Primes 2u-1 with smooth u

---

This page carries the second link of the smooth-successor route: from the Type II estimate for factorizations of 2u-1 to a count of primes $p$ with $P^+(p+1)\le x^\delta$, and from it to Hypothesis $\mathrm A^+$. The extraction is OpenAI's, from Section 7 of the weighted-dilation manuscript; running it on primes $2u-1$ is Shisheng Li and Zhong Zhuang's proposal; the derivation and checks below are Plasma AI's. The conditional-chain examination of 2026-10-15 has a distinct passing grade, examined on 2026-10-16 UTC, for the frozen balanced extraction conditional on exact P and the stated classical premises. The unrestricted dyadic assertion remains refuted-as-stated.

Besides the signed Type II estimate, which assumes premise (P), the argument uses the following source results. Together with the Section 2 estimates they are premise (E) of the combined sketch. They concern $u$ or ordinary factor tests; the only occurrence of $N_u$ is a size relation in (7.17).

• Proposition 7.1 (candidate mass) concerns $A$, $X_A$ and the prime factors of $u$ only.
• Lemma 7.3 (proxy discrepancy) concerns the factor tests and their proxies only.
• Lemma 7.4 (proxy density bounds and the Buchstab identity). Its bound (7.17) is stated for factorizations $mn=N_u$; its proof uses $N_u$ only through $N_u\asymp x$, which holds for $N_u=2u-1$.

It also uses the block sieve (the manuscript's Lemma 2.9), the Siegel–Walfisz theorem, the multiplicative large sieve and Mertens' estimates. The written reconstruction of E supplies these estimates at precisely the restrictions and parameter order used below. Its separate passing grade accepts the operative sieve fragment and exact E2–E5 scopes relative to their stated classical premises. The graded comparison checks the actual interfaces; it supplies no P acceptance. In particular, its interval-restricted proxy discrepancy is for the same original $\alpha$ required by the signed Type II estimate.

The candidate

Fix $0<\delta<1/4$. Notation is that of the Type II page. Fix a nonnegative smooth $\Psi$ supported in $[1,2]$ and bounded below by a positive constant on a closed subinterval of $(1,2)$ of positive length. Take $q=1/2$, $\ell=1$, $\lambda=2$,

$$c_1=1,\quad c_2=\tfrac65,\quad c_3=\tfrac32,\quad c_4=\tfrac95,\qquad s'=\frac1{r_0(c_1+c_2)},\qquad s_j=s'2^{-j}L,$$

with $r_0$ a fixed integer such that $c_2s'<\delta$ and $(r_0-1)c_1/2\ge c_2+1$. Let $j_x$ be the largest $j$ with $c_1s_j\ge20T$ and $\mathcal Q_j=\{p:c_1s_j\le\log p\le c_2s_j\}$. At each scale $j$ for which $[c_3s_j,c_4s_j]$ lies inside $[L^{1/10},L^{1/5}]$ or inside $[L^{3/10},L^{2/5}]$, the primes with logarithm in $[c_3s_j,c_4s_j]$ form a small or a big group. By the prime number theorem their harmonic masses lie between fixed positive bounds and each kind has a number of groups comparable to $T$, so they meet the hypotheses of the Type II page with $(a,b,c,d)=(1/10,1/5,3/10,2/5)$; they avoid the bands because $c_2<c_3<c_4<2c_1$, and the bands meet its hypotheses with $\lambda=2$ and $20T/c_1\le s_{j_x}<40T/c_1$. The exceptional slot of band $j_{**}\ge j_0$ holds an integer $v$ with $e^{c_1s_{j_{**}}}\le v\le e^{c_2s_{j_{**}}}$ and $P^-(v)>W$, where $j_0$ is a fixed lower bound depending only on the candidate geometry. Then

$$A(u)=a_Q(u_*)W_1(u),\qquad X_A=\sum_uA(u)\Psi(u/x),\qquad N_u=2u-1 .$$

These are the manuscript's choices; only $N_u$ changes. By Proposition 7.1, for constants $C,C_A,C_w$ depending only on the early choices and for large $x$,

$$A(u)\le L^C\tau(u_*),\qquad A(u)\le\exp(C_w\sqrt L)\ (u\asymp x),\qquad X_A\asymp\frac xLH_QH_P\gg xL^{-C_A},$$

where $H_Q=\sum_va_Q(v)/v$ and $H_P=\sum_{r_*=1}W_1(r)/r$; the comparison constants do not depend on $j_{**}\ge j_0$; and on the support of $A(u)\Psi(u/x)$ every prime factor of $u$ lies in $[L^{20},x^\delta]$. The upper end uses $c_2s'<\delta$: the band primes are at most $x^{c_2s'}$, the exceptional integer is at most $x^{c_2s'}$, and the group primes are below $W$.

Type I distribution in the class of 1/2

For odd squarefree $d$ put

$$r(d)=\sum_{d\mid2u-1}A(u)\Psi(u/x)-\frac{X_A}{\varphi(d)} .$$

Proposition (Type I). For every fixed $\vartheta<1/2$ and $D>0$,

$$\sum_{\substack{d\le x^\vartheta\\ d\text{ odd, squarefree}}}|r(d)| \ll xL^{-D}+X_AL^{-18}.$$

Proof. If a prime $l$ divides both $d$ and $u$, then $2u-1\equiv-1\pmod l$, so $d\nmid2u-1$. For $(u,d)=1$, $d\mid2u-1$ holds exactly when $u\equiv\bar2\pmod d$, where $\bar2=(d+1)/2$ is the inverse of $2$, a unit class. By orthogonality,

$$\sum_{d\mid2u-1}A(u)\Psi(u/x) =\frac1{\varphi(d)}\sum_{\chi\bmod d}\overline{\chi(\bar2)} \sum_uA(u)\Psi(u/x)\chi(u).$$

The principal character gives $\varphi(d)^{-1}\sum_{(u,d)=1}A(u)\Psi(u/x)$. Replacing it by $X_A/\varphi(d)$ costs in total at most

$$\sum_uA(u)\Psi(u/x)\sum_{\substack{d\le x^\vartheta\\(d,u)>1}} \frac{\mu^2(d)}{\varphi(d)} \le X_A\max_u\sum_{l\mid u}\frac1{l-1} \sum_{d'\le x}\frac{\mu^2(d')}{\varphi(d')} \ll X_AL^{-18},$$

because every prime factor of a supported $u$ is at least $L^{20}$, there are $O(L/\log L)$ of them, and the last sum is $O(L)$.

The nonprincipal characters contribute at most $\varphi(d)^{-1}\sum_{\chi\ne\chi_0}|\sum_uA(u)\Psi(u/x)\chi(u)|$, since $|\chi(\bar2)|=1$; this quantity does not involve the unit class. Extract one pure slot of band $0$ to write $A(u)=\sum_{pr=u,\ p\in\mathcal Q_0}b(r)$ with $0\le b(r)\le L^{O(1)}\tau(r)$; the extracted prime is not a group prime, so $W_1(u)=W_1(r)$. Split $p$ and $r$ into dyads $P,U$ with $PU\asymp x$; for a fixed $a_0>0$ every contributing dyad has $x^{a_0}\ll P,U\ll x^{1-a_0}$. A nonprincipal $\chi\bmod d$ is induced by a primitive $\chi^*$ modulo some $f>1$ with $d=hf$ and $(h,f)=1$, acting as $\chi^*$ together with $(p,h)=(r,h)=1$. For conductors $f\asymp R>L^{C'}$, separate $\Psi(pr/x)$ by Mellin inversion at bounded integrated cost; the multiplicative large sieve and Cauchy's inequality give

$$\sum_{f\asymp R}\frac1{\varphi(f)}\sum_{\chi^*\bmod f}^{*} \Bigl|\sum_{p\asymp P}a_p\chi^*(p)\Bigr|\Bigl|\sum_{r\asymp U}b_r\chi^*(r)\Bigr| \ll xL^{O(1)}\Bigl(\frac1R+\frac1{\sqrt P}+\frac1{\sqrt U} +\frac R{\sqrt x}\Bigr),$$

which, summed over dyads, over $h$ with weight $1/\varphi(h)$ and over $R\le x^\vartheta$, saves any fixed power of $L$ once $C'$ is large. For $1<f\le 2L^{C'}$, fix $r$ and apply the Siegel–Walfisz theorem with partial summation to the sum over $p$ with the cutoff $\Psi(pr/x)$, which saves any fixed power of $L$; the primes $p\mid h$ are $O(1)$ in number, since they exceed $x^{a_0}$ and $h\le x^\vartheta$, and cost $x^{1-a_0}L^{O(1)}$ in total. $\square$

This is the manuscript's Proposition 7.2 with the unit class $-\bar2$ replaced by $\bar2$; every estimate after orthogonality is in absolute value.

The preliminary sieve

Write $V(y)=\prod_{p\le y}(1-1/p)$, $V_1(y)=\prod_{3\le p\le y}(1-1/(p-1))$ and $\mathfrak S=2\prod_{p\ge3}(1-(p-1)^{-2})>0$, so that $V(y)\sim e^{-\gamma_E}/\log y$ and $V_1(y)\sim\mathfrak SV(y)$. Fix $0<\kappa<1/50$ and then a small fixed $b_1>0$; put $b_*=b_1/3$, choose $j_{**}\ge j_0$ with $c_2s'2^{-j_{**}}<b_*/4$ (so that $\eta_*<b_*/4$), and take $h=2\lfloor1/(40b_1)\rfloor$ and $\vartheta=1/2-\kappa/2$. For small $b_1$, $h$ is large enough for the block sieve and $b_1(4h+2)<\vartheta$.

Every $N_u=2u-1$ is odd. Apply the block sieve to the integers $u$ with weight $A(u)\Psi(u/x)$, with the bad conditions $p\mid2u-1$ for the odd primes $p\le x^{b_1}$. For squarefree odd $d$ the weight on which all conditions at $p\mid d$ hold is $X_A/\varphi(d)+r(d)$, so the density is $g(p)=1/(p-1)$, which satisfies $g(3)=1/2$ and $\sum_{w<p\le w^2}g(p)\ll1$. With the Type I bound at level $x^\vartheta$ and $D>C_A+10$,

$$\sum_{P^-(2u-1)>x^{b_1}}A(u)\Psi(u/x) \ge\frac{\mathfrak SX_A}L\Bigl\{\frac{e^{-\gamma_E}}{b_1} \bigl(1-O(e^{-c/b_1})\bigr)+o(1)\Bigr\},$$

with constants uniform for small $b_1$ and $0<\kappa<1/50$. This is the manuscript's (7.11); the density $1/(p-1)$ is the same for both signs, because $2u\pm1\equiv0\pmod p$ each fix one unit class of $u$.

Composites away from balance

Fix a finite mesh of bins $(\gamma,\gamma']$ partitioning $(b_1,1/2-\kappa]$, chosen below. Write $I_\gamma(m)=\mathbf 1_{P^-(m)>x^\gamma}$ and $I_{\rm pr}(m)=\mathbf 1_{m\text{ prime}}$, and let $B_\gamma$, $B_{\rm pr}$ be their cell proxies at precision $\Delta_L=L^{-S}$: on a cell $C_Y=(Y,e^{\Delta_L}Y]$ the proxy of a test $I$ is $c_I(C_Y)\mathbf 1_{P^-(m)>W}$, where $c_I(C_Y)$ is the number of $m\in C_Y$ with $I(m)=1$ divided by the number of $W$-rough $m\in C_Y$. By Lemma 7.3, $0\le B_I\le1$, and for any fixed $B$ the restricted differences $(I-B_I)\mathbf 1_K$ satisfy $(\mathrm D_B)$ once $S>2B+3$, uniformly for intervals $K$ inside dyads of $[x^{b_1/4},x^{1-b_1/4}]$.

A composite $2u-1$ counted above whose least prime factor $p$ lies in the bin $(x^\gamma,x^{\gamma'}]$ has $m=(2u-1)/p$ with $P^-(m)>x^\gamma$, so the composites of this bin contribute at most

$$\sum_{\substack{mp=2u-1\\ x^\gamma<p\le x^{\gamma'}\text{ prime}}} A(u)\Psi(u/x)\,I_\gamma(m).$$

All contributing dyads of $m$ and $p$ lie in $[x^{b_*},x^{1-b_*}]$. Apply the signed Type II estimate with $\alpha=(I_\gamma-B_\gamma)\mathbf 1_K$, $K$ the cofactor range within a dyad, and $\beta$ the prime indicator of the bin within a dyad: replacing $I_\gamma$ by $B_\gamma$ costs $o(X_A/L)$ in total. By the bound (7.17) of Lemma 7.4,

$$c_\gamma(m)\le\frac{\sup_{\gamma\le a\le\gamma'}D_\gamma(1-a)+o(1)}{LV(W)},$$

where $D_\gamma$ is the Buchstab density (7.15) of the manuscript. For a prime $p>W$, $P^-((2u-1)/p)>W$ if and only if $P^-(2u-1)>W$, so the bin contributes at most that constant over $LV(W)$ times

$$\sum_{x^\gamma<p\le x^{\gamma'}}\ \sum_{\substack{p\mid2u-1\\ P^-(2u-1)>W}} A(u)\Psi(u/x).$$

For each $p$, sieve the odd primes up to $W$ with base mass $X_A/(p-1)$, density $1/(l-1)$, remainders $r(pd)$ and $h=2\lceil T^2\rceil$. All products $pd$ are at most $x^{1/2-\kappa+o(1)}<x^\vartheta$, and distinct pairs $(p,d)$ give distinct products, $p$ being their only prime factor above $W$. The Type I bound therefore gives

$$\sum_{x^\gamma<p\le x^{\gamma'}}\ \sum_{\substack{p\mid2u-1\\ P^-(2u-1)>W}} A(u)\Psi(u/x)=X_AV_1(W)\sum_{x^\gamma<p\le x^{\gamma'}}\frac1{p-1} +o(X_AV(W)),$$

and since the reciprocal sum tends to $\log(\gamma'/\gamma)$, the bin contributes at most

$$\frac{\mathfrak SX_A}L\Bigl\{\sup_{\gamma\le a\le\gamma'}D_\gamma(1-a) \log(\gamma'/\gamma)+o(1)\Bigr\}.$$

By the joint continuity in Lemma 7.4, a fine enough fixed mesh makes the sum of these constants at most $\int_{b_1}^{1/2-\kappa}D_\alpha(1-\alpha)\,d\alpha/\alpha+1/10$. The identity (7.18), $\int_{b_1}^{1/2}D_\alpha(1-\alpha)\,d\alpha/\alpha=D_{b_1}(1)-1$, and the bound (7.19), $D_{b_1}(1)\le e^{-\gamma_E}b_1^{-1}(1+O(e^{-c/b_1}))$, then leave

$$\sum_{\substack{2u-1\text{ prime, or}\\ P^-(2u-1)>x^{1/2-\kappa}}}A(u)\Psi(u/x) \ge\frac{\mathfrak SX_A}L\Bigl\{1-\frac1{10}-O(b_1^{-1}e^{-c/b_1})+o(1)\Bigr\} \ge\frac{\mathfrak SX_A}{2L}$$

once $b_1$ is small enough.

Balanced composites

A composite $2u-1\ll x$ with $P^-(2u-1)>x^{1/2-\kappa}$ has exactly two prime factors, as $3(1/2-\kappa)>1$, and both lie in $\mathcal B_\kappa=[x^{1/2-2\kappa},x^{1/2+2\kappa}]$ for large $x$. Overcount by ordered pairs of primes $m,n\in\mathcal B_\kappa$ with $mn=2u-1$. Replace $I_{\rm pr}(m)$ by $B_{\rm pr}(m)$ with the signed Type II estimate, then $I_{\rm pr}(n)$ by $B_{\rm pr}(n)$ with the roles of the two factors exchanged, the remaining coefficient $B_{\rm pr}(m)$ being bounded by one. Since $B_{\rm pr}\le C/(LV(W))$ on $\mathcal B_\kappa$ by (7.16), with $C$ absolute for $0<\kappa<1/50$, the balanced composites contribute $o(X_A/L)$ plus

$$\frac C{L^2V(W)^2}\sum_{\substack{m,n\in\mathcal B_\kappa,\ mn=2u-1\\ P^-(m)>W,\ P^-(n)>W}}A(u)\Psi(u/x).$$

Balanced dyadic bound. For dyads $m\asymp M_1$ and $n\asymp M_2$ intersecting $\mathcal B_\kappa$, with $M_1M_2\asymp x$ and $0<\kappa<1/50$,

$$\sum_{\substack{m,n\in\mathcal B_\kappa,\ m\asymp M_1,\ n\asymp M_2,\ mn=2u-1\\ P^-(m)>W,\ P^-(n)>W}} A(u)\Psi(u/x)\ll X_AV(W)^2,$$

with a constant depending only on $\delta,r_0,\Psi$, the group geometry and the fixed $j',\theta$ below, not on $\kappa$, $b_1$, $j_{**}$, $S$ or the analytic parameters.

The balance restriction supplies $M_i\ll x^{0.54}$, which is used in the box-error estimate below. A product-size condition alone is not asserted to imply this bound.

The conditional-chain examination records refuted-as-stated for the frozen isolated formulation without the balanced range, using $m=1$ to lose one roughness factor. It records refutation-failed for the balanced specialization consumed here. The explicit balanced-range statement and box-error qualification above differ from the frozen wording; its exact subject and witness are retained with the examination.

Proof. Choose a large fixed $j'$ and $\theta$ with

$$c_2s'2^{-j'}<\theta<c_1s'2^{-j'+1},\qquad 10\theta\le\frac1{20},\qquad \sum_{j\ge j'}r_0c_2s'2^{-j}\le\frac1{25},$$

possible because $c_2<2c_1$, and enlarge $j_0$ so that $j_0\ge j'$; the micro bands are those with $j\ge j'$, and their harmonic mass $H_Q^{\rm mic}$ satisfies $H_Q^{\rm mic}\ll H_Q$. An assignment consists of complete ordered lists in the micro bands, the exceptional slot included, and one marked prime in each group, with weight $\lambda=\prod_{j\ge j'}(1/r_0!)\prod_g(1/V_g)$; let $D_1$ be the product of its entries. Then

$$D_1\le x^{1/20},\qquad \sum_{\rm assgn}\frac\lambda{D_1}=H_Q^{\rm mic},\qquad \sum_{\rm assgn}\lambda\le x^{1/20}H_Q^{\rm mic}.$$

Write $s=|\mathcal S|+|\mathcal B|$ and $\omega(u)=\sum_g\omega_g(u)$. For each assignment, ban every odd prime $l\le x^\theta$ that is not a group prime and does not divide $D_1$, and ban each group prime not dividing $D_1$ independently with probability $1-q$. Expanding $A(u)$ into lists and marks, every non-group prime factor of $u$ below $x^\theta$ divides $D_1$: the omitted bands contain only pure primes above $x^\theta$, and the entire exceptional integer is in $D_1$. Its prime divisors need not lie in any prime band. The group part of $D_1$ consists of exactly one mark per group, so the expected value of the permissions is the damping $q^{\omega(u)-s}$ of $W_1$. For a fixed residual product the normalized number of omitted pure-prime lists is at most one, since its contribution in each band is a product of reciprocal multiplicity factorials. Dropping those residual restrictions gives the majorant

$$A(u)\le\sum_{\rm assgn}\lambda\,\mathbf 1_{D_1\mid u}\, \mathbb E_{\rm mask}\mathbf 1_{\{l\nmid u\text{ for every banned }l\}},$$

which concerns $u$ alone and is the manuscript's (7.27).

Fix an assignment and a mask. The integer $D_1$ is odd, because its entries are primes at least $L^{20}$ and the exceptional integer, all of whose prime factors exceed $W$. So $D_1\mid u$ and $mn=2u-1$ give

$$mn\equiv-1\pmod{D_1}.$$

For an odd prime $l\le x^\theta$ with $l\nmid D_1$ the bad events are $mn\equiv0\pmod l$ when $l\le W$ (roughness of $m$ and $n$), and $mn\equiv-1\pmod l$ when $l$ is banned, which is $l\mid u$ because $l$ is odd. These residue sets are disjoint and contain $2l-1$ and $l-1$ pairs. Put $\nu(l)=(2l-1)\mathbf 1_{l\le W}+(l-1)\mathbf 1_{l\text{ banned}}$ and $g(l)=\nu(l)/l^2\le3/l$, bounded away from one for $l\ge3$. For a squarefree product $d$ of these primes, the congruence modulo $D_1$ leaves, for each unit $m\bmod D_1$, exactly one $n\bmod D_1$, also when $D_1$ has square factors; with the Chinese remainder theorem there are exactly $\varphi(D_1)\nu(d)$ admissible residue pairs modulo $D_1d$. Each pair has lattice count in the box equal to its area term with error $O(1+(M_1+M_2)/(D_1d))$, so the remainders up to $D=x^{10\theta}$ sum to

$$\ll L^{O(1)}\bigl((M_1+M_2)D+D_1D^2\bigr) \le L^{O(1)}\bigl((M_1+M_2)x^{1/20}+D_1x^{1/10}\bigr).$$

The upper-bound block sieve with $h=2$ gives the main term $\ll M_1M_2\,\varphi(D_1)D_1^{-2}\prod_l(1-g(l))$, the product over the odd primes $l\le x^\theta$ not dividing $D_1$. A banned prime below $W$ has factor $(1-1/l)(1-2/l)$ and a banned prime above $W$ has factor $1-1/l+1/l^2$, each at most $(1-1/l)^{1+2\mathbf 1_{l\le W}}(1+O(l^{-2}))$; an allowed group prime has factor $(1-1/l)^2$. Hence

$$\prod_{\substack{3\le l\le x^\theta\\ l\nmid D_1}}(1-g(l)) \le CV(W)^2V(x^\theta)\prod_{l\mid D_1}(1-1/l)^{-3} \prod_{\substack{p\text{ allowed}\\ p\nmid D_1}}(1-1/p)^{-1}.$$

The product over $l\mid D_1$ is bounded, since those primes are at least $L^{20}$ and $D_1\le x^{1/20}$, and the mask expectation of the last product is $\prod_{p\nmid D_1}(1+q/(p-1))\le H_P$, the product over group primes, by the manuscript's (7.6). Summing the main terms over assignments with $\varphi(D_1)D_1^{-2}\le D_1^{-1}$ gives $\ll xV(W)^2V(x^\theta)H_Q^{\rm mic}H_P\ll X_AV(W)^2$, and the remainders give $\ll x^{0.64}L^{O(1)}H_Q^{\rm mic}=o(X_AV(W)^2)$, using $M_i\ll x^{0.54}$ and $X_AV(W)^2\asymp xH_QH_P/L^2$. $\square$

There are $O(\kappa L+1)$ dyads for $m$ in $\mathcal B_\kappa$ and $O(1)$ for $n$ given $m$, so the balanced composites contribute at most

$$C_{\rm bal}(\kappa+L^{-1})\frac{X_A}L+o(X_A/L),$$

with $C_{\rm bal}$ depending only on $\delta,r_0,\Psi,j',\theta$ and the group geometry. The residue counts are those of the manuscript's (7.30)–(7.31): there the congruences read $mn\equiv1$ modulo $D_1$ and modulo each banned prime, with the same numbers of residue pairs.

The count and Hypothesis A+

The choices are made in the manuscript's order.

1. Fix $\delta$, $\Psi$, $r_0$ and the band and group geometry; the finite prefix of bands used for the lower bound of $X_A$; then $j'$ and $\theta$; and $j_0$ covering both.
2. Fix $\kappa\in(0,1/50)$ with $C_{\rm bal}\kappa<\mathfrak S/4$, possible because $C_{\rm bal}$ does not depend on $\kappa$.
3. Fix $b_1$ small enough for the preliminary sieve and for the subtraction above; set $b_*=b_1/3$; fix $j_{**}$ with $c_2s'2^{-j_{**}}<b_*/4$; fix the bin mesh.
4. Fix the Type II saving $D_*$ so that every replacement error, after the dyadic sums, is $o(X_A/L)$. The Type II proof then fixes $K_0$, $D_0$ and, through (P), the discrepancy exponent $B$.
5. Fix $S>2B+3$ in Lemma 7.3 and the prime-number-theorem accuracies, and let $x\to\infty$.

The coefficient bound of the proxies does not depend on $S$, so step 5 does not feed back into step 4. For all large real $x$,

$$\sum_{2u-1\text{ prime}}A(u)\Psi(u/x)\ge\frac{\mathfrak SX_A}{2L} -C_{\rm bal}\Bigl(\kappa+\frac1L\Bigr)\frac{X_A}L-o\Bigl(\frac{X_A}L\Bigr) \ge\frac{\mathfrak SX_A}{5L}.$$

After enlarging $C_w$ to absorb the fixed bound on $\Psi$, each summand is at most $\exp(C_w\sqrt L)$. Since $X_A\ge c_AxL^{-C_A}$ and $u\mapsto2u-1$ is injective, at least $(\mathfrak Sc_A/5)\,xL^{-C_A-1}\exp(-C_w\sqrt L)$ distinct primes $p=2u-1$ occur. As $\Psi(u/x)\ne0$ forces $x\le u\le2x$, they satisfy $2x-1\le p\le4x-1$, and $p+1=2u$ with every prime factor of $u$ at most $x^\delta$, so $P^+(p+1)=\max(2,P^+(u))\le x^\delta$. This proves:

Successor count, assuming (P) and (E). For every fixed $0<\delta<1/4$ there are $c,C_A,C_w>0$ such that for all sufficiently large real $x$

$$\#\{p\text{ prime}:\ 2x-1\le p\le4x-1,\ P^+(p+1)\le x^\delta\} \ge c\,x(\log x)^{-C_A-1}\exp\bigl(-C_w\sqrt{\log x}\bigr).$$

Hypothesis A+, assuming (P) and (E). Given $0<\delta<1/4$ and $\eta>0$, the quantity $\eta L-(C_A+1)\log L-C_w\sqrt L+\log c$ tends to infinity, so the successor count at $x=y$ gives at least $y^{1-\eta}$ primes $p\le4y-1<5y$ with $P^+(p+1)\le y^\delta$, for every real $y$ beyond a threshold depending on $\delta$ and $\eta$. No uniformity as $\delta\to0$ and no effective threshold is claimed, and none is needed.

The counted primes lie inside the range $2x-2<p\le5x$ of Li and Zhuang's Theorem 1.4, which states this count under the assumption that the manuscript's argument is correct.