ for the largest prime factor, its exact first-kill mass satisfies

$$qe_q=\prod_{\substack{p\in P,\ p<P^+(q)\\p\nmid q}}(1-1/p).$$

On the indicated event $P^+(q)$ exceeds every prime of $P_0$. At most $k$ primes of $P_0$ divide $q$, each greater than $k^3$. Consequently

$$\log\frac1{qe_q} \ge\sum_{\substack{p\in P_0\\p\nmid q}}\frac1p \ge k-k^{2/3}-O(k^{-2}).$$

The logarithm is nonnegative on every other first-kill cell. Integrating therefore gives $I\ge k-O(k^{2/3})=L-O(L^{2/3})$, proving (H). Zero residues make activation exact: every positive multiple of $q$ is already at least $q$.

This rules out a constant Haar-information bound and identifies its correct growth. Its universal $o(\log Q)$ bound does not establish an $o(\log X)$ arithmetic excess, or resolve (11).

Coherence does not give a uniform bound for the disjoint potential

There is no absolute $L^\infty$ bound for $W_X$, even for coherent prime-modulus systems. This does not refute a uniform finite-$L^p$ bound restricted to coherent systems, or the entropy target (11).

Choose a finite admissible $S\subseteq[1,T]$ with $H_S=\sum_{s\in S}1/s$ arbitrarily large. The elementary construction of such sets is proved in finite sharp counterexamples. Choose a prime $q>T$. For every prime $p<q$, let $r_p$ be a residue omitted by $S\bmod p$, and choose a common center by CRT with

$$\alpha\equiv qr_p\pmod p\quad(p<q),\qquad \alpha\equiv0\pmod q.$$

Install all prime moduli up to $q$. Every $qs$, $s\in S$, is first killed by $q$: it avoids each earlier prime class because $q$ is invertible modulo that prime. All these hits are active. For $X=qT$,

$$c_q(X)\ge\frac{H_S}{qH_{qT}},\qquad e_q=\frac1{qZ_{q^-}},\qquad Z_{q^-}=\prod_{p<q}(1-1/p)^{-1}\ge H_{q-1}.$$

The last bound follows by expanding the Euler product: every integer at most $q-1$ occurs. Since $T<q$ and $H_{q^2}\le2H_q$,

$$\frac{c_q(X)}{e_q}\ge\frac{H_S}{4}.$$

The nonempty $q$th first-kill cell has positive Haar measure, so this proves the asserted failure of uniform $L^\infty$ control. Its small measure prevents this calculation from establishing a moment or entropy obstruction.

Size of a nonempty first-kill cell

Proved. There is an absolute constant $C$ such that every nonempty full first-kill cell at modulus $q$ in a distinct-modulus singleton system satisfies

$$q e_q\ge \exp(-C\sqrt q). \tag{13}$$

The exponent order $\sqrt q$ is sharp. This is a pointwise size bound, not a summable entropy or activated trace estimate.

First consider any nonempty survivor set formed by singleton moduli at most $N$. Set

$$y=2\sqrt N,\qquad L=\prod_{p\le y}p^{\lfloor\log_pN\rfloor}.$$

The elementary prime-counting bound gives $\log L\le\pi(y)\log N=O(\sqrt N)$, with bounded $N$ absorbed in the constant. Fix the residue modulo $L$ of one surviving point. All conditions at moduli supported on primes at most $y$ are now avoided. Every other modulus is $pm$ for a unique prime $p>y$, with $m\le N/p<y$. Its small-prime part is fixed, so it either imposes no condition or excludes one digit modulo $p$.

Distinctness allows at most $\lfloor N/p\rfloor$ such excluded digits. Different large prime coordinates are independent. Since $N/p^2<1/4$, the full surviving Haar measure is at least

$$\frac1L\prod_{y<p\le N}(1-N/p^2) \ge\frac1L\exp\left(-\frac43N\sum_{n>y}\frac1{n^2}\right) \ge\exp(-C\sqrt N). \tag{14}$$

This proves the corresponding lower bound for any nonempty survivor set.

For a nonempty $F_q$, run the same argument with $N=q$ inside $C_q=[a_q]_q$, choosing a point of $F_q$ to fix the small-prime digits. The chosen fiber has conditional measure at least $1/L$. Any large prime dividing $q$ is already fixed; all earlier conditions at that prime are avoided, since the chosen point survives them. The remaining large primes obey the same product bound. The resulting conditional survival probability is $qe_q$, proving (13).

For sharpness, the translated digit sieve proved above has $q=M^2+1$ and

$$qe_q=\frac{\gcd(q,L_{M^2})}{L_{M^2}} \le q(M+1)2^{-M}\le\exp(-c\sqrt q)$$

for all sufficiently large $M$, with an absolute $c>0$. Thus a uniformly better exponent order is impossible. Summing the pointwise consequence $\log(1/(qe_q))\ll\sqrt q$ gives no useful uniform bound on the actual first-kill entropy; that global question remains open.

A small exponential rank error: later refuted

Superseded target. The prime-only amplification proves that the additive excess is unbounded even for $\Omega(q)=1$. Thus no finite rank-only additive error exists, irrespective of its growth with rank. The sufficiency calculation below is logically correct but has a false hypothesis; it records an approach that this obstruction rules out, not an open target.

Before the prime-only strengthening, the subexponential obstruction left the following possibility. For fixed $1<z<2$, the proof of (9) gives the valid, more flexible truncation bound

$$H_{B_{>r}}(X) \ll_z z^{-r}(\log(3X))^{z+1}. \tag{15}$$

Indeed $\sum_{q\le X}z^{\Omega(q)}/q$ is at most $\prod_{p\le X}(1-z/p)^{-1}\ll_z(\log(3X))^z$; comparison of local factors with $(1-1/p)^{-z}$ proves the final estimate.

Consequently the following now-refuted finite additive bound

$$H_B(X)\le\beta H_{\lfloor X\rfloor}+O(e^{cr}) \quad\text{when }\Omega(q)\le r, \qquad 0<c<\tfrac12\log2, \tag{16}$$

would settle the full problem if true with an implied constant independent of $r$, $X$, and the family. Choose $z<2$ sufficiently close to $2$ that $c<\log z/z$, then choose $z/\log z<a<1/c$ and take $r=\lceil a\log\log(3X)\rceil$. Equation (15) is $o(\log X)$ because $a\log z>z$, and the error in (16) is $o(\log X)$ because $ac<1$. The low-rank Haar footprint is no larger than the whole footprint, so finite approximation gives the conditional conclusion.

This identifies a range not excluded by the lower bound $C_r\ge\exp((1/168+o(1))r)$. The prime-only construction excludes that entire range: even the rank-one additive constant is infinite. The constant coprime entropy theorem is a different statement and remains valid.