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These are Plasma AI's author-recorded deductions supporting the shorter-span argument for Problem 251. They have no independent verification warrant. The canonical sketch retains the full conditional contradiction. In particular, this note proves neither its centered-moment input (VM) nor its truncated phase bound.

Use the parameters of (S1):

$$K=\lceil12\log_2\log X\rceil,\qquad \lambda=\frac65,\qquad D_*=\lfloor\lambda K\log X\rfloor,\qquad J=\lfloor D_*/2\rfloor,\qquad z=X^{1/2-1/K^2}.$$

Put $\mathcal I=(X,2X+D_*]$, $N_X=\pi(2X)-\pi(X)$, $T_z=\{m\in\mathcal I:P^-(m)>z\}$ and $\mathbb P_X=\mathbb P\cap\mathcal I$. For a finite odd set $T\subseteq\mathcal I$, write

$$V_T(x)=\#\{1\le a\le J:x+2a\in T\},\qquad W(T)=\sum_{x\in T\cap(X,2X]}(V_T(x)-\lambda K)^2.$$

All constants below are absolute once the displayed parameters are fixed. The sieve argument uses a bounded number of linear factors, at most three. It does not give a sieve estimate uniform in the growing number $K$ of factors needed by the phase count.

A uniform sieve for at most three linear factors

Write $F(n)=n/\varphi(n)$ for positive integers, with $F(1)=1$. Let $1\le k\le3$, let $b_0=0,b_1,\ldots,b_{k-1}$ be distinct integers, and put

$$\Delta=\prod_{0\le i<j<k}(b_i-b_j),\qquad \Psi_k(\Delta)=F(|\Delta|)^{k-1},$$

with empty product $\Delta=1$ when $k=1$.

Lemma M1. Let $Y\to\infty$ and $w=Y^{1/10}$. Suppose $L=1$ or $L$ is a prime greater than $w$. The number of positive integers $r\le Y$ for which none of

$$r,\quad Lr+b_1,\quad\ldots,\quad Lr+b_{k-1}$$

is divisible by a prime less than $w$ is

$$\ll_k\Psi_k(\Delta)\frac{Y}{(\log Y)^k}. \tag{M1}$$

The constant is independent of $L$ and the offsets. If their residue classes cover all classes at some prime below $w$, the count is zero.

Proof. For a prime $p<w$, multiplication by $L$ identifies the forbidden residues with $0,-b_1,\ldots,-b_{k-1}$ modulo $p$. Let $\nu(p)$ be their number. In the nonzero case $1\le\nu(p)<p$ for every such prime. On squarefree numbers supported on these primes define

$$a(d)=\prod_{p\mid d}\frac{\nu(p)}p,\qquad h(d)=\prod_{p\mid d}\frac{a(p)}{1-a(p)},\qquad G(w)=\sum_{d<w}\mu^2(d)h(d),\qquad W_0(u)=\prod_{p<u}(1-a(p)).$$

Divisor sums involving $a$ or $h$ below range over squarefree integers.

The Chinese remainder theorem counts $r\le Y$ for which $d$ divides $r\prod_{1\le i<k}(Lr+b_i)$ as $Ya(d)+O(\nu(d))$, where $\nu(d)=\prod_{p\mid d}\nu(p)$. Choose squarefree-supported weights

$$\alpha_d= \frac{\mu(d)}{G(w)\prod_{p\mid d}(1-a(p))} \sum_{\substack{m<w/d\\(m,d)=1}}\mu^2(m)h(m) \quad(d<w),$$

and zero otherwise. Then $\alpha_1=1$ and $|\alpha_d|\le W_0(w)^{-1}$. Squaring the divisor sum gives an upper bound for the desired count. Its main quadratic form is

$$\sum_{d,e<w}\alpha_d\alpha_e a([d,e])=\frac1{G(w)}. \tag{M2}$$

Indeed, $a([d,e])=a(d)a(e)\sum_{l\mid(d,e)}h(l)^{-1}$, so the form is $\sum_l y_l^2/h(l)$, where $y_l=\sum_{l\mid d}a(d)\alpha_d$. Möbius inversion of the displayed weights gives $y_l=\mu(l)h(l)/G(w)$ for squarefree $l<w$, proving (M2). The remainder has absolute value at most

$$W_0(w)^{-2} \left(\sum_{d<w}\mu^2(d)\nu(d)\right)^2 \ll_k w^2(1+\log w)^{4k-2}. \tag{M3}$$

Here $\nu(d)\le k^{\omega(d)}\le\tau_k(d)$, whose elementary divisor sum is at most $w(1+\log w)^{k-1}$. Also $W_0(w)\gg_k(\log w)^{-k}$: the finitely many primes $p\le k$ have $1-a(p)\ge1/p$, and for $p>k$ the lower product $\prod(1-k/p)$ differs from $\prod(1-1/p)^k$ by bounded positive factors. Mertens' product estimate applies.

For completeness a short lower bound for $G$ suffices. Set $u=w^{1/(4k)}$. The sum of $h(d)$ over divisors of $\prod_{p<u}p$ is $W_0(u)^{-1}$. Relative to this finite positive sum, the mean of $\log d$ is

$$\sum_{p<u}a(p)\log p\le k\log u+O_k(1) =\tfrac14\log w+O_k(1).$$

Thus at least half its mass has $d<w$ for large $w$, and $G(w)\ge1/(2W_0(u))$. This is an inequality for a finite Euler product, with no independence assumption about the primes being counted. If $p\nmid\Delta$, then $\nu(p)=k$ and $(1-k/p)/(1-1/p)^k\le1$. At the other primes $\nu(p)\ge1$, so their normalized factor is at most $(1-1/p)^{1-k}$. Consequently

$$W_0(u)\le\Psi_k(\Delta)\prod_{p<u}(1-1/p)^k \ll_k\frac{\Psi_k(\Delta)}{(\log w)^k}.$$

The main term $Y/G(w)$ now has the bound (M1); (M3), with $w=Y^{1/10}$, is smaller. This proves the lemma. $\square$

The required averages of the factor $\Psi_k$ have no growing loss. For every fixed positive integer $m$,

$$\sum_{n\le D}F(n)^m\ll_m D. \tag{M4}$$

To see this, expand $F(n)^m=\sum_{d\mid n}f_m(d)$, where $f_m$ is nonnegative, multiplicative, squarefree-supported and $f_m(p)=(1-1/p)^{-m}-1=O_m(1/p)$. The Euler product $\sum_d f_m(d)/d$ converges. In particular, summing over signed offsets $0<|b_i|\le D$ with distinct $b_i$ gives

$$\sum_{b_1}\Psi_2(b_1)\ll D,\qquad \sum_{b_1,b_2}\Psi_3(b_1b_2(b_1-b_2))\ll D^2. \tag{M5}$$

For the second inequality use $F(|b_1b_2(b_1-b_2)|)^2 \le F(|b_1|)^2F(|b_2|)^2F(|b_1-b_2|)^2$ and Hölder's inequality. Each of the three sums of sixth powers over the pairs is $O(D^2)$ by (M4); every nonzero difference occurs at most $O(D)$ times. Restricting the offsets to even integers preserves these upper bounds.

Near-square-root composites have negligible local mass

Let $E=T_z\setminus\mathbb P_X$. For $j=0,1,2$ define $B_j$ as follows: for each $m\in E$, count ordered, distinct, nonzero even offsets $b_1,\ldots,b_j$, each of absolute value at most $D_*$, for which every $m+b_i\in T_z$; then sum over $m$. Thus $B_0=|E|$.

Proposition M2. At the displayed parameters,

$$B_j\ll\frac{N_X}{K^2} \left(\frac{D_*}{\log X}\right)^j \quad(0\le j\le2). \tag{M6}$$

Proof. Since $z^3>3X$, every $m\in E$ has a unique expression $m=\ell r$ with primes $z<\ell\le r$ and $\ell\le\sqrt{3X}$. For fixed $\ell$ and offsets, take $Y=3X/\ell$. Uniformly in this range $\log Y\asymp\log X$, $\ell>Y^{1/10}$ and $z>Y^{1/10}$. Every counted $r$ and every $\ell r+b_i$ avoids all primes below $Y^{1/10}$. Lemma M1 with $k=j+1$ bounds their number by

$$\ll_j\Psi_{j+1}(\Delta) \frac{X}{\ell(\log X)^{j+1}}.$$

Enlarging the range to $r\le Y$ only increases this bound, as does discarding the restrictions $r\ge\ell$ and $m+b_i\in\mathcal I$. Mertens' prime reciprocal estimate gives

$$\sum_{z<\ell\le\sqrt{3X}}\frac1\ell \ll\log\frac{\log\sqrt{3X}}{\log z}+\frac1{\log z} \ll\frac1{K^2}.$$

Apply (M5), or its empty-offset version when $j=0$, and use $N_X\sim X/\log X$. This proves (M6). Squares $m=\ell^2$ are included, and both ends of $\mathcal I$ are covered. $\square$

Corollary M3. Suppose $f_T(x)\in[0,1]$ depends only on $T\cap[x,x+D_*]$ and on fixed parameters, with the same rule for each $T$. Then

$$\left|\sum_{x\in T_z\cap(X,2X]}f_{T_z}(x) -\sum_{x\in\mathbb P_X\cap(X,2X]}f_{\mathbb P_X}(x)\right| \ll\frac{N_X}{K}. \tag{M7}$$

Also

$$|W(T_z)-W(\mathbb P_X)|\ll N_X. \tag{M8}$$

Proof. A root can change a bounded summand only if it is composite, or it is a prime with a composite of $E$ in its forward window. These possibilities affect at most $B_0+B_1\ll N_X/K$ roots. This proves (M7), even if $f$ is nonmonotone.

For (M8), write $A_0(T)=|T\cap(X,2X]|$, $A_1(T)=\sum_{x\in T\cap(X,2X]}V_T(x)$ and $A_2(T)=\sum_{x\in T\cap(X,2X]}V_T(x)(V_T(x)-1)$. Each pair or ordered triple counted in the difference of these quantities contains a composite $m$. All other sites are within distance $D_*$ of $m$. Therefore, with $\Delta A_i=A_i(T_z)-A_i(\mathbb P_X)$,

$$0\le\Delta A_0\le B_0,\qquad 0\le\Delta A_1\le2B_1,\qquad 0\le\Delta A_2\le3B_2.$$

The exact expansion $W(T)=A_2(T)+(1-2\lambda K)A_1(T)+\lambda^2K^2A_0(T)$ then yields $|\Delta W|\ll B_2+KB_1+K^2B_0\ll N_X$. In particular the negative coefficient is kept; no matching main terms have been inferred from upper sieve bounds. $\square$

It follows that (VM) is equivalent, along the same scales, to

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

This prime-rooted centered estimate remains unproved. The monotone count $\mathcal C_d$ and the overfull count $\mathcal D$ also satisfy (M7), uniformly in their parameters. Thus composites can be removed from the bounded upper event with error $o(N_X)$. This does not supply an upper bound for that event on the primes.

The same argument does not remove composites from the required signed phase statistic $E_d(T)$ of S6. Its summand has absolute value at most $(5/4)^K$, since $\binom vK(1/5)^{v-K}(4/5)^K\le1$ for $v\ge K$ and $|\kappa_d|\le1$. The bound from affected roots is consequently $O((5/4)^KN_X/K)$, whereas S16 needs a bound of order $N_X5^{-(\lfloor5K/4\rfloor-K)}$. The former is much larger. Likewise the cutoff count $\mathcal Q_d$ can be as large as $\binom{\lfloor5K/4\rfloor}{K}$ per affected root. These statistics are not covered by the bounded-observable conclusion (M7).

The stopped base has a nonzero second moment

Let $y=X^{2/5}$ and $T_y=\{m\in\mathcal I:P^-(m)>y\}$. Put $c=1+\log(3/2)$. The PNT product calculation and the near-square-root remainder bound in U12 and U10 give

$$|T_y\cap(X,2X]|=(c+o(1))N_X. \tag{M9}$$

Only this one-point census, not a correlation asymptotic, is used in the following lower bound.

Proposition M4.

$$\left(\frac{\lambda^2(c-1)^3}{3}+o(1)\right)K^2N_X \le W(T_y)\ll K^2N_X. \tag{M10}$$

Proof. List the infinite sequence of $y$-rough positive integers as $s_i$. For each fixed $C>0$, telescoping gives, uniformly for $1\le m\le CK$,

$$\sum_{X<s_i\le2X}(s_{i+m}-s_i)=(1+o(1))mX. \tag{M11}$$

Indeed the sum is the sum of the $m$ terms immediately after the final root minus the first $m$ root terms. These lie respectively at $2X+o(X)$ and $X+o(X)$, uniformly over that range of $m$: the rough integers include every prime above $y$, and PNT supplies more than $CK$ primes in each fixed relative neighborhood of these endpoints. Letting the relative neighborhood tend to zero proves the uniform statement. There are more than $CK$ roots by (M9).

For $1\le u\le c$, take $m=\lfloor u\lambda K\rfloor+1$. If $V_{T_y}(s_i)\le u\lambda K$, then $s_{i+m}-s_i>2J$. The window lies in $\mathcal I$, so the finite and infinite definitions of this count agree. Markov's inequality and (M11) give, uniformly in $u$,

$$\#\{X<s_i\le2X:V_{T_y}(s_i)\le u\lambda K\} \le(u+o(1))N_X.$$

Using (M9), the number with $V_{T_y}(s_i)>u\lambda K$ is at least $(c-u-o(1))N_X$. Integration of this lower bound gives

$$\begin{aligned} W(T_y) &\ge\int_{\lambda K}^{c\lambda K} 2(v-\lambda K)\, \#\{x\in T_y\cap(X,2X]:V_{T_y}(x)>v\}\,dv\\ &\ge\left(2\lambda^2\int_1^c(u-1)(c-u)\,du+o(1)\right)K^2N_X\\ &=\left(\frac{\lambda^2(c-1)^3}{3}+o(1)\right)K^2N_X. \end{aligned}$$

For the upper bound apply M1 with $L=1$, $Y=3X$, and respectively zero, one and two positive even offsets. Its sifting level $Y^{1/10}$ is below $y$. Averaging by (M5) gives $A_0(T_y)\ll N_X$, $A_1(T_y)\ll KN_X$ and $A_2(T_y)\ll K^2N_X$. The expansion used in M3 now proves the upper bound. $\square$

In the exact Buchstab identity S13,

$$W(T_z)=W(T_y)+\text{the signed deletion sum},$$

the base is therefore of order $K^2N_X$. If (VM) holds, the deletion sum must cancel a positive contribution of at least $(\lambda^2\log(3/2)^3/3+o(1))K^2N_X$. Proposition M4 does not compute the leading constant of the base and gives no matching asymptotic for the signed sum. Bounding each correlation from above still cannot supply this cancellation.

Position in the whole contradiction

Rationality with any fixed odd denominator part $d$ forces the shorter consecutive-block count to be at least $(1/24-o(1))N_X$. The same roots belong to the bounded subset event $\mathcal C_d(\mathbb P_X)$ and to $\mathcal C_d(T_z)$, whose difference is $O(N_X/K)$ by M3. To finish by the two-cost split still requires both a bound for overfull roots and a bound for compatible subsets at the other roots, on the same unbounded scales for every fixed odd $d$. Neither has been proved, even for $d=1$.

The failed step is an arithmetic estimate, not removal of the rare near-square-root composites. Reopening this route requires either the prime-rooted estimate (PVM), or a sufficient direct overfull-root bound, together with the truncated phase estimate; a direct estimate for the bounded event could replace both. The fixed-dimensional sieve above has only constant-factor upper bounds for the relevant prime correlations. It supplies neither their signed cancellation nor the growing-order phase information.