---

This is Plasma AI's author-recorded alternative to step 5 of the [[research/prime_binary_new_attack/proof_sketch|canonical contradiction sketch]] for Problem 251. It proves the reductions and the reference calculation below, not the required estimate for primes or sieve survivors. No independent review is claimed.

The change of span has a specific purpose. At span $16K\log X$, the [[research/prime_binary_new_attack/consecutive_block_counts|nonnegative majorant analysis, U13--U14]], shows that almost every reference window contains some compatible $K$-subset. At the shorter span used here, that existence event has reference probability tending to zero. It is monotone under adding sites, unlike the event concerning the first $K$ sites. This permits an upper bound by an actual sifted set. The arithmetic estimate for that monotone event remains open. Counting all its certificates with multiplicity still fails, even at the shorter span.

The smaller forced count

Use $h_n=(p_{n+1}-p_n)/2$, $H_n=G_n/2$, $I_X=\{n:X<p_n\le2X\}$ and $N_X=|I_X|$ from the sketch. Fix an odd $d$, including $d=1$, and put

$$K=\lceil12\log_2\log X\rceil,\quad M=2^K,\quad R=\lfloor4d\log X\rfloor,\quad D_*=\left\lfloor\frac65K\log X\right\rfloor,\quad J=\lfloor D_*/2\rfloor. \tag{S1}$$

Let $r_d(u)$ be the least positive residue of $-du$ modulo $M$. Thus an ordinary zero residue means $r_d(u)=M$, and is excluded when $R<M$. Write $U_{n,K}=\sum_{j<K}2^{K-1-j}h_{n+j}$, and define

$$B_d^*(X)=\#\{n\in I_X:p_{n+K}-p_n\le D_*,\ r_d(U_{n,K})\le R\}.$$

Proposition S1. If $S=a/(2^sd)$, then $B_d^*(X)\ge(1/24-o(1))N_X$ as $X\to\infty$.

Proof. Past $n\ge\max(s+2,3)$, the sketch's lattice gives positive integers $c_n=dH_n$ with $c_{n+K}+dU_{n,K}=Mc_n$. The uniform mean-tail bound (T) excludes at most $(1/8+o(1))N_X$ starts with $c_{n+K}>R$. At every other start, $r_d(U_{n,K})=c_{n+K}\le R$, since $R<M$ for large $X$. PNT telescoping, exactly as in U3, gives

$$\sum_{n\in I_X}(p_{n+K}-p_n)=(1+o(1))KX.$$

Consequently at most $(5/6+o(1))N_X$ starts have span above $D_*$. The union bound leaves $1-1/8-5/6=1/24$. The scales eventually pass every fixed preperiod $s$. $\square$

More generally, span $\lambda K\log X$ and cutoff $\tau d\log X$ give the same lower bound with coefficient $1-1/\lambda-1/(2\tau)$, when it is positive. In particular this first-moment argument requires $\lambda>1$. Proposition S1 uses only PNT and the retained lattice identities, not a local prime model.

An event preserved when sites are added

For a $K$-set $A=\{a_1<\cdots<a_K\}\subseteq\{1,\ldots,J\}$ put $a_0=0$ and

$$U(A)=\sum_{i=1}^K2^{K-i}(a_i-a_{i-1}).$$

Call $A$ compatible when $r_d(U(A))\le R$. These are half-offsets: at root $x$, its selected sites are $x+2a_i$. For a finite set $T$ of odd integers and $x\in T$, let

$$\begin{aligned} Y_T(x)&=\{a\in\{1,\ldots,J\}:x+2a\in T\},\\ Z_T(x)&=\#\{A\subseteq Y_T(x):|A|=K,\ A\text{ compatible}\},\\ \mathcal C_d(T)&=\sum_{x\in T\cap(X,2X]}\min(1,Z_T(x)). \end{aligned} \tag{S2}$$

Each root is counted once if it has any compatible subset. Both $Z_T(x)$ and $\mathcal C_d(T)$ are nondecreasing when $T$ is enlarged. For $\mathbb P_X=\mathbb P\cap(X,2X+D_*]$, every root counted by $B_d^*(X)$ has its first $K$ prime successors as such a subset. Therefore

$$B_d^*(X)\le\mathcal C_d(\mathbb P_X). \tag{S3}$$

Consecutiveness is used in the rationality lower bound S1. The upper event in (S3) allows other prime subsets explicitly; it does not identify them with consecutive blocks.

For example, take $z=X^{1/2-1/K^2}$ and

$$T_z=\{m\in(X,2X+D_*]:P^-(m)>z\}.$$

For large $X$, $T_z$ consists of odd integers and contains $\mathbb P_X$. Thus $\mathcal C_d(\mathbb P_X)\le\mathcal C_d(T_z)$. The following actual-arithmetic estimate would finish the contradiction:

$$\mathcal C_d(T_z)\le\frac1{48}N_X \quad\text{along arbitrarily large }X,\quad\text{for every fixed odd }d. \tag{MS}$$

Together with S1--S3, (MS) contradicts rationality for each $d$ and every preperiod. It is unproved, even for $d=1$. The choice of sieve level is part of this proposed estimate. Lowering it adds survivors and can only increase the upper event.

The unweighted composite count in U10 implies $|T_z\cap(X,2X]|=N_X+O(N_X/K^2)$. Its bound of $K+1$ per deletion does not apply to (S2): deleting one site can destroy certificates at every earlier root within $D_*$, not just its first $K$ predecessors. The fixed-dimensional sieve in [[research/prime_binary_new_attack/survivor_moments|Corollary M3 of the survivor-moment note]] instead counts the surviving neighbors of composite sites. It gives $\mathcal C_d(T_z)-\mathcal C_d(\mathbb P_X)=O(N_X/K)$, uniformly in $d$. This does not bound either event; it permits the bounded upper estimate to be sought directly on the primes, with an $o(N_X)$ error.

An exact census of compatible subsets

Lemma S2. For integers $J\ge2K$, $K\ge1$, odd $d$, and $1\le R<M=2^K$, let $L_d(J,K,R)$ count compatible $K$-sets in $\{1,\ldots,J\}$. Then

$$R\binom{\lfloor J/2\rfloor}{K} \le L_d(J,K,R) \le R\binom{\lfloor(J+K)/2\rfloor}{K}. \tag{S4}$$

If $K^2/J\to0$, this gives uniformly in odd $d$ and $1\le R<M$

$$\frac{L_d(J,K,R)}{\binom JK} =\frac R{2^K}\bigl(1+O(K^2/J)\bigr). \tag{S5}$$

Proof. A $K$-set is equivalently a positive gap vector $v_i=a_i-a_{i-1}$ with $\sum_i v_i\le J$. Write uniquely $v_i=2w_i+1+b_i$, where $w_i\ge0$ and $b_i\in\{0,1\}$. For fixed $(w_i)$ the sum $U$ is a fixed integer plus $\sum_i2^{K-i}b_i$, so the $2^K$ bit vectors run once through all its residues modulo $M$. Exactly $R$ are compatible, since $d$ is odd.

If $\sum_iw_i\le\lfloor(J-2K)/2\rfloor$, every bit vector meets the span cutoff. Stars and bars therefore gives the lower bound in (S4). Conversely a vector meeting the cutoff has $\sum_iw_i\le\lfloor(J-K)/2\rfloor$, giving the upper bound, even if some of its $R$ compatible bit vectors fail the cutoff. The numbers of quotient vectors are the two binomial coefficients displayed in (S4). Comparing their falling-factorial products with $\binom JK$ proves (S5). $\square$

This is a census of integer subsets, not a theorem about the primes. It keeps the positive-residue convention: with $K=1$, $J=4$ and $d=R=1$, only the two odd singleton offsets qualify. Testing the ordinary residue as at most $R$ would incorrectly count all four.

Why the monotone event can be small

Proposition S3 (reference only). Include each of the $J$ half-offsets independently with probability $\rho=2/\log X$, obtaining a set $Y$. With (S1), the probability that $Y$ contains a compatible $K$-subset tends to zero, for every fixed odd $d$.

Proof. Put $V=|Y|$ and $t=\lfloor5K/4\rfloor$. Conditional on $V=v$, $Y$ is uniform among all $v$-subsets. Hence, by (S5), the expected number of compatible $K$-subsets in $Y$ is

$$\binom vK\frac{L_d(J,K,R)}{\binom JK} =\binom vK\frac R{2^K}(1+o(1)),$$

with zero when $v<K$. The error is uniform in $v$. A union bound, used only after restricting $V$, gives

$$\Pr(Z_Y>0) \le\Pr(V>t)+(1+o(1))\frac R{2^K}\binom tK. \tag{S6}$$

Here $J\rho=(6/5)K+o(1)$. The exponential Markov inequality gives

$$\Pr(V>t)\le\exp(-(I+o(1))K),\qquad I=\frac54\log\frac{25}{24}-\frac1{20}>0. \tag{S7}$$

For example use $\mathbb E e^{uV}=(1-\rho+\rho e^u)^J \le\exp(J\rho(e^u-1))$ with $e^u=25/24$. For the other term, $\binom tK4^K\le5^t\le5^{5K/4}$, so

$$\frac R{2^K}\binom tK \le R\left(\frac{5^{5/4}}8\right)^K \ll_d(\log X)^{15\log_2 5-35}=o(1). \tag{S8}$$

The exponent is negative because $5^{15}<2^{35}$. Both bounds apply at the growing length in (S1). $\square$

The saturation at one per root is essential. At the same parameters the unrestricted expected number of certificates is, by S2 and Stirling,

$$\mathbb E Z_Y =\rho^K L_d(J,K,R) \sim\frac R{\sqrt{2\pi K}}\left(\frac{3e}5\right)^K \longrightarrow\infty. \tag{S9}$$

Thus replacing $\min(1,Z_Y)$ by $Z_Y$ loses the desired bound already in this reference, before any prime-tuple sieve loss. Rare windows with many sites dominate (S9). This concerns that sum of certificates, not every nonnegative majorant. U13's monotonicity lemma still applies here; U14's near-one reference conclusion has a different span and does not. No part of S3 transfers the reference law to $T_z$ or to primes.

The arithmetic attempt and its two costs

For the actual set $T_z$ write $V_z(x)=|Y_{T_z}(x)|$ and define

$$\begin{aligned} \mathcal D_z&=\#\{x\in T_z\cap(X,2X]:V_z(x)>t\},\\ \mathcal Q_{d,z}&=\sum_{\substack{x\in T_z\cap(X,2X]\\V_z(x)\le t}}Z_{T_z}(x). \end{aligned}$$

Exactly as a pointwise inequality, without a probabilistic assumption,

$$\mathcal C_d(T_z)\le\mathcal D_z+\mathcal Q_{d,z}. \tag{S10}$$

An upper bound $N_X/96$ for each term, along the same unbounded scales, would give (MS). Neither is proved. The reference calculations S7 and S8 identify different obligations: unusually many survivors around a surviving root, and compatible subsets among the remaining roots. In particular a cluster bound alone would not prove irrationality.

The first concrete arithmetic test is a centered second moment. Put $\lambda=6/5$, $\chi=\mathbf1_{T_z}$ and

$$\mathcal V_z=\sum_{X<x\le2X}\chi(x)(V_z(x)-\lambda K)^2.$$

Since $t< V_z(x)$ implies $V_z(x)-\lambda K>K/20$, $\mathcal D_z\le400\mathcal V_z/K^2$. In particular $\mathcal V_z=o(K^2N_X)$ would handle this first cost. Expanding gives

$$\begin{aligned} \mathcal V_z={}& \sum_{\substack{1\le a,b\le J\\a\ne b}} \sum_{X<x\le2X}\chi(x)\chi(x+2a)\chi(x+2b)\\ &+(1-2\lambda K)\sum_{1\le a\le J} \sum_{X<x\le2X}\chi(x)\chi(x+2a) +\lambda^2K^2\sum_{X<x\le2X}\chi(x). \end{aligned} \tag{VM}$$

The negative pair term must be retained. Upper estimates for both the pair and triple counts do not prove the required cancellation. The required estimate concerns these averaged survivor sums together, at the displayed level $z$, with an error $o(K^2N_X)$, or a direct bound $\mathcal D_z\le N_X/96$. It is an unproved arithmetic task, not an application of PNT in progressions or Bombieri--Vinogradov to their stated counting functions; Proposition S4 below shows that it implies a short-interval law of large numbers of unknown status. Success would leave $\mathcal Q_{d,z}$ to be bounded with its local-count cutoff intact; removing that cutoff returns to the loss in (S9).

The inspected prime moment theorem does not settle the first test. [[../library/primes/kuperberg_2023_sums_singular_series_large_sets_tail/_index|Kuperberg, arXiv:2210.09775v2]], Theorem 1.8 on printed p. 4, bounds uncentered moments for intervals of length $h=\Lambda\log X$. Corollary 1.9 on printed pp. 4--5 requires the threshold $k$ to satisfy $k/\Lambda\to\infty$. Here $\Lambda\sim6K/5$ and $k\sim5K/4$, so their ratio tends to $25/24$. Moreover (VM) is rooted at a survivor and concerns survivors, not all integer roots with a prime count. The displayed moments and the corollary's hypotheses were checked on the page images; the proof of Theorem 1.8 and its use of the tuple sieve were read in the transcription, without an independent audit of the sieve. The available uncentered upper bounds do not provide the negative pair contribution or the centered error in (VM).

What the two costs require

This section proves implications of the two costs in (S10), restates the second cost in a damped signed form, and compares both with the inspected source estimates. It refutes neither (VM), (MS) nor the phase bound. The separate survivor-moment note proves that rough composites change $\mathcal V_z$ by only $O(N_X)$, without estimating the remaining prime-rooted moment.

The moment forces a law of large numbers

Proposition S4. Keep the parameters (S1) and $\lambda=6/5$. Suppose $\mathcal V_z=o(K^2N_X)$ along a set of scales $X$. Then along those scales, for every fixed $\varepsilon>0$,

$$\#\{x\in T_z\cap(X,2X]:|V_z(x)-\lambda K|>\varepsilon K\}=o(N_X) \tag{S11}$$

and

$$\#\{n\in I_X:\pi(p_n+2J)-\pi(p_n)>(\lambda+\varepsilon)K\}=o(N_X). \tag{S12}$$

Consequently, with $L=\lfloor(\lambda+\varepsilon)K\rfloor+1$, all but $o(N_X)$ starts $n\in I_X$ have $p_{n+L}-p_n>2J$.

Proof. Each root counted in (S11) contributes more than $\varepsilon^2K^2$ to $\mathcal V_z$, so there are at most $\mathcal V_z/(\varepsilon^2K^2)$ of them. For large $X$ every prime $p\in(X,2X]$ lies in $T_z$. Every prime in $(p,p+2J]$ is odd, so it is $p+2a$ with $1\le a\le J$; it is at most $2X+D_*$, because $2J\le D_*$, and it exceeds $z$. So it lies in $T_z$, and $\pi(p+2J)-\pi(p)\le V_z(p)$. Hence (S11) gives (S12). Finally, if $\pi(p_n+2J)-\pi(p_n)\le(\lambda+\varepsilon)K<L$, then $p_{n+L}>p_n+2J$. $\square$

Here $2J=(6/5+o(1))K\log X$ and $L=(6/5+\varepsilon+o(1))K$. So (VM) implies that after almost every prime the next $L\asymp\log\log X$ gaps sum to at least $(\lambda/(\lambda+\varepsilon)-o(1))L\log X$: near almost every prime there is no cluster of relative density above $1+\varepsilon/\lambda$ at this scale. This lower law of large numbers for sums of about $\log\log X$ consecutive gaps is a consequence of the proposed first input, not of rationality.

In fact Corollary M3 of the survivor-moment note gives the stronger exact interface

$$\mathcal V_z= \sum_{X<p\le2X}\bigl(\pi(p+2J)-\pi(p)-\lambda K\bigr)^2+O(N_X).$$

Thus (VM) is equivalent to the prime-rooted estimate (PVM) there, along the same scales. The sieve proving this comparison has at most three linear factors; its upper bounds remove the composite contamination, but do not evaluate the signed pair/triple combination on the primes.

We know no proof of (S12), unconditionally or under the Riemann hypothesis. For comparison, Selberg's mean-square bound under the Riemann hypothesis, $\int_X^{2X}|\psi(x+h)-\psi(x)-h|^2\,dx\ll hX(\log X)^2$, read here in the Saffari--Vaughan form restated in Gorodetsky, Section 1.1, gives by Chebyshev's inequality that the set of $x\in[X,2X]$ with $|\psi(x+h)-\psi(x)-h|>\varepsilon h$ has measure $\ll X(\log X)^2/(\varepsilon^2h)$. This is $o(X)$ only when $h/(\log X)^2\to\infty$; here $h\asymp\log X\log\log X$, and (S12) roots each interval at a prime as well. The pair term of (VM) is the survivor pair correlation averaged over the shifts $2a\le2J$. Matomäki, Radziwiłł and Tao, Correlations of the von Mangoldt and higher divisor functions I. Long shift ranges, Proc. London Math. Soc. 118 (2019), arXiv:1707.01315v3, prove the expected asymptotic for $\sum_{X<n\le2X}\Lambda(n)\Lambda(n+h)$ for almost all $|h|\le H$ when $X^{8/33+\varepsilon}\le H\le X^{1-\varepsilon}$; only the abstract was read here. Shift ranges of size $O(K\log X)$ lie far outside that theorem.

The survivor indicator $\mathbf 1[P^-(n)>z]$ is multiplicative, and [[../library/primes/matomaki_2016_multiplicative_functions_short_intervals/theorem_1|Theorem 1 of Matomäki and Radziwiłł]] compares short and long averages of bounded multiplicative functions in almost all intervals of any growing length. Its error is absolute, at least $C'\log\log h/\log h$, while the survivors have density about $1/\log X$. Relative precision would need $\delta\ll1/\log X$, and then its exceptional-set bound exceeds $X$. It does not give (S11).

Gorodetsky's The variance of integers without small prime factors in short intervals, Math. Z. 308, 59 (2024), Theorem 1.1, also misses this range. Its unrooted variance theorem requires, among other hypotheses,

$$(1+a)\frac{\log H}{\log\log H} \le(1-\varepsilon)\frac{\log X}{\log y}, \qquad a=\frac{\log\log H}{\log y}.$$

At $H=2J$ and $y=z$, the left side tends to infinity and the right side tends to $2(1-\varepsilon)$. Its centering is also the periodic density $H\prod_{p\le y}(1-1/p)$. The published statement and these definitions were inspected; the proof was not used.

Tuple upper bounds do not reach (S12) either. Markov's inequality applied to the rooted factorial moments $\sum_p\binom{\pi(p+2J)-\pi(p)}k$ bounds the proportion in (S12) by roughly $B(1+\varepsilon/\lambda)^{-k}$ when those moments are at most $B$ times their Poisson values. The inspected tuple bound has $B$ of size $(2+\delta)^kk!$ for fixed $\delta>0$, which defeats every choice of $k$.

The Buchstab transport rearranges the moment

For a finite odd set $T$ put $W(T)=\sum_{x\in T\cap(X,2X]}(V_T(x)-\lambda K)^2$, so that $\mathcal V_z=W(T_z)$. Deleting one site $m$ from $T$ removes the root $m$ and lowers $V_T$ by one at each earlier root whose window contains $m$, so $W$ changes by

$$\Omega_T(m)=-\mathbf 1_{X<m\le2X}(V_T(m)-\lambda K)^2 +\sum_{\substack{x\in T\cap(X,2X]\\m=x+2a,\ 1\le a\le J}} (1+2\lambda K-2V_T(x)).$$

Let $y=X^{2/5}$, let $T_y$ be the set of $m\in(X,2X+D_*]$ with $P^-(m)>y$, and for a prime $\ell\in(y,z]$ let $T_{<\ell}$ be the set of those $m$ with no prime factor below $\ell$. Since $\ell>D_*$, two multiples of $\ell$ are more than $D_*$ apart, so no window meets two sites deleted at the same prime and their changes add. Since $\ell^3>3X$, every site of $T_{<\ell}$ divisible by $\ell$ is $\ell r$ with $r\ge\ell$ prime. Deleting the multiples of each prime $\ell\in(y,z]$ in increasing order therefore gives the exact identity

$$W(T_z)=W(T_y)+\sum_{\substack{y<\ell\le z\\\ell\text{ prime}}} \sum_{\substack{r\ge\ell\text{ prime}\\\ell r\in(X,2X+D_*]}} \Omega_{T_{<\ell}}(\ell r). \tag{S13}$$

The sum of the two terms is needed with error $o(K^2N_X)$. Their orders alone do not supply that cancellation.

• The base. By the prime number theorem applied to the products $\ell r$ with $X^{2/5}<\ell\le r$, the set $T_y\cap(X,2X]$ has $(1+\log(3/2)+o(1))N_X$ elements. The pair and triple sums in $W(T_y)$ are correlations of $X^{2/5}$-rough numbers at shifts up to $D_*$. Proposition M4 of the survivor-moment note proves that $W(T_y)\gg K^2N_X$ by this census and a telescoping span bound, and that $W(T_y)\ll K^2N_X$ by a fixed-dimensional upper sieve. More precisely its lower coefficient is $\lambda^2\log(3/2)^3/3+o(1)>0$. No concentration hypothesis is used for these bounds. The leading constant is unestimated, and the signed transport would have to cancel a nonzero term of order $K^2N_X$.
• The transport. $\Omega_{T_{<\ell}}(\ell r)$ is a polynomial of degree two in the indicators of $T_{<\ell}$ within $D_*$ of $\ell r$. The double sum is thus a bilinear sum over $\ell\in(X^{2/5},z]$ and $r\ge\ell$, weighted by short-shift correlations of the intermediate sifted sets around $\ell r$. Applying a Type II estimate would require a bound for this specified coefficient class, including its dependence on $\ell$. No such bound is supplied by the inspected estimates; the short-shift correlations remain unestimated.

So (S13) is an exact rearrangement and supplies no centered estimate. It has the same unresolved need for a signed correction as the [[research/prime_binary_new_attack/consecutive_block_counts|upper-count note]] records for its consecutive count in Proposition U12. The base moment's order is known; the matching cancellation over the deleted semiprimes is not.

Nonnegative polynomial cutoff bounds

For a finite odd set $T$ write $\mathcal Q_d(T)=\sum_{x\in T\cap(X,2X],\,V_T(x)\le t}Z_T(x)$, so that $\mathcal Q_{d,z}=\mathcal Q_d(T_z)$.

Lemma S5. Let $P(v)=\sum_{k=0}^{k_0}c_k\binom vk$ with every $c_k\ge0$, and suppose $P(v)\ge1$ for the integers $0\le v\le t$. Then $P(v)\ge1$ for every integer $v\ge0$, and for every finite odd set $T$

$$\sum_{x\in T\cap(X,2X]}Z_T(x)P(V_T(x)) \ \ge\ \sum_{x\in T\cap(X,2X]}Z_T(x)\ \ge\ \mathcal Q_d(T).$$

The left side is a nonnegative combination of the tuple indicators $\mathbf 1_T(x)\prod_{a\in A\cup B}\mathbf 1_T(x+2a)$, with $A$ a compatible $K$-set and $|B|\le k_0$.

Proof. Each $\binom vk$ is nondecreasing on the integers $v\ge0$, so $P$ is, and $P(v)\ge P(t)\ge1$ for $v>t$. Moreover $Z_T(x)\binom{V_T(x)}k$ counts the pairs $(A,B)$ of subsets of $Y_T(x)$ with $A$ compatible, $|A|=K$ and $|B|=k$. $\square$

So a bound for $\mathcal Q_{d,z}$ obtained by majorizing the cutoff with such a polynomial and then bounding each tuple count from above is never smaller than the untruncated certificate count $\sum_xZ_{T_z}(x)$. Its reference expectation diverges, by (S9). Such a bound could succeed only if the actual untruncated count fell below its reference value by a factor tending to infinity, a deficit that the Hardy–Littlewood heuristics do not predict. Within this polynomial cutoff approach, an improvement therefore requires negative coefficients and estimates retaining their signs. This conclusion does not exclude a direct arithmetic bound for the truncated event or every other majorant.

A damped signed form of the second cost

Put $\eta=1/5$. For a finite odd set $T$ let $A_0(T)=|T\cap(X,2X]|$ and

$$F_d(T)=\sum_{x\in T\cap(X,2X]}\eta^{V_T(x)-K}Z_T(x),$$

with summands zero when $V_T(x)<K$. Let $C=2^{\lfloor\log_2R\rfloor}$, so that $C\le R<2C$ and $C$ divides $M$ once $R<M$. Define

$$P_{d,C}(u)=\frac CM\#\{1\le a\le R:a\equiv-du\pmod C\},\qquad \kappa_d(u)=\mathbf 1[r_d(u)\le R]-P_{d,C}(u),$$

and the signed statistic

$$E_d(T)=\sum_{x\in T\cap(X,2X]}\eta^{V_T(x)-K} \sum_{\substack{A\subseteq Y_T(x)\\|A|=K}}\kappa_d(U(A)). \tag{S14}$$

Lemma S6. For every finite odd set $T$, and $X$ so large that $R<M$,

$$\mathcal Q_d(T)\le2R\Bigl(\frac{5^{5/4}}8\Bigr)^KA_0(T)+5^{t-K}E_d(T). \tag{S15}$$

The function $\kappa_d$ has zero Fourier coefficient modulo $M$ at every frequency $r$ with $(M/C)\mid r$, that is, at every character whose conductor divides $C$.

Proof. On roots with $V_T(x)\le t$ one has $\eta^{V_T(x)-K}\ge\eta^{t-K}$, so $\mathcal Q_d(T)\le5^{t-K}F_d(T)$. As $i$ runs over $0\le i<M/C$, the residues $-d(u+iC)$ modulo $M$ run once through the class of $-du$ modulo $C$, because $d$ is odd; their least positive representatives are the $a\in[1,M]$ in that class. Hence $P_{d,C}(u)$ is the average of $\mathbf 1[r_d(\cdot)\le R]$ over $u+iC$, and $0\le P_{d,C}(u)\le (R/C+1)C/M\le2R/M$. Averaging over the subgroup $C\mathbb Z/M\mathbb Z$ keeps exactly the frequencies $r$ with $(M/C)\mid r$, which gives the Fourier statement. Splitting $\mathbf 1[r_d\le R]=P_{d,C}+\kappa_d$ in $F_d$ gives $F_d(T)\le(2R/M)\sum_x\eta^{V_T(x)-K}\binom{V_T(x)}K+E_d(T)$. For every integer $v\ge K$, $\binom vK(4/5)^K(1/5)^{v-K}\le1$, being one value of a binomial distribution, so the last sum is at most $(5/4)^KA_0(T)$. Finally $t-K\le K/4$, so $5^{t-K}(2R/M)(5/4)^K\le2R(5^{5/4}/8)^K$. $\square$

At $T=T_z$, $A_0(T_z)=(1+O(1/K^2))N_X$ by the survivor census of Lemma U10, and $R(5^{5/4}/8)^K\ll_d(\log X)^{15\log_2 5-35}=o(1)$, as in (S8). Therefore the one-sided bound

$$E_d(T_z)\le\frac{N_X}{192\cdot5^{t-K}} \tag{S16}$$

on the scales of the first cost would give $\mathcal Q_{d,z}\le N_X/96$. Since $5^{t-K}=(\log X)^{3\log_25+o(1)}$, the budget is about $N_X(\log X)^{-6.97}$. The trivial bound is $|E_d(T_z)|\le(5/4)^KA_0(T_z)=(\log X)^{12\log_2(5/4)+o(1)}N_X$, about $(\log X)^{3.86}N_X$, so (S16) asks for a saving of about $(\log X)^{10.8}$ in a signed sum. Since

$$\eta^{V_T(x)-K}\prod_{a\in A}\mathbf 1_T(x+2a) =\prod_{a\in A}\mathbf 1_T(x+2a) \prod_{b\in[1,J]\setminus A}\bigl(1-\tfrac45\mathbf 1_T(x+2b)\bigr),$$

$E_d(T_z)$ expands into survivor correlations of every order from $K+1$ to $J+1$ with alternating signs. Lemma S6 removes the coarse part of the phase deterministically; it supplies no estimate for the high-conductor residual. The survivor-moment note's $O(N_X/K)$ bound for affected roots does not transfer this signed sum at the required precision: its summand can have size $(5/4)^K$, and the resulting error is larger than the budget in (S16). The bounded event $\mathcal C_d$, in contrast, does transfer with an $o(N_X)$ error.

Where this leaves the route

The proposed moment and damped estimates ask for correlation information that the inspected theorems do not supply at this scale. The first would imply (S12), a law of large numbers for sums of about $\log\log X$ consecutive gaps. The selected approach to the second leaves the signed sum (S16), whose expansion contains correlations of order exceeding $K$. Uniform prime-tuple hypotheses, such as those assumed by the conditional claims on the problem page, concern growing patterns; no implication from the bounds required here to those hypotheses is asserted. The shorter span supplies no unconditional substitute for the missing estimates. This account concerns the proposed estimates for the two costs of (S10); it does not prove that every bound for either cost must pass through the same correlation expansion.

Band deletions below the sieve level

A different attempt on (MS) lowers the sieve level and asks each band of deleted semiprimes to destroy a fixed share of the counted roots. This section proves the exact deletion identities and states a sufficient lower bound for the losses. Such a bound also controls the prime-supported roots. That implication is a necessary consequence of the proposed route, not a reason an arithmetic proof of the loss bound cannot work.

Write $I=(X,2X+D_*]\cap\mathbb Z$ and let $T\subseteq I$ be a finite set of odd integers. For $m\in T$ and a root $x\in T\cap(X,2X]$ with $0<m-x\le2J$ and $m-x$ even, call $m$ decisive at $x$ when $Z_T(x)\ge1$ and every compatible $K$-subset of $Y_T(x)$ contains $(m-x)/2$. Put

$$L_T(m)=\mathbf 1[X<m\le2X]\min(1,Z_T(m)) +\#\{x:m\text{ is decisive at }x\}.$$

Lemma S17. For every such $T$ and $m\in T$,

$$\mathcal C_d(T)-\mathcal C_d(T\setminus\{m\})=L_T(m)\ge0. \tag{S17}$$

Summing over the sites, $\sum_{m\in T}L_T(m)$ equals $\mathcal C_d(T)$ plus, for each root with $Z_T(x)\ge1$, the number of half-offsets common to all its compatible $K$-subsets; hence it is at most $(K+1)\mathcal C_d(T)$. If $E\subseteq T$ and any two members of $E$ differ by more than $2J$, then $\mathcal C_d(T)-\mathcal C_d(T\setminus E)=\sum_{m\in E}L_T(m)$.

Proof. Deleting $m$ removes its own summand. Another root $x$ can change only if $(m-x)/2\in Y_T(x)$, that is, $0<m-x\le2J$ with $m-x$ even. Its compatible subsets after the deletion are exactly those not containing $(m-x)/2$, so its summand falls from one to zero exactly when $m$ is decisive at $x$. This proves (S17). In the sum over $m$, each counted root is counted once for itself and once for each half-offset common to all its compatible subsets, each of which has $K$ elements. For separated $E$, the set $\{x\}\cup(x+2Y_T(x))$ of each root lies in an interval of length $2J$ and so meets at most one member of $E$; the summand of $x$ changes exactly as in the single deletion of that member from $T$. $\square$

The sum over sites is an upper bound on total influence; it gives no lower bound for any loss. A root whose compatible subsets have no common half-offset has no decisive site, and a site can become decisive only after other sites have been deleted.

Now take $K\ge10^4$ and, for $0\le i\le8$,

$$\alpha_i=\frac{49}{100}+\frac{8i}K,\qquad T_i=\{m\in I:P^-(m)>X^{\alpha_i}\},\qquad \mathcal C_i=\mathcal C_d(T_i).$$

For $0\le i\le7$, put $L_i=\mathcal C_i-\mathcal C_{i+1}$. Then $1/3<\alpha_i<1/2-1/K^2$. For large $X$ every member of $T_i$ is a prime or a product of two primes greater than $X^{\alpha_i}$, and $\mathbb P_X\subseteq T_z\subseteq T_8\subseteq\cdots\subseteq T_0$. Sieve by the primes $\ell\in(X^{\alpha_i},X^{\alpha_{i+1}}]$ in increasing order. At the step of $\ell$ the removed sites are the products $\ell r\in I$ with $r\ge\ell$ prime; since $\ell>2J$, they are separated by more than $2J$. Lemma S17 therefore writes $L_i$ exactly as the sum of $L_{T_{<\ell}}(\ell r)$ over these products, where $T_{<\ell}$ is the set just before the step of $\ell$. The product count in the proof of Proposition U12 of the [[research/prime_binary_new_attack/consecutive_block_counts|upper-count note]], with $a=49/100$ and $b=1/2$, gives $(\log(51/49)+o(1))N_X$ such products in $(X,2X]$ with $\ell\le\sqrt X$; those with $\ell>\sqrt X$ number $O(X/(\log X)^2)$. Hence $\mathcal C_0\le|T_0\cap(X,2X]|<2N_X$.

The bands would suffice if each lost half its counted roots up to a small error. The inequalities

$$L_i\ge\frac12\mathcal C_i-\frac{N_X}{1024}\qquad(0\le i\le7), \tag{PB}$$

on one unbounded set of scales, give $\mathcal C_8\le2^{-8}\mathcal C_0+(N_X/1024)\sum_{j<8}2^{-j}<5N_X/512$. Since $\mathcal C_d(T_z)\le\mathcal C_8$ and $5/512<1/48$, this is (MS). The following proposition identifies the roots that no band can remove.

Proposition S18. For $0\le i\le7$,

$$L_i\le\mathcal C_i-\mathcal C_d(\mathbb P_X). \tag{S18}$$

Consequently (PB) implies $\mathcal C_d(\mathbb P_X)<5N_X/512$ on its scales. If $S=a/(2^sd)$, then for all large $X$ at least one inequality of (PB) fails.

Proof. Monotonicity and $\mathbb P_X\subseteq T_{i+1}$ give $\mathcal C_{i+1}\ge\mathcal C_d(\mathbb P_X)$, which is (S18). The bound $\mathcal C_8<5N_X/512$ obtained from (PB) is at least $\mathcal C_d(\mathbb P_X)$. Under rationality, (S3) and S1 give $\mathcal C_d(\mathbb P_X)\ge B_d^*(X)\ge(1/24-o(1))N_X$, which exceeds $5N_X/512$ for large $X$. $\square$

Deleting composites never removes a compatible subset of primes. Only composite roots, or prime roots whose compatible subsets all contain a composite, can be lost. Thus (S18) converts the proposed loss bounds into upper bounds for prime roots certified by primes alone. The bound supplied by (PB), together with (S3) and S1, contradicts rationality. The band identities alone rearrange $\mathcal C_0-\mathcal C_8$ and supply no estimate. This persistent prime contribution exists for every order or range of deletions ending at a set containing $\mathbb P_X$, since the total loss is at most $\mathcal C_d(T)-\mathcal C_d(\mathbb P_X)$. It does not refute (PB): the conclusion that (PB) must fail under rationality is precisely what makes an unconditional proof of (PB) sufficient. The remaining question is an arithmetic lower bound for the actual losses.

An arithmetic expansion of the band loss

The following calculation attempts that lower bound by exposing the prime cofactor of each deleted semiprime. It gives an exact sum to estimate, but the available progression theorem does not estimate its full coefficient. The failure is in the proposed application of the theorem, not a disproof of (PB).

Fix the parameters of (S1), and put $\mathcal B=\{-J,\ldots,-1,1,\ldots,J\}$. For $v\in\{0,1\}^{\mathcal B}$, translate the deleted site to zero and set $T(v)=\{0\}\cup\{2b:b\in\mathcal B,\ v_b=1\}$. The even coordinates here are translations of odd sites. Write $F_T(x)=\min(1,Z_T(x))$ using the same half-offset convention as (S2), and define the local loss

$$\mathcal L(v)=F_{T(v)}(0)+ \sum_{a=1}^J v_{-a} \bigl(F_{T(v)}(-2a)-F_{T(v)\setminus\{0\}}(-2a)\bigr).$$

Thus $0\le\mathcal L(v)\le J+1$. It vanishes if fewer than $K$ of the neighbor sites are present: every counted loss needs a root and $K$ successors, one of these $K+1$ distinct sites being the deleted site. The unique multilinear expansion on the Boolean cube is consequently

$$\begin{aligned} \mathcal L(v)&=\sum_{\substack{B\subseteq\mathcal B\\|B|\ge K}} a_B\prod_{b\in B}v_b,\\ a_B&=\sum_{E\subseteq B}(-1)^{|B|-|E|}\mathcal L(\mathbf1_E). \end{aligned} \tag{S19}$$

The second identity is finite inclusion--exclusion; it also proves the first by evaluating at every $\mathbf1_E$. In particular $|a_B|\le(J+1)2^{|B|}$ and $\sum_B|a_B|\le(J+1)3^{2J}$. These are crude bounds, not lower bounds for the loss from any cancellation. Nonnegativity of $\mathcal L$ does not make its coefficients nonnegative. For example, at $K=1$, $J=3$ and $d=R=1$, only offsets $1$ and $3$ qualify, and direct evaluation gives

$$\mathcal L(v)=v_1+v_3-v_1v_3+ v_{-1}(1-v_2)+v_{-3}(1-v_{-2}).$$

Let $\chi_{<\ell}(n)=\mathbf1[P^-(n)\ge\ell]$. In a band of (PB), the set immediately before sieving by $\ell$ has this indicator on $I$. For $m\in(X+2J,2X]$, the coefficient of its deletion in S17 is therefore $\mathcal L((\chi_{<\ell}(m+2b))_{b\in\mathcal B})$. All earlier roots and all their forward sites lie in the permitted interval. Sites within $2J$ of the left boundary, and deleted sites in the upper buffer $(2X,2X+D_*]$, contribute a total error $O(J^2)$: there are $O(J)$ such integer sites, each with loss at most $J+1$. This error is bounded before expanding (S19), so it is not multiplied by $\sum|a_B|$.

It follows that the exact arithmetic sum for a band is

$$\begin{aligned} L_i={}& \sum_{\substack{B\subseteq\mathcal B\\|B|\ge K}}a_B \sum_{\substack{X^{\alpha_i}<\ell\le X^{\alpha_{i+1}}\\\ell\ {\rm prime}}} \ \sum_{\substack{X/\ell<r\le2X/\ell\\r\ {\rm prime}}} \prod_{b\in B}\chi_{<\ell}(\ell r+2b) +O(J^2). \end{aligned} \tag{S20}$$

Here $r>\ell$, since $\alpha_{i+1}<1/2$; thus $\ell$ is indeed the least prime factor of each removed site. The separated-deletion part of S17 justifies summing the coefficients computed before removing all multiples of a fixed $\ell$. Formula (S20) keeps the original rooted count. Its error is $o(N_X)$, and every nonzero monomial involves at least $K$ neighboring survival conditions. Fixed-dimensional estimates such as M1 do not evaluate this growing sum.

To try a progression estimate in the variable $r$, expand each factor exactly as

$$\chi_{<\ell}(n)= \sum_{\substack{q\mid n\\q\mid\prod_{p<\ell}p}}\mu(q).$$

Each argument is at most $3X$ for large $X$, so one may restrict every $q_b$ to at most $3X$ without changing this exact sum. For a chosen divisor $q_b$ at each site, let $Q=\mathop{\rm lcm}_{b\in B}q_b$. The resulting congruences are $\ell r\equiv-2b\pmod {q_b}$. They are consistent exactly when $\gcd(q_b,q_c)\mid2(b-c)$ for every pair $b,c$. When consistent they specify a single class $r\equiv a\pmod Q$, because $\ell$ is coprime to $Q$. A nonprimitive class contributes no prime $r$: every prime divisor of $Q$ is below $\ell$, whereas $r>\ell$. Each remaining term is a prime count in a primitive progression over $(Y,2Y]$, where $Y=X/\ell$.

This last statement is an exact reduction, not a license to replace all those counts by their main terms. The Bombieri--Vinogradov theorem as stated in the Polymath source, arXiv v4, Section 2, Claim 2.2 and Theorem 2.3, supplies averaged errors for moduli at most $Y^\theta$ for each fixed $\theta<1/2$. At the cofactor scale here this is at most $X^{51\theta/100}$, strictly below $X^{51/200}$. The exact divisor expansion includes larger moduli; even a single divisor $q_b<\ell$ need not be in that range. There is no proved truncation or cancellation of the out-of-range terms. Nor does the pointwise bound $\mathcal L\le J+1$ control the sum of the absolute divisor coefficients required for a termwise error estimate.

One may instead try to apply a convolution theorem at the product scale $X$. Claim 2.6 and Theorem 2.8 of the same source give the corresponding generalized Bombieri--Vinogradov statement for exponents below $1/2$, with divisor-bounded factor coefficients and a stated Siegel--Walfisz condition on one factor. A fixed divisor choice in (S20) can be reorganized into a semiprime progression sum, with a least-factor cutoff depending on its divisors. To use that theorem one still has to handle these cutoffs, sum the signed coefficients, and control the moduli beyond its range. Neither statement is an estimate for an arbitrary bounded function of the neighboring sifted sites. The cited statements and hypotheses are inspected; their external proofs are not independently audited here.

There is an actual arithmetic test for the simplest proposed disposal of the omitted divisors. At the level $z$ of (MS), U10 and Mertens' product formula give, respectively,

$$\begin{aligned} |T_z\cap(X,2X]|&=(1+o(1))N_X,\\ X\prod_{p\le z}(1-1/p)&=(2e^{-\gamma}+o(1))N_X. \end{aligned} \tag{S21}$$

Here $\gamma$ is Euler's constant and $2e^{-\gamma}>1$. The first identity counts the actual primes and the $O(N_X/K^2)$ rough composites. The second uses $\log z=(1/2-1/K^2)\log X$ and $N_X\sim X/\log X$. Thus replacing the exact one-site divisor sum by its periodic density has error $(1-2e^{-\gamma}+o(1))N_X$, already of order $N_X$. An $o(N_X)$ remainder for that replacement is false. This calculation does not determine the sign or size of the corresponding weighted remainder in (S20), and it does not refute (PB). It shows why the large-divisor contribution needs an arithmetic treatment even before addressing the phase condition.

The missing input for this attack is a lower estimate of (S20) by $\mathcal C_i/2-N_X/1024$, simultaneously for the eight bands on unbounded scales for each fixed odd $d$. A usable progression or switching argument must prove both the summed error control, including the out-of-range divisors, and the comparison of its retained main term with $\mathcal C_i/2$. Neither is supplied by (S19)--(S21). Obtaining such an estimate would finish the whole contradiction by (PB), S1 and S3; no independent-gap distribution would then be needed.

A bounded target and its aggregate error

The numerical target in (MS) is stronger than the contradiction needs. Keeping every parameter of (S1), it suffices, for each fixed odd $d$, to prove on an unbounded set of scales

$$\mathcal C_d(\mathbb P_X) \le\left(\frac1{24}-\varepsilon_d\right)N_X, \qquad \varepsilon_d>0. \tag{BC}$$

The scales and the saving may depend on $d$. The estimate may also be proved under the rationality assumption: a contradiction does not require an unconditional upper theorem. That assumption must supply a valid additional argument, rather than the desired upper estimate itself.

Proposition S22. Condition (BC) contradicts $S=a/(2^sd)$ for every fixed $s$. The concrete upper bound $\mathcal C_d(\mathbb P_X)\le N_X/32$ suffices. With the eight bands of (PB), put

$$e_i=\mathcal C_{i+1}-\frac12\mathcal C_i, \qquad E=\sum_{i=0}^7 2^{-(7-i)}e_i.$$

The single bound $E\le N_X/64$ on unbounded scales also suffices. Under rationality, in contrast, these same quantities satisfy $E\ge(13/384-o(1))N_X$.

Proof. S1 and S3 force $\mathcal C_d(\mathbb P_X)\ge(1/24-o(1))N_X$ past every fixed preperiod. This contradicts (BC), and $1/24-1/32=1/96>0$. Iterating the definition of $e_i$ gives the exact identity

$$\mathcal C_8=2^{-8}\mathcal C_0+E. \tag{S22}$$

Since $\mathcal C_0<2N_X$ and $\mathcal C_d(\mathbb P_X)\le\mathcal C_8$, the proposed bound on $E$ gives

$$\mathcal C_d(\mathbb P_X) \le\mathcal C_8 <\left(\frac1{128}+\frac1{64}\right)N_X =\frac3{128}N_X<\frac1{32}N_X.$$

Conversely, under rationality the forced lower count and (S22) give $E\ge(1/24-1/128-o(1))N_X=(13/384-o(1))N_X$. $\square$

Corollary M3 transfers a fixed saving below $1/24$ between the actual primes and $T_z$, with an $o(N_X)$ change and, if necessary, a smaller positive saving. Neither the prime bound $N_X/32$ nor the aggregate bound above is asserted to imply the specific bound $N_X/48$ in (MS). They contradict S1 directly. No centered moment or unbounded certificate count is needed if (BC) itself can be estimated.

The individual inequalities (PB) imply $e_i\le N_X/1024$ and hence $E<N_X/512$, a stronger sufficient bound. The aggregate target permits errors in different bands to compensate. It does not yet supply a new arithmetic estimate: $E$ is identically $\mathcal C_8-2^{-8}\mathcal C_0$. In particular, the rationality lower bound on $E$ is part of the proposed contradiction, not a refutation of an unconditional upper bound for it.

The attempted use of fixed-dimensional sieve bounds on (S20) stops at this aggregate too. Every monomial there has at least $K$ neighboring survival factors, without a coefficient-positivity guarantee. Upper bounds for separate correlations provide neither the required signed remainder nor its comparison with the bounded count. The pointwise size of the local loss does not give the small absolute coefficient sum required by a termwise progression estimate. S21 also prevents discarding the full arithmetic correction to periodic density as $o(N_X)$, already for one site. That test does not determine the weighted correction. To finish this route one must estimate (BC), or prove the stated bound for $E$ with its actual coefficients and all out-of-range divisor contributions included. Both estimates remain unknown, even for $d=1$.

Failure account

• The shorter-span rationality lower bound and the monotone upper enlargement are proved. (MS) for actual survivors is unknown for every odd $d$. It is a stronger sufficient upper target, not an equivalence to the original consecutive-block upper bound. Proposition S22 instead needs only a fixed saving below $1/24$ for the bounded prime event, for example $N_X/32$, or the aggregate band bound $E\le N_X/64$. These weaker sufficient estimates are also unproved.
• S3 proves rarity only in the stated independent-site reference. Its parameter choice avoids U14's large-span saturation, but gives no arithmetic comparison. The underlying sieve level is near $\sqrt X$, not the small periodic exclusion of U9.
• Bounding the monotone event by the untruncated sum of compatible tuple indicators fails to be small in the reference, by (S9). Lemma S5 extends this to every cutoff majorant that is a nonnegative combination of binomial coefficients in the local count, followed by tuple upper bounds: such a bound is at least the untruncated count. No assertion is made about signed or other arithmetic majorants.
• The centered moment (VM) is unknown and does not depend on $d$. Corollary M3 of the survivor-moment note identifies it, up to $O(N_X)$, with the prime-rooted moment (PVM). Proposition S4 proves that it implies (S12), a law of large numbers for sums of about $(6/5)K$ consecutive gaps after almost every prime. No proof of (S12) is known to this note, even under the Riemann hypothesis, whose mean-square bound needs intervals longer than $(\log X)^2$. Its pair term is an average over shifts of size $O(K\log X)$, far below the shift ranges of the known averaged pair-correlation theorem. Tuple upper bounds cannot give it: their loss $(2+\delta)^kk!$ defeats the factorial-moment route, and the inspected uncentered moment theorem has the wrong threshold range and root convention. These inspected bounds do not establish (VM), which remains a short-interval distribution statement of unknown status.
• The signed Buchstab transport (S13) is exact. Its base $W(T_y)$ is bounded above and below by positive constants times $K^2N_X$ in Proposition M4; no leading constant is computed. Its correction is a bilinear sum over semiprimes weighted by short-shift correlations, and no cancellation to $o(K^2N_X)$ is proved. Reopens with an estimate for that signed combination, or directly for (PVM); another bound for the number of deleted semiprimes is not enough.
• The truncated phase cost $\mathcal Q_{d,z}$ is unknown for every odd $d$. Lemma S6 reduces it to the one-sided bound (S16) for the damped high-conductor statistic $E_d(T_z)$, with budget about $N_X(\log X)^{-6.97}$ against a trivial size about $(\log X)^{3.86}N_X$. That statistic expands into survivor correlations of every order from $K+1$ to $J+1$ with alternating signs. No required estimate is proved here or supplied by the inspected sources. Reopens with a cancellation argument for (S16) that does not pass through asymptotics for each correlation, or with a direct estimate for the bounded event $\min(1,Z_{T_z})$.
• The [[research/prime_binary_new_attack/rational_deletion_signs|conditional deletion-sign bound]] concerns the consecutive-block count at an enlarged span. Its negligible negative mass under rationality estimates neither arithmetic cost in (S10); it supplies no transfer from the independent reference to the actual survivors here.
• Lowering the sieve level to $X^{49/100}$ and asking each of eight bands of deleted semiprimes to destroy half of the counted roots, (PB), would finish the contradiction; Lemma S17 gives each band loss exactly. The band composition and Proposition S18 prove that (PB) implies (MS) and the prime bound $\mathcal C_d(\mathbb P_X)<5N_X/512$, because deleting composites never removes a prime root certified by primes alone. This necessary consequence does not refute (PB) or prohibit an arithmetic proof of it. The exact expansion (S20) attempts such a proof through the prime cofactors of removed semiprimes. The termwise progression argument fails to control its signed coefficients, the moduli outside the available range, and the comparison of the remaining contribution with $\mathcal C_i/2$. Those weighted estimates are unknown even for $d=1$. The simpler replacement of the one-site count by periodic density with an $o(N_X)$ remainder is false by (S21). Reopens with an estimate of the full weighted arithmetic sum, including that correction, or with a direct upper bound for the bounded prime event.