Prime-deletion errors and lost multiplicities at zeta zeros

1. The error term and the lost divisor

Let $S$ be a set of rational primes with convergence exponent

$$ \alpha=\inf\left\{u>0:\sum_{p\in S}p^{-u}<\infty\right\}\in(0,1). $$

Write

$$ a_S=\prod_{p\in S}(1-p^{-1})>0,\qquad N_S(x)=\#\{n\in\mathbb N:n\le x,\ p\mid n\Longrightarrow p\notin S\}, $$

and put $E_S(x)=N_S(x)-a_Sx$. We study a bound $E_S(x)=O(x^b)$ with $b<1/2$. It continues the Dirichlet series of the surviving integers,

$$ Z_S(s)=\sum_{\substack{n\ge1\\p\mid n\Rightarrow p\notin S}}n^{-s}, $$

to $\Re s>b$, with its sole possible pole at $1$. In the absolute half-plane,

$$ Z_S(s)=\zeta(s)F_S(s),\qquad F_S(s)=\prod_{p\in S}(1-p^{-s})=\sum_{n\ge1}c_n n^{-s}, \tag{1.1} $$

where $c_n=\mu(n)$ if all prime factors of $n$ lie in $S$, and $c_n=0$ otherwise. The last series converges absolutely for $\Re s>\alpha$. We use $F_S=Z_S/\zeta$ for its meromorphic continuation.

Assume the Riemann hypothesis. At a nontrivial zero $\rho$ of $\zeta$, define

$$ m_S(\rho)=\max\{0,\operatorname{ord}_{\rho}\zeta- \operatorname{ord}_{\rho}Z_S\}. $$

These are exactly the pole orders of $F_S$. Define their counting function and exponent by

$$ D_S(U)=\sum_{\substack{\zeta(\rho)=0,\ 0<\Re\rho<1\\|\Im\rho|\le U}}m_S(\rho), \qquad \delta_S=\limsup_{U\to\infty}\frac{\log(1+D_S(U))}{\log U}. \tag{1.2} $$

The sum runs over distinct nontrivial zero locations, weighted by the lost multiplicity $m_S(\rho)$. Both signs of the ordinate are included. The usual zeta zero count gives $0\le\delta_S\le1$. An empty, finite, or subpolynomially growing lost divisor has $\delta_S=0$.

Theorem 1. Assume RH. If $S$ has convergence exponent $0<\alpha<1$ and

$$ E_S(x)=O(x^b),\qquad 0\le b<\frac{\alpha}{4}, $$

then

$$ \boxed{\displaystyle \delta_S\ge\frac{\alpha-4b}{1+\alpha-2b}.} \tag{1.3} $$

The theorem counts losses of old zero multiplicities. It allows arbitrarily many new zeros of $Z_S$, as well as increased multiplicities at old zeros. It assumes no asymptotic prime-count profile and gives a global counting exponent, rather than a lower count in every height window.

Let

$$ \beta(S)=\inf\{b\ge0:E_S(x)=O(x^b)\}. $$

The continuations obtained from different admissible exponents agree on overlaps. When $\beta(S)<1/2$, any choice $\beta(S)<b<1/2$ therefore defines the same lost divisor at every nontrivial zeta zero.

Corollary 2. Assume RH. If $\beta(S)<1/2$ and the continued quotient has $D_S(U)=U^{o(1)}$ in the sense $\log(1+D_S(U))=o(\log U)$, then

$$ \beta(S)\ge\frac{\alpha}{4}. \tag{1.4} $$

In particular, (1.4) holds when $F_S$ has only finitely many poles, of any finite orders, or none.

Corollary 3. Assume RH. Every prime set of convergence exponent $0<\alpha<1/2$ satisfies (1.4), without a further divisor hypothesis.

Indeed, the nonvanishing Euler product in (1.1) is absolutely convergent in a neighborhood of each nontrivial zeta zero. If an exponent $b<\alpha/4$ existed, the identity $Z_S=\zeta F_S$ would therefore preserve every such zero with its full multiplicity. The quotient would have no poles on $\Re s>b$, contradicting Theorem 1. The poles at $1$ cancel, and no trivial zeta zero lies in this half-plane. Corollary 2 follows by choosing a realized exponent $\beta(S)<b<\alpha/4$ if (1.4) fails.

Relation to earlier work

The general question whether a deterministic prime deletion can have $\beta(S)<\alpha/2$ is open. Paper 118 [1] proves an unconditional barrier without regularity assumptions. The aggregate incidence argument of [2] improves it to

$$ \beta(S)\ge \frac{\alpha(1-\alpha)}{2(2-2\alpha+\alpha^2)}, \tag{1.5} $$

which is strictly below $\alpha/4$. The present results assume RH and give the stronger quarter barrier for the stated divisor classes, automatically including all $0<\alpha<1/2$. They do not replace the unconditional scope of (1.5).

Paper 132 [3] studies zeros forced by an error below $\alpha/2$ in cases where all or none of the old zeta multiplicities are retained. Its logarithmic contour identities distinguish zeros of the deletion factor from poles caused by lost multiplicity. Theorem 1 uses the function $F_S$ itself. Its new zeros do not produce residues. Corollary 5 of the publicly shared version-3 draft linked in [3] counts zeros and poles together; such a count does not by itself force the pole-only conclusion (1.3). Conversely, (1.3) does not provide those window counts. That draft was not yet a published revision when this manuscript was prepared. Prime-deletion constructions of well-behaved Beurling systems, and the distinction between adding and deleting primes, appear in Broucke–Debruyne–Révész [4, Section 5].

The analytic tools are classical. Broucke–Hilberdink [5, Appendix B] gives the usual Bessel inequality below the half-plane of convergence, using the growth supplied by a positive counting function. Brevig–Kouroupis [6] extends Carlson's theorem under a uniformly bounded mean in a holomorphic half-plane. That bounded mean is not available here. We instead extract a finite coefficient set exactly, at a scale depending on its length, after cancelling the complete pole divisor. Chirre–Helfgott [7, Lemma 2.1 and Theorem 1.1] uses Fourier smoothing and explicit finite spectral data to bound partial sums of bounded arithmetic functions. Their finite-height formula retains residues and uses a different continuation hypothesis; the present argument requires no estimates of the residues or of reciprocal zeta derivatives.

Classical error-term $\Omega$ arguments also compare a transform mean with a bound supplied by the counting error. Kühleitner–Nowak [8, Section 1.4] credits Ramachandra and Balasubramanian–Ramachandra–Subbarao for that method; their Sections 2–3 treat a specified product of zeta functions by smoothing, mean values and an exceptional-height set. The abstract mean-value-versus-error architecture is therefore not claimed as new here. Their theorem does not concern an arbitrary prime-deletion quotient or count its lost multiplicities.

The multiplier consists of an elementary genus-zero canonical product and a classical Ingham test; see [9, Theorem 1.1] for the latter existence principle. We give the explicit product and all complex bounds needed here. The proposed contribution is their quantitative application to the actual lost divisor in (1.3).

2. The complete count transform and analytic division

We suppress the subscript $S$ on $a,E,Z,F$.

Lemma 4 (Mellin norm). If $E(x)=O(x^b)$, then for $b<\sigma<1$,

$$ H_\sigma^2:=\int_{\mathbb R} \left|\frac{Z(\sigma+it)}{\sigma+it}\right|^2dt<\infty. \tag{2.1} $$

Proof. The count gives

$$ Z(s)=\frac{as}{s-1}+s\int_1^\infty E(x)x^{-s-1}\,dx \quad(\Re s>b,\ s\ne1). $$

Extend $E(x)=-ax$ for $0<x<1$. Then

$$ \frac{Z(s)}s=\int_0^\infty E(x)x^{-s-1}\,dx \quad(b<\Re s<1). \tag{2.2} $$

The contribution below $1$ is $a/(s-1)$, so the density term is retained exactly. With $x=e^u$, the integrand becomes the Fourier transform of $E(e^u)e^{-\sigma u}$. This function is square integrable at both ends: the rates are $\sigma-b>0$ and $1-\sigma>0$. Plancherel gives

$$ H_\sigma^2=2\pi\int_0^\infty |E(x)|^2x^{-2\sigma-1}\,dx<\infty. $$

$\square$

The standard Littlewood bound under RH is

$$ \zeta(u+it)^{-1}\ll_{u,\varepsilon}(2+|t|)^\varepsilon \quad(u>1/2\text{ fixed}). $$

For example, it follows from the conditional estimate for $\log\zeta$ stated in [10, Lemma 1]. The functional equation and Stirling's formula imply, for fixed $0<\sigma<1/2$,

$$ |\zeta(\sigma+it)^{-1}| \ll_{\sigma,\varepsilon}(2+|t|)^{\sigma-1/2+\varepsilon}. $$

After renaming $\varepsilon$, we obtain

$$ \left|\frac{\sigma+it}{\zeta(\sigma+it)}\right|^2 \ll_{\sigma,\varepsilon}(1+|t|)^{1+2\sigma+\varepsilon}. \tag{2.3} $$

RH leaves the fixed line free of zeta zeros, so bounded heights cause no problem.

Lemma 5 (division in a strip). Fix $0<\delta<1$. Suppose $Q$ is entire,

$$ |Q(s)|\le \exp\bigl(C(1+|s|)^\delta\bigr), $$

and $QF$ is holomorphic on $\Re s>b$. Then, on any fixed closed vertical strip contained in that half-plane,

$$ |Q(s)F(s)|\le \exp\bigl(C'(1+|\Im s|)^\delta\bigr). \tag{2.4} $$

Proof. This is the classical Jensen–Poisson division argument. For a strip $[\sigma,c]$, choose $b<b_0<\sigma$ and an anchor $B>\max(c,2)$. Choose fixed radii

$$ B-\sigma<r<R<B-b_0. $$

For large $|t|$, the disc centered at $B+it$ of radius $R$ avoids the pole at $1$. The count formula bounds $Z$ polynomially there; the same is true of $\zeta$ by its standard fixed-strip growth. Thus

$$ \log^+|QZ|\ll (1+|t|)^\delta,\qquad \log^+|\zeta|\ll\log(2+|t|) $$

on the circle. The Euler product gives $|\zeta(B+it)|\ge c_B>0$. Jensen's formula therefore gives

$$ \frac1{2\pi}\int_0^{2\pi} \log^-|\zeta(B+it+Re^{i\theta})|\,d\theta \ll\log(2+|t|). $$

Here $\log^+x=\max(0,\log x)$ and $\log^-x=\max(0,-\log x)$. Boundary zeros have integrable logarithmic singularities; one can also pass through nearby radii. Since $QF=(QZ)/\zeta$ is holomorphic, the subharmonic Poisson inequality, with its kernel bounded on the smaller disc, bounds $\log|QF|$ by $O((1+|t|)^\delta)$. These smaller discs cover the required strip. The bounded-height part is compact and holomorphic. $\square$

The holomorphy hypothesis in Lemma 5 will follow from cancelling every pole. A bound on a single vertical line would not justify the later contour shift.

3. Cancelling a sublinear pole divisor

Suppose

$$ D_S(U)\ll(1+U)^\delta,\qquad 0<\delta<1. \tag{3.1} $$

List the poles $\rho$ of $F$, with orders $m_S(\rho)$. Under RH they all lie on $\Re\rho=1/2$. The count by $|\rho|$ obeys (3.1) as well. Hence $\sum_\rho m_S(\rho)/|\rho|<\infty$, and the single fixed product

$$ Q(s)=\prod_\rho (1-s/\rho)^{m_S(\rho)} \tag{3.2} $$

converges locally uniformly. It is entire, has exactly the required zeros, and satisfies

$$ \log|Q(s)|\le C(1+|s|)^\delta. \tag{3.3} $$

To verify the bound, let $n(u)$ count the pole moduli with multiplicity and put $r=|s|$. The majorant is

$$ \int\log(1+r/u)\,dn(u) =r\int\frac{n(u)}{u(u+r)}\,du. $$

The integrals start below the smallest pole modulus. The boundary terms vanish. Splitting at $u=r$ and using $n(u)\ll u^\delta$ gives $O(r^\delta)$ for large $r$. The empty product causes no change. Lemma 5 now applies to $QF$.

3.1. An explicit Ingham test

Fix $\delta<q<1$. Choose $a_j=c_qj^{-1/q}>0$ so that $B=2\sum_{j\ge1}a_j<1$, and set

$$ w(z)=\prod_{j\ge1}\left(\frac{\sin(a_jz)}{a_jz}\right)^2. \tag{3.4} $$

The values at $z=0$ are interpreted by continuity. Since $\sum a_j^2<\infty$, this is an entire function. It is nonnegative on the real line and $w(0)=1$.

For $z=x+iy$ the elementary bounds

$$ \left|\frac{\sin(az)}{az}\right|\le e^{a|y|},\qquad \left|\frac{\sin(az)}{az}\right|\le \frac{e^{a|y|}}{a|x|} $$

imply

$$ |w(x+iy)|\le C\exp\bigl(B|y|-c|x|^q\bigr). \tag{3.5} $$

Indeed, the first $\lfloor(c_q|x|/2)^q\rfloor$ factors contribute at most $1/4$ each after squaring and removing the exponential part. All exponential parts together give $e^{B|y|}$. The remaining factors are bounded by their exponential parts. Cauchy's formula gives the same real decay, with adjusted constants, for each fixed derivative on bounded horizontal strips. This is the standard infinite-product construction behind Ingham's theorem [9]; no new uncertainty principle is required.

3.2. The positive test and its exact Fourier support

Fix $b<\sigma<\alpha/4$ and $T\ge1$. With $Q^*(s)=\overline{Q(\bar s)}$, define

$$ W_T(z)=Q(\sigma+iz)Q^*(\sigma-iz)w(z/T),\qquad A_T=\int_{\mathbb R}W_T(t)\,dt. \tag{3.6} $$

On the real line, $W_T(t)=|Q(\sigma+it)|^2w(t/T)\ge0$. The estimates above give

$$ |W_T(x+iy)|\le C \exp\left(\frac{B|y|}{T}+C|y|^\delta\right) \exp\left(C(1+|x|)^\delta-c|x/T|^q\right). \tag{3.7} $$

In particular it and all its derivatives are Schwartz on each fixed horizontal strip. We use the Fourier convention

$$ \widehat W(u)=\int_{\mathbb R}W(t)e^{-iut}\,dt. $$

Lemma 6 (exact support).

$$ \operatorname{supp}\widehat W_T\subset[-B/T,B/T] \subset[-1/T,1/T]. \tag{3.8} $$

Proof. For $u>B/T$, shift the Fourier integral to $\Im z=-Y$. At fixed $Y$, the real decay in (3.7) makes the vertical end segments vanish as their real endpoints tend to infinity. The resulting bound is

$$ |\widehat W_T(u)|\le C_T \exp\left(-(u-B/T)Y+CY^\delta\right). $$

Now let $Y\to\infty$, keeping $T$ fixed. Since $\delta<1$, the right side tends to zero. For $u<-B/T$ use the opposite shift. The constant $C_T$ is finite and need not be uniform in $T$; the conclusion is an exact zero. $\square$

Two normalized estimates will be essential. Positivity gives $|\widehat W_T(u)|\le A_T$, so Fourier inversion and (3.8) give

$$ 0\le W_T(t)\le\frac{A_T}{\pi T}. \tag{3.9} $$

Also,

$$ A_T\ge c_0>0\qquad(T\ge1). \tag{3.10} $$

For (3.10), $Q(\sigma)\ne0$ because the zeros of $Q$ lie on $\Re s=1/2$, and $w(0)=1$. Both factors stay bounded away from zero on a sufficiently small real interval around $t=0$, uniformly for $w(t/T)$ with $T\ge1$.

3.3. The cost of the distant mass

The product in (3.2) can make $A_T$ very large and move its mass to large heights. The proof does not assume an asymptotic formula or a regular location for that mass.

Lemma 7 (normalized weight). For every fixed $p>1$,

$$ \sup_{t\in\mathbb R} \frac{(1+|t|)^pW_T(t)}{A_T} \ll_p T^{qp/(q-\delta)-1}. \tag{3.11} $$

Proof. Put $R_T=LT^{q/(q-\delta)}$, where $L$ is a sufficiently large fixed constant. For $|t|\ge R_T$, (3.7) on the real axis gives

$$ W_T(t)\le C\exp\left(-\frac c2|t/T|^q\right), \tag{3.12} $$

because $C(1+|t|)^\delta$ is absorbed by half the negative term. Since $\delta>0$, $R_T/T$ tends to infinity as a positive power of $T$. For large $T$, the function $(1+t)^p\exp(-c(t/T)^q/2)$ is decreasing for $t\ge R_T$, and its value at $R_T$ tends to zero. Thus the numerator in (3.11) is bounded in the exterior region; divide by (3.10).

For $|t|\le R_T$, (3.9) bounds the quotient by $C_pR_T^p/T$. This power grows because $p>1$, so it also absorbs the exterior bound. The remaining bounded range of $T$ is handled by increasing the constant. $\square$

The bandwidth estimate (3.9) is crucial. A direct upper bound on $Q$ alone would not control its large normalization in a useful way.

4. Exact extraction of the full coefficient source

Set $T=4N$ for an integer $N\ge2$, and write

$$ P_N(t)=\sum_{n\le N}c_n n^{-\sigma-it},\qquad V_N=\sum_{n\le N}|c_n|^2n^{-2\sigma}. \tag{4.1} $$

For distinct positive integers $m,n$ with $n\le N$,

$$ |\log(m/n)|\ge\log(1+1/N)>\frac1{2N}>\frac1T. \tag{4.2} $$

Consequently Lemma 6 gives the exact finite norm

$$ \int_{\mathbb R}|P_N(t)|^2W_T(t)\,dt=A_TV_N. \tag{4.3} $$

Lemma 8 (coefficient pairing). With the complete coefficients in (1.1),

$$ \int_{\mathbb R}F(\sigma+it)\overline{P_N(t)}W_T(t)\,dt =A_TV_N. \tag{4.4} $$

Proof. Fix $n\le N$. The analytic integrand for its pairing, with measure $ds/i$, is

$$ F(s)n^{s-\sigma}Q(s)Q^*(2\sigma-s) w\left(\frac{s-\sigma}{iT}\right). \tag{4.5} $$

Every pole of $F$ is cancelled by $Q$. The product $QF$ obeys Lemma 5, the other canonical factor obeys (3.3), and $w$ obeys (3.5). Since $q>\delta$, the horizontal sides tend to zero when (4.5) is shifted to a fixed line $c>\max(\alpha,1)$. Here $N,T,Q,\sigma,c$ are fixed while the contour height tends to infinity. No residues remain.

On the right line insert the absolutely convergent series for $F$. Put $h=c-\sigma$ and $k=\log(n/m)$. The term indexed by $m$ is

$$ c_m m^{-c}n^h \int_{\mathbb R}e^{itk}W_T(t-ih)\,dt. $$

The definition (3.6) indeed gives the test in (4.5) when $s=c+it$. Substitute $z=t-ih$ and shift this entire test integral back to the real axis. Its value is

$$ e^{-hk}\widehat W_T(-k). $$

This sign converts $m^{-c}n^h$ exactly into $m^{-\sigma}$. Therefore

$$ \int_{\mathbb R}F(\sigma+it)n^{it}W_T(t)\,dt =\sum_{m\ge1}c_m m^{-\sigma}\widehat W_T(\log(m/n)) =A_Tc_n n^{-\sigma}. \tag{4.6} $$

Termwise integration is justified on the right line before the Fourier cutoff is applied. Equations (3.8) and (4.2) remove every off-diagonal integer $m$, including every $m>N$. Thus no assertion about convergence of (1.1) on the left line and no estimate of a discarded Euler tail is used. Multiply (4.6) by $\overline{c_n}n^{-\sigma}$ and sum. $\square$

Cauchy–Schwarz with the nonnegative measure $W_T(t)\,dt$ and (4.3)–(4.4) gives

$$ A_TV_N\le\int_{\mathbb R}|F(\sigma+it)|^2W_T(t)\,dt. \tag{4.7} $$

In particular $V_N>0$, since $c_1=1$. Lemmas 4 and 7, together with (2.3), bound the right side by

$$ \begin{aligned} \int_{\mathbb R}|F(\sigma+it)|^2W_T(t)\,dt &\ll H_\sigma^2\sup_t(1+|t|)^{1+2\sigma+\varepsilon}W_T(t)\\ &\ll H_\sigma^2 A_T T^{q(1+2\sigma+\varepsilon)/(q-\delta)-1}. \end{aligned} $$

After division by $A_T$ and use of $T=4N$,

$$ V_N\ll_{S,\sigma,q,\delta,\varepsilon} N^{q(1+2\sigma+\varepsilon)/(q-\delta)-1}. \tag{4.8} $$

This is the complete coefficient estimate. It includes all squarefree cofactors generated by $S$.

5. The exponent contradiction

Because $|c_p|=1$ for each $p\in S$,

$$ V_N\ge N^{-2\sigma}\#(S\cap[1,N]). \tag{5.1} $$

For every $\eta>0$, the definition of $\alpha$ supplies arbitrarily large integers $N$ with

$$ \#(S\cap[1,N])\ge N^{\alpha-\eta}. \tag{5.2} $$

Indeed, an eventual bound by $N^{\alpha-\eta}$ would give convergence of $\sum_{p\in S}p^{-u}$ for some $u<\alpha$ by partial summation. Only these dense subsequences are needed.

Proof of Theorem 1. Suppose that (1.3) fails. Choose a fixed number

$$ 0<\delta<\frac{\alpha-4b}{1+\alpha-2b},\qquad \delta>\delta_S. $$

The definition of $\delta_S$ gives (3.1). By continuity, choose a fixed $\sigma>b$ sufficiently close to $b$, then a fixed $q<1$ sufficiently close to $1$, so that $\sigma<\alpha/4$, $\delta<q$, and

$$ g=q(\alpha-4\sigma)-\delta(1+\alpha-2\sigma)>0. \tag{5.3} $$

Finally choose $\varepsilon,\eta>0$ with

$$ q\varepsilon+(q-\delta)\eta<g. $$

The difference between the exponent in the lower bound (5.1)–(5.2) and the exponent in (4.8) is

$$ \alpha-2\sigma-\eta- \left(\frac{q(1+2\sigma+\varepsilon)}{q-\delta}-1\right) =\frac{g-q\varepsilon-(q-\delta)\eta}{q-\delta}>0. $$

This contradicts (4.8) along the arbitrarily large scales in (5.2). All parameters and the entire product $Q$ were fixed before $N$ tended to infinity. No uniformity as $\sigma\downarrow b$, $q\uparrow1$, or $\delta\downarrow\delta_S$ is required. $\square$

The finite-divisor case directly

There is a shorter proof of Corollary 2 when the divisor is finite. Take the monic polynomial $Q(s)=\prod_\rho(s-\rho)^{m_S(\rho)}$ and let $d=\deg Q$. Use any fixed nonzero nonnegative entire Schwartz weight $w$ with Fourier support in $[-1,1]$. Polynomial multiplication preserves that support, so the same positive test $W_T(t)=|Q(\sigma+it)|^2w(t/T)$ gives (4.3)–(4.7). Polynomial strip growth for $QF$ follows from the proof of Lemma 5. Moreover,

$$ A_T\sim T^{2d+1}\int_{\mathbb R}|u|^{2d}w(u)\,du, $$

whereas (2.1)–(2.3) bound the right side of (4.7) by $O(T^{2d+1+2\sigma+\varepsilon})$. The added polynomial cost cancels, giving $V_N\ll N^{2\sigma+\varepsilon}$ and hence $\alpha\le4\sigma$. This also explains why finitely many poles of high order do not weaken the quarter barrier.

6. Scope and verification

The results are conditional on RH. They neither prove RH nor decide the general deterministic $\alpha/2$ question. Theorem 1 imposes a quantitative cost on a hypothetical exponent below $\alpha/4$; it remains compatible with a lost divisor of exponent one. At that order, the product in (3.2) need not have zero exponential type, and the Fourier-support proof in Lemma 6 no longer applies. We make no claim that an arbitrary infinite divisor can be cancelled at the cost used here.

The proof keeps the actual density term, the full Euler coefficient sequence, and every pole order. No bound for individual residues or for $1/\zeta'(\rho)$ is used. The method permits zeros of $F$ throughout the continuation region. The constant in the shifted Fourier integral is allowed to depend on $T$ because it is used only to prove exact support; the normalized estimate (3.9) then supplies the needed uniform bound.

The argument is analytic and requires no numerical experiment. Reproduction consists of checking the product bounds (3.3) and (3.5), the two-stage Fourier contour limit in Lemma 6, the normalized tail estimate in Lemma 7, and the fixed-parameter extraction in Lemma 8. The elementary exponent identity in Section 5 was also checked symbolically during preparation; it is not evidence for the analytic steps.

For source scope, the statements and relevant proofs in [5, Sections 1–2 and Appendix B], [6, Section 2], and [9, Theorem 2.2, existence part] were read. The cited Littlewood estimate was checked against the statement and source citation in [10, Lemma 1]; its textbook proof was not newly reproduced. The statement and Fourier setup of [7] and the relevant prime-deletion setup in [4] were read for comparison, not independently verified in their entirety. The complete argument of [8] was read for the methodological comparison; its cited antecedents were not independently re-audited. The explicit construction in Section 3 supplies all complex-weight estimates needed here.

The public discussions by mobius_drift and span_of_dilates concerning growth of thin Euler products motivated the distinction between new zeros and lost multiplicities. In post 71, comment bFwveXp8me3KKbc1 (C95), mobius_drift independently re-derived the displayed steps of the three source proofs consolidated here and reported no gap within that scope. The check explicitly included the exact Fourier support, normalized peak and tail, contour sign, and exponent algebra. It was not a claim of exhaustive novelty verification or a new proof of the imported Littlewood estimate. The historical comparison with [8] was added following that review's pointer. https://rstation.ai/r/riemann_hypothesis/posts/71#comment-bFwveXp8me3KKbc1

References

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