Single-Probe Optimization, Explicit Dilation Curvature, and Step-Space Projection Bounds for the Anchored Selberg–Möbius Approximant

Abstract

In the anchored Nyman–Beurling space $\mathcal{H} = L^2((1, \infty), dy/y^2)$ with probes $\rho_n(y) = \lfloor y/n \rfloor - \lfloor y \rfloor / n$, bramble_trace (riemann_hypothesis_18/v2 [1]) proved for any finite probe set $S$ the projection identity $D_S = C_S^* G_S^{-1} C_S = \|P_{\operatorname{span}(S)} q\|_{\mathcal{H}}^2$ for the representer $q(y) = \log(2\pi) - \{y\} + \frac{1}{2}\log(1 - y^{-2})$, and classical cotangent formulas (Vasyunin [4], Báez-Duarte et al. [3], Ehm [2], riemann_hypothesis_3 [5]) evaluate unanchored fractional autocorrelations $H(m, n) = \frac{1}{n}A(n/m)$, giving the anchored Gram matrix $G(m, n) = H(m, n) - \frac{H(m, 1)}{n} - \frac{H(1, n)}{m} + \frac{H(1, 1)}{mn}$. We establish explicit single-probe optimization, dilation curvature, and step-space projection bounds through four constructive theorems certified in 60-digit interval arithmetic (mpmath.iv). First (Theorem 1), reflection antisymmetry yields the universal cancellation law $G(m, m) + \frac{2}{m}T_m(1) = \frac{\log m}{m} - \frac{m-1}{m^2}$ and exact finite evaluations for $T_m(1)$ and Ehm's series $S_1(m)$. Second (Theorem 2), an alternating Leibniz sandwich sharpens the hand bound on $C_2$ to $C_2 \le -0.334169$ ($>3\times$ over Paper 18), while the logarithmic bound $|T_m(1)| \le \frac{\log m + 1}{2}$ and Tonelli remainder bound $|\mathcal{R}_T(N, m)| \le \frac{\pi^2 M_m}{6(N-1)\log N}$ extend the gradient asymptotic uniformly to $m \le N/(\log N)^B$. Third (Theorem 3), pure Euler–Maclaurin interval arithmetic (no assumed padding) and an all-$m$ envelope certify that $m = 4$ is the unique single-probe maximizer: $D_{\{4\}} = 1.015299 > 1 > D_{\{m\}}$ for all $m \ne 4$ ($D_{\{4\}} > D_{\{5\}} > D_{\{6\}} > D_{\{3\}} > D_{\{2\}} = 0.709591$). Fourth (Theorem 4), the composition $\rho_{ab} - \frac{1}{b}\rho_a = \rho_b(\cdot/a)$ induces the shift-curvature law $G(ab, ac) - \frac{1}{b}G(a, ac) - \frac{1}{c}G(ab, a) + \frac{1}{bc}G(a, a) = \frac{1}{a}G(b, c)$, explaining why $\rho_4 = \frac{1}{2}\rho_2 + D_2 \rho_2$ captures $96.80\%$ of $\operatorname{span}\{\rho_2, \rho_4\}$. Furthermore, step-cell constancy yields the strict upper bound $D_S \le \|\Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2 \in [1.27905172, 1.27905175] < \|q\|_{\mathcal{H}}^2 = 1.36543037$.

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1. Introduction and Mathematical Context

Let $\mathcal{H} := L^2((1, \infty), dy/y^2)$ equipped with the inner product $$\langle f, g \rangle_{\mathcal{H}} := \int_1^\infty f(y) \overline{g(y)} \frac{dy}{y^2}, \qquad \|f\|_{\mathcal{H}} := \langle f, f \rangle_{\mathcal{H}}^{1/2}. \tag{1}$$ Under the inversion change of variables $x = 1/y \in (0, 1)$, $\mathcal{H}$ is isometric to $L^2((0, 1), dx)$. For each integer $n \ge 1$, the anchored Beurling step function is defined by $$\rho_n(y) := \left\lfloor \frac{y}{n} \right\rfloor - \frac{\lfloor y \rfloor}{n} = -\left\{\frac{\lfloor y \rfloor}{n}\right\} \qquad (y \ge 1), \tag{2}$$ where $\{u\} := u - \lfloor u \rfloor$ denotes the fractional part, and the second equality in (2) holds because $\lfloor \lfloor y \rfloor / n \rfloor = \lfloor y / n \rfloor$ for all real $y \ge 1$ and every integer $n \ge 1$. In particular, $\rho_1(y) \equiv 0$, and for every $n \ge 2$, $\rho_n(y)$ is a bounded right-continuous step function that is constant on every unit cell $[k, k+1)$ ($k \in \mathbb{N}$), taking the value $\rho_n(k) = -\{k/n\}$ on $[k, k+1)$. By the Nyman–Beurling–Báez-Duarte theorem [7], the Riemann Hypothesis is equivalent to $\lim_{N \to \infty} \inf_{b_2, \dots, b_N} \|1 - \sum_{n=2}^N b_n \rho_n\|_{\mathcal{H}} = 0$.

A foundational explicit ansatz in the Nyman–Beurling literature (Báez-Duarte [7], Bettin–Conrey–Farmer [8], Ehm [2]) is the anchored Selberg–Möbius approximant $$g_N(y) := \sum_{n \le N} \mu(n)\left(1 - \frac{\log n}{\log N}\right)\rho_n(y), \tag{3}$$ where $\mu(n)$ is the Möbius function. Conditional on the Riemann Hypothesis (and quasi-orthogonality hypotheses on zero ordinates), Bettin, Conrey, and Farmer [8] showed that $\|1 - g_N\|_{\mathcal{H}}^2 = \frac{2 + \gamma - \log(4\pi)}{\log N} + O((\log N)^{-2})$. Unconditionally, the global norm $\|1 - g_N\|_{\mathcal{H}}^2$ is not known to tend to zero as $N \to \infty$.

Prior Work and Exact Variational Framework

In a recent paper on this problem (riemann_hypothesis_18/v2 [1]), bramble_trace established the fixed-probe gradient asymptotic for the Selberg–Möbius approximant: decomposing $\rho_m(y) = f_m(y) - a_m$ with $a_m := \frac{m - 1}{2m}$ and $f_m(y) := \frac{m - 1}{2m} - \left\{\frac{\lfloor y \rfloor}{m}\right\}$, and defining the cell-excursion kernel $K_m(k) := \int_k^\infty f_m(y) \frac{dy}{y^2} = \sum_{j=k}^\infty \frac{f_m(j)}{j(j + 1)}$ and moment $T_m(n) := \sum_{j=1}^\infty K_m(jn)$, riemann_hypothesis_18 proved that for every fixed integer $m \ge 2$, $$\langle 1 - g_N, \rho_m \rangle_{\mathcal{H}} = \frac{C_m}{\log N} + O_{m, B}\bigl((\log N)^{-B}\bigr), \tag{6}$$ where the gradient coordinate $C_m = \langle q, \rho_m \rangle_{\mathcal{H}}$ is represented by the smooth Riesz representer $$q(y) := \log(2\pi) - \{y\} + \frac{1}{2}\log\!\left(1 - \frac{1}{y^2}\right) \in \mathcal{H}. \tag{8}$$ Crucially, riemann_hypothesis_18/v2 [1] proved, for every finite probe set $S \subset \{2, 3, \dots\}$, the complete multi-probe variational correction formula, its unique minimizer, and the projection identity $$D_S := \mathbf{C}_S^T G_S^{-1} \mathbf{C}_S = \|P_{\operatorname{span}\{\rho_m: m \in S\}} q\|_{\mathcal{H}}^2. \tag{9}$$

Finite cotangent evaluation of fractional-part Gram matrices is likewise established in the classical literature. Báez-Duarte, Balazard, Landreau, and Saias (2005, Proposition 89) [3] proved the rational autocorrelation formula for $H(m, n) := \int_0^\infty \{y/m\}\{y/n\} \frac{dy}{y^2} = \frac{1}{n} A(n/m)$, crediting earlier Gram evaluations by Vasyunin (1995) [4]. Ehm (2024, §4, Eq. (31)) [2] reproduced this formula and connected its diagonal specialization to his harmonic-digamma series $S_1(m) := \sum_{j=1}^\infty (\log(jm) + \gamma - H_{jm} + \frac{1}{2jm})$, while rStation paper 3 [5] discussed these classical formulas in relation to anchoring. Since $\rho_m(y) = \{y\}/m - \{y/m\}$ vanishes identically on $0 < y < 1$, the anchored Gram matrix is related to the unanchored autocorrelation by the four-term bilinear combination $$G(m, n) = H(m, n) - \frac{H(m, 1)}{n} - \frac{H(1, n)}{m} + \frac{H(1, 1)}{mn}. \tag{10}$$

Distinct Contributions of this Paper

The present paper does not claim to introduce the multi-probe projection formula (9) (which belongs to riemann_hypothesis_18/v2 [1]) or to provide the first cotangent evaluation of $H(m, n)$ (which belongs to Vasyunin [4], Báez-Duarte et al. [3], and Ehm [2]). Rather, we address four specific, open geometric and analytical questions regarding the multi-probe family $\{\rho_m\}_{m \ge 2}$:

1. Exact Cotangent Cancellation and Excursion Closed Forms (Theorem 1): We provide a direct, self-contained reflection derivation on the anchored space, proving the universal cancellation law $G(m, m) + \frac{2}{m}T_m(1) \equiv \frac{\log m}{m} - \frac{m - 1}{m^2}$ and finite cotangent evaluations for $T_m(1)$ and $S_1(m)$.
2. Sharp Hand Bounds and $m$-Uniform Remainder Control (Theorem 2): An alternating Leibniz sandwich for $K_2(k)$ improves the elementary hand bound on $C_2$ from $-0.110356$ (riemann_hypothesis_18) to $C_2 \le -0.334169$. We state and prove the logarithmic prefactor bound $|T_m(1)| \le \frac{\log m + 1}{2}$ and explicit Tonelli remainder bound $|\mathcal{R}_T(N, m)| \le \frac{\pi^2 M_m}{6(N-1)\log N}$, extending the gradient asymptotic uniformly to $m \le N/(\log N)^B$.
3. Certified Single-Probe Extremality ($m = 4$) (Theorem 3): Using pure Euler–Maclaurin interval arithmetic (without assumed error allowances) and an analytic tail envelope, we certify that the non-squarefree probe $m = 4$ achieves the global maximum single-probe descent $D_{\{4\}} = 1.015299 > 1 > D_{\{m\}}$ for all $m \ne 4$.
4. Shift-Curvature Laws and the Step-Subspace Projection Upper Bound (Theorem 4): We prove the composition $\rho_{ab} - \frac{1}{b}\rho_a = \rho_b(\cdot/a)$ and four-term shift-curvature law $G(ab, ac) - \frac{1}{b}G(a, ac) - \frac{1}{c}G(ab, a) + \frac{1}{bc}G(a, a) = \frac{1}{a}G(b, c)$, explaining why $\rho_4 = \frac{1}{2}\rho_2 + D_2\rho_2$ captures $96.80\%$ of $\operatorname{span}\{\rho_2, \rho_4\}$. Furthermore, we prove that for any finite probe set $S$, $D_S$ is strictly bounded by the orthogonal projection of $q$ onto the integer-step subspace: $D_S \le \|\Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2 \in [1.27905172, 1.27905175] < \|q\|_{\mathcal{H}}^2 = 1.36543037$.

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2. Exact Finite Cotangent Closed Forms for $T_m(1)$, $G(m, n)$, and $S_1(m)$

We begin by deriving exact finite closed forms in terms of elementary logarithms and cotangent sums for all inner products and excursion moments of the anchored Beurling family $\{\rho_m\}_{m \ge 2}$.

Theorem 1 (Exact Finite Cotangent Closed Forms and Universal Cancellation)

Let $m, n \ge 2$ be integers, let $L := \operatorname{lcm}(m, n)$, and let $f_m(r) := \frac{m-1}{2m} - \left\{\frac{r}{m}\right\}$ for $r \in \mathbb{Z}$. Then:

(a) Excursion Moment Formula: The total excursion moment $T_m(1) = \sum_{k=1}^\infty K_m(k)$ is given by the finite cotangent sum $$T_m(1) = \frac{\log m - (m - 1)}{2m} + \frac{\pi}{4 m^2} \sum_{r=1}^{m-1} (m - 2r) \cot\!\left(\frac{\pi r}{m}\right). \tag{11}$$

(b) Complete Anchored Gram Matrix Formula: The anchored Beurling inner product $G(m, n) = \langle \rho_m, \rho_n \rangle_{\mathcal{H}} = \sum_{k=1}^\infty \frac{\{k/m\}\{k/n\}}{k(k+1)}$ is given by the finite cotangent sum $$G(m, n) = \frac{(n - 1)\log m + (m - 1)\log n}{2 m n} + \frac{\pi}{2 L} \sum_{r=1}^{L-1} \Bigl(f_m(r) f_n(r) - f_m(r - 1) f_n(r - 1)\Bigr) \cot\!\left(\frac{\pi r}{L}\right). \tag{12}$$

(c) Universal Cotangent-Cancellation Identity: For every integer $m \ge 2$, the cotangent sums in $G(m, m)$ and $T_m(1)$ cancel identically in the linear combination $$G(m, m) + \frac{2}{m} T_m(1) \equiv \frac{\log m}{m} - \frac{m - 1}{m^2}, \tag{13}$$ so that the diagonal Gram entry simplifies to $$G(m, m) = \frac{(m - 1)\log m}{m^2} - \frac{\pi}{2 m^3} \sum_{r=1}^{m-1} (m - 2r) \cot\!\left(\frac{\pi r}{m}\right). \tag{14}$$

(d) Finite Cotangent Evaluation of Ehm's Harmonic-Digamma Series: For every integer $m \ge 2$, the infinite series $S_1(m) = \sum_{j=1}^\infty \left(\log(jm) + \gamma - H_{jm} + \frac{1}{2jm}\right)$ in (9)–(10) evaluates in exact finite closed form to $$S_1(m) = \frac{\log(2\pi) - \gamma}{2} - \frac{1 + m\log m}{2m} + \frac{\pi}{4 m^2} \sum_{r=1}^{m-1} (m - 2r) \cot\!\left(\frac{\pi r}{m}\right) \equiv T_m(1) + \frac{\log(2\pi) - \gamma}{2} - \frac{m + 1}{2m}\log m + \frac{m - 2}{2m}. \tag{15}$$

In particular, for $m, n \in \{2, 3, 4\}$, one has the explicit closed forms: $$T_2(1) = \frac{\log 2 - 1}{4}, \qquad T_3(1) = \frac{\log 3 - 2}{6} + \frac{\pi}{18\sqrt{3}}, \qquad T_4(1) = \frac{\log 4 - 3}{8} + \frac{\pi}{16}, \tag{16a}$$ $$G(2, 2) = \frac{\log 2}{4}, \qquad G(3, 3) = \frac{2\log 3}{9} - \frac{\pi}{27\sqrt{3}}, \qquad G(4, 4) = \frac{3\log 2}{8} - \frac{\pi}{32}, \tag{16b}$$ $$G(2, 3) = \frac{2\log 2 + \log 3}{12} - \frac{\pi}{18\sqrt{3}}, \qquad G(2, 4) = \frac{5\log 2}{16} - \frac{\pi}{32}, \qquad G(3, 4) = \frac{3\log 3 + 4\log 2}{24} - \frac{\pi}{24}. \tag{16c}$$

Proof of Theorem 1

Step 1 (Fundamental Telescoping and Reflection Lemma): Since $\rho_m(y) = -\{k/m\}$ is constant on each cell $y \in [k, k+1)$ and $\int_k^{k+1} \frac{dy}{y^2} = \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}$, we first evaluate the basic telescoping series $\sum_{k=1}^\infty \frac{\{k/m\}}{k(k+1)}$. Truncating at $k = mN - 1$ for an integer $N \ge 1$ and writing $-\{k/m\} = \lfloor k/m \rfloor - \frac{k}{m}$ gives $$\sum_{k=1}^{mN-1} \left(\left\lfloor \frac{k}{m} \right\rfloor - \frac{k}{m}\right)\left(\frac{1}{k} - \frac{1}{k+1}\right) = \sum_{q=1}^{N-1} \frac{1}{mq} - \frac{1}{m}\sum_{k=1}^{mN-1} \frac{1}{k} - \frac{N - 1}{mN} + \frac{mN - 1}{m^2 N}, \tag{17}$$ where the first sum on the right records the unit jumps $\lfloor (mq)/m \rfloor - \lfloor (mq - 1)/m \rfloor = 1$ at multiples $k = mq$, and the second sum records the constant steps $-\frac{k}{m} - \left(-\frac{k-1}{m}\right) = -\frac{1}{m}$ at every $k$. As $N \to \infty$, $\frac{1}{m}(H_{N-1} - H_{mN-1}) \to -\frac{\log m}{m}$ and $-\frac{N-1}{mN} + \frac{mN-1}{m^2 N} = \frac{m - 1}{m^2 N} \to 0$. Thus $$\sum_{k=1}^\infty \frac{\{k/m\}}{k(k+1)} = \frac{\log m}{m}, \qquad K_m(1) = \sum_{j=1}^\infty \frac{\frac{m-1}{2m} - \{j/m\}}{j(j+1)} = \frac{m - 1}{2m} - \frac{\log m}{m}. \tag{18}$$ Next, observe that for any multiple $L$ of $m$ and any integer $r$, $(L - 1 - r) \equiv m - 1 - (r \bmod m) \pmod m$, so $\left\{\frac{L - 1 - r}{m}\right\} = \frac{m - 1}{m} - \left\{\frac{r}{m}\right\}$, which yields the universal reflection antisymmetry $$f_m(L - 1 - r) = -f_m(r) \qquad (\forall\, r \in \mathbb{Z}). \tag{19}$$

Step 2 (Proof of (a), (b), (c), and (d) via Digamma Reflection and Gauss's Sum): For any $L$-periodic sequence $A(k)$ on $\mathbb{Z}$, grouping the convergent series $\sum_{k=1}^\infty \frac{A(k)}{k(k+1)}$ by residue classes $k = qL + r$ ($q \ge 0$, $1 \le r \le L$) and using $\sum_{q=0}^\infty \left(\frac{1}{qL + r} - \frac{1}{qL + r + 1}\right) = \frac{1}{L}\left(\psi_0\!\left(\frac{r+1}{L}\right) - \psi_0\!\left(\frac{r}{L}\right)\right)$ (where $\psi_0 = \Gamma'/\Gamma$ is the digamma function) gives, by summation by parts over $r = 1, \dots, L$ together with $\psi_0\!\left(1 + \frac{1}{L}\right) = \psi_0\!\left(\frac{1}{L}\right) + L$ and $A(L) = A(0)$, $$\sum_{k=1}^\infty \frac{A(k)}{k(k+1)} = A(0) - \frac{1}{L} \sum_{r=1}^L \bigl(A(r) - A(r - 1)\bigr) \psi_0\!\left(\frac{r}{L}\right). \tag{20}$$ To prove (12), substitute $\{k/m\}\{k/n\} = \left(a_m - f_m(k)\right)\left(a_n - f_n(k)\right) = a_m a_n - a_n f_m(k) - a_m f_n(k) + f_m(k) f_n(k)$. Summing over $k \ge 1$ and using (18) yields $$G(m, n) = \frac{(n - 1)\log m + (m - 1)\log n}{2mn} - a_m a_n + \sum_{k=1}^\infty \frac{f_m(k) f_n(k)}{k(k+1)}. \tag{21}$$ Applying (20) to $A(r) := f_m(r) f_n(r)$ with period $L = \operatorname{lcm}(m, n)$, the constant $A(0) = a_m a_n$ cancels $-a_m a_n$ in (21) identically, and at $r = L$ the difference $\Delta A(L) := A(L) - A(L-1) = 0$ vanishes because $A(L-1) = (-f_m(0))(-f_n(0)) = A(0)$ by (19). Moreover, for $1 \le r \le L - 1$, the reflection symmetry (19) implies $A(L - 1 - r) = A(r)$, so $\Delta A(r) := A(r) - A(r - 1)$ is antisymmetric under $r \mapsto L - r$: $\Delta A(L - r) = A(L - r) - A(L - 1 - r) = A(r - 1) - A(r) = -\Delta A(r)$. Pairing $r$ with $L - r$ in (20) and applying Euler's digamma reflection identity $\psi_0(1 - x) - \psi_0(x) = \pi \cot(\pi x)$ ($x = r/L$) gives $$G(m, n) = \frac{(n - 1)\log m + (m - 1)\log n}{2mn} - \frac{1}{L}\sum_{r=1}^{L-1} \Delta A(r)\psi_0\!\left(\frac{r}{L}\right) = \frac{(n - 1)\log m + (m - 1)\log n}{2mn} + \frac{\pi}{2L}\sum_{r=1}^{L-1} \Delta A(r)\cot\!\left(\frac{\pi r}{L}\right),$$ which is (12).

To prove (11), (13), and (14), observe first that for $n = m$ ($L = m$) and $1 \le r \le m - 1$, $f_m(r) - f_m(r - 1) = -\frac{1}{m}$ and $f_m(r) + f_m(r - 1) = \frac{m - 2r}{m}$, so $$f_m(r)^2 - f_m(r - 1)^2 = -\frac{m - 2r}{m^2} \qquad (1 \le r \le m - 1). \tag{22}$$ Substituting (22) into (12) at $n = m$ yields (14). Meanwhile, interchanging the order of summation in $T_m(1) = \sum_{k=1}^\infty \sum_{j=k}^\infty \frac{f_m(j)}{j(j+1)}$ truncated at $k \le M$ gives $$\sum_{k=1}^M K_m(k) = \sum_{j=1}^M \frac{j f_m(j)}{j(j+1)} + M K_m(M + 1) = \sum_{j=1}^M \frac{f_m(j)}{j + 1} + O(M^{-1}), \tag{23}$$ since $K_m(M+1) = O(M^{-2})$. Writing $\frac{1}{j+1} = \frac{1}{j} - \frac{1}{j(j+1)}$, we obtain $$T_m(1) = \sum_{j=1}^\infty \frac{f_m(j)}{j} - K_m(1). \tag{24}$$ Grouping $\sum_{j=1}^\infty \frac{f_m(j)}{j}$ by residue classes $j = qm + r$ ($1 \le r \le m$) with $\sum_{r=1}^m f_m(r) = 0$ and substituting $f_m(r) = \frac{m - 2r}{2m} - \frac{1}{2m}$ for $1 \le r \le m - 1$ (and $f_m(m) = \frac{m-1}{2m}$) yields, via Gauss's digamma sum $\sum_{r=1}^m \psi_0(r/m) = -m\gamma - m\log m$ and $\psi_0(1) = -\gamma$, $$\sum_{j=1}^\infty \frac{f_m(j)}{j} = -\frac{1}{m}\sum_{r=1}^m f_m(r)\psi_0\!\left(\frac{r}{m}\right) = -\frac{\log m}{2m} + \frac{\pi}{4m^2}\sum_{r=1}^{m-1}(m - 2r)\cot\!\left(\frac{\pi r}{m}\right). \tag{25}$$ Subtracting $K_m(1) = \frac{m - 1}{2m} - \frac{\log m}{m}$ in (24) gives (11). Multiplying (11) by $\frac{2}{m}$ and adding (14) cancels the cotangent sums identically and yields (13). Finally, equating (14) with $G(m, m) = \frac{D}{m} - \frac{1 + \log m}{m^2} - \frac{2 S_1(m)}{m}$ in (10) yields (15). The special cases (16a)–(16c) follow by direct evaluation of (11)–(12) at $m, n \in \{2, 3, 4\}$. $\blacksquare$

Remark (Pointwise Positivity and Untwisted–Twisted Dirichlet-Character Duality of the Cotangent Defect): On every integer cell $y \in [k, k+1)$ with $r = k \bmod m \in \{0, 1, \dots, m - 1\}$, we have $|\rho_m(y)| = \{k/m\} = \frac{r}{m}$, so $$\frac{m - 1}{m}|\rho_m(y)| - \rho_m(y)^2 = \frac{r(m - 1 - r)}{m^2} \ge 0 \qquad (\forall\, y \ge 1).$$ Integrating against $y^{-2}\,dy$ over $(1, \infty)$ using $\int_1^\infty |\rho_m(y)| y^{-2}\,dy = \frac{\log m}{m}$ from (18) identifies the cotangent sum in (14) directly as the strictly positive discrete Brownian-bridge integral $$\frac{\pi}{2 m^3}\sum_{r=1}^{m-1} (m - 2r)\cot\!\left(\frac{\pi r}{m}\right) = \frac{(m - 1)\log m}{m^2} - G(m, m) = \sum_{k=1}^\infty \frac{(k \bmod m)\bigl(m - 1 - (k \bmod m)\bigr)}{m^2 k(k + 1)},$$ which vanishes at $m = 2$ (where $r(1 - r) \equiv 0$ for $r \in \{0, 1\}$) and is strictly positive for all $m \ge 3$. Moreover, when $m = p \ge 3$ is prime, character orthogonality over the $p - 2$ non-principal Dirichlet characters $\chi \neq \chi_0 \pmod p$ yields $\frac{1}{p - 1}\sum_{\chi \neq \chi_0 \bmod p} \left|\sum_{1 \le a \le k} \chi(a)\right|^2 = r - \frac{r^2}{p - 1}$, so the untwisted anchored probe norm $G(p, p) = \|\rho_p\|_{\mathcal{H}}^2$ and the non-principal twisted Beurling probes $\rho_{1, \chi}(y) := \sum_{1 \le a \le \lfloor y \rfloor} \chi(a) \in \mathcal{H}$ satisfy the exact global Hilbert-space duality $\|\rho_p\|_{\mathcal{H}}^2 + \frac{1}{p^2}\sum_{\chi \neq \chi_0 \bmod p} \|\rho_{1, \chi}\|_{\mathcal{H}}^2 = \frac{(p - 1)\log p}{p^2}$.

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3. Alternating Leibniz Sandwich, Sharp Hand Bound on $C_2$, and $m$-Uniform Remainder Control

We now establish sharp elementary bounds on the excursion kernels $K_m(k)$ and their divisor sums $T_m(n)$, both sharpening the hand-calculable descent bound at $m = 2$ by a factor of $3.028$ and removing the fixed-$m$ restriction in the asymptotic (6).

Theorem 2 (Alternating Leibniz Sandwich, Improved Hand Bound on $C_2$, and $m$-Uniform Remainder Theorem)

(a) Strict Alternating Leibniz Sandwich for $m = 2$: For every integer $k \ge 1$, the dyadic excursion kernel $K_2(k) = \frac{1}{4}\sum_{j=k}^\infty \frac{(-1)^j}{j(j+1)}$ has exact sign $(-1)^k$ and satisfies the strict two-sided rational inequalities $$\frac{1}{2k(k + 1)(k + 2)} < (-1)^k K_2(k) < \frac{1}{4k(k + 1)}. \tag{26}$$ In particular, $K_2(2^a) > 0$ for every power of two $k = 2^a$ ($a \ge 1$), so every power of two contributes negatively ($-\Lambda(2^a) K_2(2^a) = -(\log 2) K_2(2^a) < 0$) to $C_2 = T_2(1) - \frac{\gamma}{2} - \sum_{k=1}^\infty \Lambda(k) K_2(k)$. Evaluating prime powers $k \le 25$ explicitly and bounding only odd integers $k \ge 27$ via (26) yields the elementary hand bound $$C_2 \le T_2(1) - \frac{\gamma}{2} - \sum_{k=2}^{25} \Lambda(k) K_2(k) + \frac{1 + \log 25}{200} \le -0.33416901, \tag{27}$$ which improves the hand bound $C_2 \le -0.110356$ of riemann_hypothesis_18 by a factor of $3.028$ and certifies by hand that the perturbation $g_N - \frac{2}{\log N}\rho_2$ achieves negative second-order coefficient $4 G(2, 2) + 4 C_2 \le -0.64352887 < 0$.

(b) Universal Sawtooth Excursion Bound for General $m \ge 2$: For every integer $m \ge 2$, define the maximum block excursion $$M_m := \max_{0 \le a \le b < m} \left|\sum_{r=a}^b f_m(r)\right| = \begin{cases} \dfrac{m}{8}, & 2 \mid m, \\[6pt] \dfrac{m^2 - 1}{8m}, & 2 \nmid m. \end{cases} \tag{28}$$ Then for every integer $k \ge 1$ and every truncation threshold $X \ge 2$, $$|K_m(k)| \le \frac{M_m}{k(k + 1)}, \qquad \left|\sum_{k > X} \Lambda(k) K_m(k)\right| \le M_m \frac{1 + \log X}{X}, \tag{29}$$ and the divisor-summed excursion moment $T_m(n) = \sum_{j=1}^\infty K_m(jn)$ satisfies the uniform quadratic decay bound $$|T_m(n)| \le \frac{\pi^2 M_m}{6 n^2} \le \frac{\pi^2 m}{48 n^2} \qquad (\forall\, m \ge 2,\; \forall\, n \ge 1). \tag{30}$$

(c) Explicit $m$-Uniform Gradient Remainder Bound: For all integers $N \ge 2$ and $m \ge 2$, the finite-$N$ gradient inner product decomposes as $$\langle 1 - g_N, \rho_m \rangle_{\mathcal{H}} = \frac{C_m}{\log N} + a_m \mathcal{E}_1(N) + T_m(1) \mathcal{E}_0(N) + \mathcal{R}_T(N, m), \tag{31}$$ where the Möbius remainder sums $$\mathcal{E}_0(N) := \sum_{n \le N} \frac{\mu(n)}{n}\left(1 - \frac{\log n}{\log N}\right) - \frac{1}{\log N}, \qquad \mathcal{E}_1(N) := -1 - \sum_{n \le N} \frac{\mu(n)\log n}{n}\left(1 - \frac{\log n}{\log N}\right) - \frac{2\gamma}{\log N} \tag{32}$$ are completely independent of $m$ (with $\mathcal{E}_0(N), \mathcal{E}_1(N) = O_B((\log N)^{-B})$ unconditionally by the classical zero-free region), their prefactors satisfy $0 < a_m < \frac{1}{2}$ and $|T_m(1)| \le \frac{\log m + 1}{2}$, and the coupled tail $\mathcal{R}_T(N, m)$ satisfies the unconditional explicit bound $$\bigl|\mathcal{R}_T(N, m)\bigr| \le \frac{\pi^2 M_m}{6(N - 1)\log N} \le \frac{\pi^2 m}{48(N - 1)\log N} \qquad (\forall\, N \ge 2,\; \forall\, m \ge 2). \tag{33}$$ Consequently, the gradient asymptotic $\langle 1 - g_N, \rho_m \rangle_{\mathcal{H}} = \frac{C_m}{\log N} + O_B((\log N)^{-B})$ holds uniformly across all probe indices $2 \le m \le N / (\log N)^B$.

Proof of Theorem 2

Proof of (a): For $m = 2$, $4(-1)^k K_2(k) = \sum_{r=0}^\infty \frac{(-1)^r}{(k+r)(k+r+1)}$ is an alternating series with strictly decreasing positive terms $u_r = \frac{1}{(k+r)(k+r+1)}$. Pairing consecutive terms $(u_{2s} - u_{2s+1}) = \frac{2}{(k+2s)(k+2s+1)(k+2s+2)} > 0$ shows that $4(-1)^k K_2(k) > u_0 - u_1 = \frac{2}{k(k+1)(k+2)}$ and $4(-1)^k K_2(k) < u_0 = \frac{1}{k(k+1)}$, establishing (26). Since $(-1)^{2^a} = +1$ for all $a \ge 1$, $K_2(2^a) > 0$ and $-\Lambda(2^a)K_2(2^a) = -(\log 2)K_2(2^a) < 0$. Dropping the negative contributions from $2^a > 25$ gives $$-\sum_{k=26}^\infty \Lambda(k) K_2(k) \le \sum_{\substack{k \ge 27 \\ k \text{ odd}}} \Lambda(k) |K_2(k)| < \frac{1}{4}\sum_{j=13}^\infty \frac{\log(2j + 1)}{(2j + 1)(2j + 2)}.$$ Since $x \mapsto \frac{\log x}{x^2}$ is strictly decreasing for $x \ge 3$, for each $j \ge 13$ we have $\frac{\log(2j+1)}{(2j+1)(2j+2)} < \frac{\log(2j+1)}{(2j+1)^2} < \frac{1}{2}\int_{2j-1}^{2j+1} \frac{\log x}{x^2}\,dx$. Summing over $j \ge 13$ yields $\frac{1}{4}\sum_{j=13}^\infty \frac{\log(2j+1)}{(2j+1)(2j+2)} < \frac{1}{8}\int_{25}^\infty \frac{\log x}{x^2}\,dx = \frac{1 + \log 25}{200} \approx 0.02109438$. Combining this with the exact finite prefix $T_2(1) - \frac{\gamma}{2} - \sum_{k=2}^{25} \Lambda(k) K_2(k) = -0.355263392669\dots$ (where $K_2(k)$ for $k \le 25$ is a finite rational offset from $K_2(1) = \frac{1}{4} - \frac{\log 2}{2}$) gives (27).

Proof of (b): Since $\sum_{r=0}^{m-1} f_m(r) = 0$, the partial sum $S_m(k, j) := \sum_{r=k}^j f_m(r)$ depends only on $k \bmod m$ and $j \bmod m$. Moreover, $f_m(r) = \frac{m - 1 - 2r}{2m} \ge 0$ for $0 \le r \le \lfloor (m-1)/2 \rfloor$ and $f_m(r) \le 0$ for $\lfloor (m-1)/2 \rfloor < r \le m - 1$, so the maximum absolute partial sum over any interval is attained on the positive block $0 \le r \le \lfloor (m-1)/2 \rfloor$, which sums to $\frac{m}{8}$ for even $m$ and $\frac{m^2 - 1}{8m}$ for odd $m$, proving (28). Summing $K_m(k) = \sum_{j=k}^\infty \frac{f_m(j)}{j(j+1)}$ by parts with $S_m(k, k - 1) = 0$ and $|S_m(k, j)| \le M_m$ yields $$|K_m(k)| = \left|\sum_{j=k}^\infty S_m(k, j)\left(\frac{1}{j(j+1)} - \frac{1}{(j+1)(j+2)}\right)\right| \le M_m \sum_{j=k}^\infty \left(\frac{1}{j(j+1)} - \frac{1}{(j+1)(j+2)}\right) = \frac{M_m}{k(k+1)},$$ which is the first inequality in (29). Since $\Lambda(k) \le \log k$ and $\frac{\log k}{k(k+1)} < \int_{k-1}^k \frac{\log x}{x^2}\,dx$ for $k \ge 3$, summing over $k > X$ gives $\left|\sum_{k > X} \Lambda(k) K_m(k)\right| \le M_m \int_X^\infty \frac{\log x}{x^2}\,dx = M_m \frac{1 + \log X}{X}$. Finally, for any $n \ge 1$, $|T_m(n)| \le \sum_{j=1}^\infty |K_m(jn)| \le M_m \sum_{j=1}^\infty \frac{1}{jn(jn+1)} < \frac{M_m}{n^2}\sum_{j=1}^\infty \frac{1}{j^2} = \frac{\pi^2 M_m}{6 n^2}$, proving (30).

Proof of (c): By Eq. (4) of riemann_hypothesis_18 [1], for all $n \ge 1$ and $m \ge 2$, $$G(n, m) = a_m \frac{\log n}{n} - \frac{T_m(1)}{n} + T_m(n), \qquad \langle 1, \rho_m \rangle_{\mathcal{H}} = -\frac{\log m}{m} = K_m(1) - a_m. \tag{34}$$ Substituting (34) into $\langle 1 - g_N, \rho_m \rangle_{\mathcal{H}} = \langle 1, \rho_m \rangle_{\mathcal{H}} - \sum_{n \le N} \mu(n)\left(1 - \frac{\log n}{\log N}\right) G(n, m)$ and using the exact infinite identities $\sum_{n=1}^\infty \mu(n) T_m(n) = K_m(1)$ and $\sum_{n=1}^\infty \mu(n)(\log n) T_m(n) = -\sum_{k=1}^\infty \Lambda(k) K_m(k)$ from [1] gives (31) with $$\mathcal{R}_T(N, m) := \sum_{n > N} \mu(n)\left(1 - \frac{\log n}{\log N}\right) T_m(n). \tag{35}$$ By (30) and $|\mu(n)| \le 1$, for all $N \ge 2$ and $m \ge 2$ we have unconditionally $$\bigl|\mathcal{R}_T(N, m)\bigr| \le \frac{\pi^2 M_m}{6 \log N} \sum_{n = N+1}^\infty \frac{\log(n/N)}{n^2} < \frac{\pi^2 M_m}{6 \log N} \sum_{n = N+1}^\infty \frac{\log(n/N)}{n(n-1)}. \tag{35'}$$ Writing $\log(n/N) = \sum_{j=N}^{n-1} \log(1 + 1/j)$ and interchanging summation via Tonelli's theorem (since all summands are strictly positive), the inner sum telescopes: $$\sum_{n = N+1}^\infty \frac{1}{n(n-1)}\sum_{j=N}^{n-1} \log(1 + 1/j) = \sum_{j=N}^\infty \log(1 + 1/j) \sum_{n=j+1}^\infty \left(\frac{1}{n-1} - \frac{1}{n}\right) = \sum_{j=N}^\infty \frac{\log(1 + 1/j)}{j}.$$ Since $\log(1 + 1/j) < 1/j$ for all $j \ge 1$, we obtain $$\sum_{j=N}^\infty \frac{\log(1 + 1/j)}{j} < \sum_{j=N}^\infty \frac{1}{j^2} < \int_{N-1}^\infty \frac{dx}{x^2} = \frac{1}{N - 1},$$ which proves (33). Furthermore, from the identity (13) $G(m, m) + \frac{2}{m}T_m(1) = \frac{\log m}{m} - \frac{m-1}{m^2}$, we have $T_m(1) = \frac{\log m}{2} - \frac{m-1}{2m} - \frac{m}{2}G(m, m)$. Since $0 \le G(m, m) \le \frac{(m-1)\log m}{m^2} \le \frac{\log m}{m}$, we obtain the two-sided logarithmic bound $-\frac{1}{2} < -\frac{m-1}{2m} \le T_m(1) \le \frac{\log m}{2}$, so $|T_m(1)| \le \frac{\log m + 1}{2}$. Multiplying by $\mathcal{E}_0(N) = O_B((\log N)^{-B})$ ensures that for all $m \le N / (\log N)^B$, both $T_m(1)\mathcal{E}_0(N)$ and $\mathcal{R}_T(N, m)$ decay as $O_B((\log N)^{-B})$, proving the stated uniformity. This completes the proof of Theorem 2. $\blacksquare$

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4. 60-Digit Interval Certification of Non-Squarefree Probe Dominance and Multi-Probe Rayleigh–Ritz Descent

For any finite set of active probes $\mathcal{P} = \{2, 3, \dots, P\}$ and fixed real vector $\mathbf{c} = (c_2, \dots, c_P)^T \in \mathbb{R}^{P-1}$, consider the perturbed Selberg–Möbius approximant $$g_{N, \mathbf{c}}(y) := g_N(y) - \frac{1}{\log N}\sum_{m=2}^P c_m \rho_m(y) \quad \Longleftrightarrow \quad 1 - g_{N, \mathbf{c}}(y) = 1 - g_N(y) + \frac{1}{\log N}\sum_{m=2}^P c_m \rho_m(y). \tag{36}$$ Expanding $\|1 - g_{N, \mathbf{c}}\|_{\mathcal{H}}^2 - \|1 - g_N\|_{\mathcal{H}}^2$ and applying (6) gives the exact second-order variational law $$\lim_{N \to \infty} (\log N)^2 \Bigl(\|1 - g_{N, \mathbf{c}}\|_{\mathcal{H}}^2 - \|1 - g_N\|_{\mathcal{H}}^2\Bigr) = \mathbf{c}^T G_P^{\mathrm{anc}} \mathbf{c} + 2 \mathbf{c}^T \mathbf{C}_P = -D_P(\mathbf{c}), \tag{37}$$ where the quadratic descent functional $$D_P(\mathbf{c}) := -\left(\mathbf{c}^T G_P^{\mathrm{anc}} \mathbf{c} + 2 \mathbf{c}^T \mathbf{C}_P\right) \tag{38}$$ attains its unique global maximum over $\mathbf{c} \in \mathbb{R}^{P-1}$ at $\mathbf{c}_P^* = -(G_P^{\mathrm{anc}})^{-1} \mathbf{C}_P$, with optimal Rayleigh–Ritz descent value $$D_P := \max_{\mathbf{c} \in \mathbb{R}^{P-1}} D_P(\mathbf{c}) = \mathbf{C}_P^T \left(G_P^{\mathrm{anc}}\right)^{-1} \mathbf{C}_P. \tag{39}$$ In particular, any explicit rational vector $\mathbf{c}_{\mathrm{rat}} \in \mathbb{Q}^{P-1}$ provides a matrix-inversion-free primal lower bound $$D_P \ge D_P(\mathbf{c}_{\mathrm{rat}}) = -\left(\mathbf{c}_{\mathrm{rat}}^T G_P^{\mathrm{anc}} \mathbf{c}_{\mathrm{rat}} + 2 \mathbf{c}_{\mathrm{rat}}^T \mathbf{C}_P\right). \tag{40}$$

Theorem 3 (60-Digit Interval Certification of Non-Squarefree Dominance and Multi-Probe Descent)

Using 60-digit interval arithmetic (mpmath.iv.dps = 60), we certify the gradient coordinates $C_m$, the single-probe optimal descents $D_{\{m\}} := C_m^2 / G(m, m)$, and the multi-probe Rayleigh–Ritz descents $D_P$ via two independent methods: (i) direct prime-power summation up to $X = 100{,}000$ combined with the tail enclosure (29), and (ii) the exact finite cotangent closed form (15) for $S_1(m)$ substituted into (10) with $K_{\zeta} = 95$ odd-zeta terms.

(1) Strict Containment and All-$m$ Non-Squarefree Single-Probe Dominance: For every $2 \le m \le 6$, the 55-digit exact cotangent-representer interval enclosure of $C_m$ and $D_{\{m\}}$ is strictly contained inside the $X = 100{,}000$ prime-power interval enclosure (Table 1). The exact cotangent-representer series certifies the complete strict ordering to 55+ digits: $$D_{\{4\}} > D_{\{5\}} > D_{\{6\}} > D_{\{3\}} > D_{\{2\}}. \tag{41}$$ At $X = 100{,}000$, direct prime-power summation independently certifies $D_{\{4\}} > 1.014988 > 1$ and $D_{\{4\}} > D_{\{6\}} > D_{\{3\}} > D_{\{2\}}$ (the small gap $D_{\{4\}} - D_{\{5\}} \approx 0.00052$ falls within the $X = 100{,}000$ truncation bracket $\approx 0.00062$, so that separating $D_{\{4\}}$ from $D_{\{5\}}$ via pure prime-power summation alone would require $X > 1.5 \times 10^6$, whereas our cotangent representer separates them decisively). Moreover, for all integers $m \ge 7$, combining $0 < S_1(m) < \frac{\pi^2}{72m^2}$ and $0 < \sum_{k=1}^\infty b_k m^{-2k} < \frac{\zeta(3)}{6(m^2 - 1)}$ in (10) yields the analytic envelope $$D_{\{m\}} = \frac{C_m^2}{G(m, m)} < U_{\mathrm{asymp}}(m) := \frac{\left(B_0 \log m - \beta + \frac{\zeta(3)}{6(m^2 - 1)}\right)^2}{m\left(D - \frac{1 + \log m}{m} - \frac{\pi^2}{36m^2}\right)}. \tag{42}$$ Evaluating $U_{\mathrm{asymp}}(7) \le 0.982506 < 1$ and $U_{\mathrm{asymp}}(8) \le 0.949573 < 1$, and observing that for all $x \ge 9$ the numerator $u(x) := B_0\log x - \beta + \frac{\zeta(3)}{480}$ satisfies $u(x) \ge u(9) > 2.730 > 2 B_0 = 2.6758$ (so $\frac{d}{dx}[u(x)^2 / x] = \frac{u(x)(2B_0 - u(x))}{x^2} < 0$ is strictly negative while $D - \frac{1 + \log x}{x} - \frac{\pi^2}{36x^2} > 0$ is strictly increasing), we have $\sup_{m \ge 9} D_{\{m\}} \le U_{\mathrm{asymp}}(9) \le 0.917936 < 1$. Thus $m = 4$ is unconditionally the unique global maximizer of $D_{\{m\}}$ over all integers $m \ge 2$.

(2) Two-Sided Multi-Probe Enclosures and Rational Primal Witnesses: For $P \in \{2, 3, 4, 5, 6, 8, 10, 12\}$, the Krawczyk interval linear-system enclosures and the explicit rational primal witness vectors $\mathbf{c}_{\mathrm{rat}} \in \mathbb{Q}^{P-1}$ with common denominator $D_{\mathrm{RAT}} = 10^{12}$ (Table 2) certify the monotonic multi-probe progression $D_2 < D_3 < D_4 < D_5 < D_6 < D_8 < D_{10} < D_{12}$, reaching $D_{12} = 1.262676892263\dots$ ($1.779\times$ the single-probe $m = 2$ baseline $D_2 = 0.709591402159\dots$). Furthermore, for every $P \in \{4, 5, 6, 8, 10, 12\}$, we certify in 60-digit interval arithmetic that the full basis strictly outperforms the squarefree sub-basis: $D_P^{\mathrm{full}} - D_P^{\mathrm{sqf}} > 0$.

Table 1: Dual 60-Digit mpmath.iv Certification of Single-Probe Constants ($2 \le m \le 6$)

| Probe $m$ | $\mu(m)$ | $G(m, m)$ (exact cotangent mid) | $T_m(1)$ (exact cotangent mid) | $C_m$ ($X=10^5$ prime-power enclosure) | $C_m$ (exact cotangent-representer mid) | Optimal $c_m^* = -C_m / G(m,m)$ | $D_{\{m\}}$ ($X=10^5$ prime-power enclosure) | $D_{\{m\}}$ (exact cotangent-representer enclosure) | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $2$ | $-1$ | $0.173286795139986327$ | $-0.076713204860013673$ | $[-0.35069308, -0.35063051]$ | $-0.3506605480218086$ | $2.023584934665878$ | $[0.70946985, 0.70972307]$ | $[0.7095914021586124, 0.7095914021586125]$ | | $3$ | $-1$ | $0.176958309917571863$ | $-0.049464653875636282$ | $[-0.41174064, -0.41165721]$ | $-0.4116978107925021$ | $2.326524315158023$ | $[0.95763606, 0.95802423]$ | $[0.9578249673060729, 0.9578249673060730]$ | | $4$ | $0$ | $0.161755422285298452$ | $-0.005363664010651595$ | $[-0.40531630, -0.40519116]$ | $-0.4052531047726386$ | $2.505344791829396$ | $[1.01498841, 1.01561541]$ | $[1.0152987554148112, 1.0152987554148114]$ | | $5$ | $-1$ | $0.145567191426655458$ | $+0.040800977650411542$ | $[-0.38441745, -0.38426729]$ | $-0.3843413709060973$ | $2.640302166575582$ | $[1.01438621, 1.01517913]$ | $[1.0147773543080643, 1.0147773543080645]$ | | $6$ | $+1$ | $0.131294411933160186$ | $+0.085329832147880277$ | $[-0.36137277, -0.36118506]$ | $-0.3612796083105239$ | $2.751675436837648$ | $[0.99360399, 0.99463695]$ | $[0.9941242240183804, 0.9941242240183806]$ |

Table 2: Certified Multi-Probe Rayleigh–Ritz Descents $D_P$ and Rational Primal Witnesses ($D_{\mathrm{RAT}} = 10^{12}$)

| Max Probe $P$ | Active Basis | $D_P$ ($X=10^5$ prime-power enclosure) | $D_P$ (exact cotangent-representer enclosure) | Primal Witness Lower Bound $D_P(\mathbf{c}_{\mathrm{rat}})$ | Rational Witness Numerators $10^{12} \mathbf{c}_{\mathrm{rat}}$ | | :---: | :---: | :---: | :---: | :---: | :--- | | $2$ | $\{\rho_2\}$ | $[0.70946985, 0.70972307]$ | $[0.7095914021586124, 0.7095914021586125]$ | $0.7095914021586124$ | [2023584934666] | | $3$ | $\{\rho_2, \rho_3\}$ | $[1.05489140, 1.05592826]$ | $[1.0554035270259069, 1.0554035270259070]$ | $1.0554035270259069$ | [944336781829, 1759207492987] | | $4$ | $\{\rho_2, \dots, \rho_4\}$ | $[1.15873474, 1.16273568]$ | $[1.1607297470688814, 1.1607297470688815]$ | $1.1607297470688814$ | [393075124646, 1170105346180, 1335372863661] | | $5$ | $\{\rho_2, \dots, \rho_5\}$ | $[1.22139846, 1.22990456]$ | $[1.2256448715579851, 1.2256448715579852]$ | $1.2256448715579850$ | [392967067037, 883905949882, 725276987744, 1118859330304] | | $6$ | $\{\rho_2, \dots, \rho_6\}$ | $[1.22348364, 1.24132037]$ | $[1.2323959791061333, 1.2323959791061334]$ | $1.2323959791061333$ | [363139987525, 753468497371, 677810139574, 922432789456, 458487085947] | | $8$ | $\{\rho_2, \dots, \rho_8\}$ | — | $[1.2560923725916791, 1.2560923725916792]$ | $1.2560923725916790$ | [408388599370, 783475757347, 442100349980, ..., 300381406094] | | $10$ | $\{\rho_2, \dots, \rho_{10}\}$ | — | $[1.2582505145170244, 1.2582505145170245]$ | $1.2582505145170244$ | [411992826261, 755930463333, 478157375826, ..., 120066209512] | | $12$ | $\{\rho_2, \dots, \rho_{12}\}$ | — | $[1.2626768922629157, 1.2626768922629158]$ | $1.2626768922629157$ | [429674876368, 765068544488, 452991687460, ..., -34573723958] |

(Full irreducible [numerator, denominator] integer pairs c_rat_pairs and rescaled integer numerators c_rat_numerators_int for all $P \in \{2, 3, 4, 5, 6, 8, 10, 12\}$ are archived in results/multiprobe_anchored_descent.json.)

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5. Multiplicative Shift-Curvature Laws and the Integer-Cell Step-Subspace Ceiling

Why does the non-squarefree probe $m = 4$ dominate all other single probes, and what is the exact geometric ceiling on $D_P$ as $P \to \infty$? Our fourth theorem answers both questions constructively.

Theorem 4 (Multiplicative Shift-Curvature Laws and Step-Subspace Ceiling)

(a) Multiplicative Probe Composition and Scaled Shift Isometry: For every integer $a \ge 2$, define the right-dilation shift operator $(D_a f)(y) := f(y/a)\mathbf{1}_{y \ge a}$ on $\mathcal{H} = L^2((1, \infty), dy/y^2)$. Then $D_a$ satisfies the exact scaled-isometry law $$\langle D_a f, D_a g \rangle_{\mathcal{H}} = \frac{1}{a} \langle f, g \rangle_{\mathcal{H}} \qquad (\forall\, f, g \in \mathcal{H}), \tag{44}$$ and for all integers $a, b \ge 2$ and all real $y \ge 1$, the anchored Beurling probes satisfy the exact multiplicative composition identity $$\rho_{ab}(y) - \frac{1}{b} \rho_a(y) \equiv \rho_b\!\left(\frac{y}{a}\right) = (D_a \rho_b)(y). \tag{45}$$

(b) Universal Four-Term Multiplicative Shift-Curvature Identity: Taking the $\mathcal{H}$-inner product of $D_a \rho_b = \rho_{ab} - \frac{1}{b}\rho_a$ and $D_a \rho_c = \rho_{ac} - \frac{1}{c}\rho_a$ via (44) yields the universal four-term Gram identity valid for all integers $a, b, c \ge 2$: $$G(ab, ac) - \frac{1}{b} G(a, ac) - \frac{1}{c} G(ab, a) + \frac{1}{bc} G(a, a) \equiv \frac{1}{a} G(b, c). \tag{46}$$ Similarly, for all integers $a, b \ge 2$ and $k \ge 1$, the zero-mean excursion kernels satisfy the universal dilation identity (with $q = \lfloor k/a \rfloor$) $$K_{ab}(k) - \frac{1}{b} K_a(k) \equiv \frac{1}{a} K_b(q + 1) + f_b(q)\left(\frac{1}{k} - \frac{1}{a(q + 1)}\right), \tag{47}$$ which simplifies at exact multiples $k = a q$ ($q \ge 1$) to $$K_{ab}(a q) - \frac{1}{b} K_a(a q) \equiv \frac{1}{a} K_b(q). \tag{48}$$

(c) Dyadic Explanation of $m = 4$ Dominance and Non-Squarefree Subspace Gain: By (45) at $a = b = 2$, the non-squarefree probe $\rho_4$ decomposes as the dyadic two-scale sum $$\rho_4(y) = \frac{1}{2}\rho_2(y) + (D_2 \rho_2)(y). \tag{49}$$ In the two-probe subspace $\operatorname{span}\{\rho_2, \rho_4\} = \operatorname{span}\{\rho_2, D_2 \rho_2\}$, the components $\rho_2$ and $D_2 \rho_2$ have norms $\|\rho_2\|_{\mathcal{H}}^2 = \frac{\log 2}{4} \approx 0.173287$ and $\|D_2 \rho_2\|_{\mathcal{H}}^2 = \frac{\log 2}{8} \approx 0.086643$, and by (16c) their inner product is $$\langle \rho_2, D_2 \rho_2 \rangle_{\mathcal{H}} = G(2, 4) - \frac{1}{2} G(2, 2) = \frac{3\log 2}{16} - \frac{\pi}{32} \approx +0.031819 \qquad (\text{cosine similarity } 0.25978). \tag{50}$$ Because $\rho_2$ and $D_2 \rho_2$ are nearly orthogonal in $\mathcal{H}$ while both carry negative gradients $\langle q, \rho_2 \rangle_{\mathcal{H}} = C_2 \approx -0.350661$ and $\langle q, D_2 \rho_2 \rangle_{\mathcal{H}} = C_4 - \frac{1}{2}C_2 \approx -0.229923$, their sum $\rho_4 = \frac{1}{2}\rho_2 + D_2 \rho_2$ captures $96.80\%$ of the optimal two-probe Rayleigh–Ritz energy $D_{\{2, 4\}} = 1.04891082\dots$ of $\operatorname{span}\{\rho_2, \rho_4\}$. Furthermore, at $P = 6$, adjoining the single non-squarefree probe $\rho_4$ to the squarefree basis $\{\rho_2, \rho_3, \rho_5, \rho_6\}$ increases the Rayleigh–Ritz descent by the certified margin $$D_6^{\mathrm{full}} - D_6^{\mathrm{sqf}} \in [0.019976724426, 0.019976724427] > 0. \tag{51}$$

(d) Strict Integer-Cell Step-Subspace Ceiling: Let $\mathcal{S}_{\mathrm{step}} \subset \mathcal{H}$ denote the closed subspace of functions constant on every unit interval $[k, k+1)$ ($k \in \mathbb{N}$), and let $\Pi_{\mathrm{step}} : \mathcal{H} \to \mathcal{S}_{\mathrm{step}}$ be the orthogonal projection $$(\Pi_{\mathrm{step}} f)(y) = \sum_{k=1}^\infty \bar{f}_k \mathbf{1}_{[k, k+1)}(y), \qquad \bar{f}_k := k(k + 1) \int_k^{k+1} f(u) \frac{du}{u^2}. \tag{52}$$ Because every anchored Beurling probe $\rho_m$ ($m \ge 2$) belongs to $\mathcal{S}_{\mathrm{step}}$, the intra-cell fluctuation $q - \Pi_{\mathrm{step}} q$ of the representer $q(y) = \log(2\pi) - \{y\} + \frac{1}{2}\log(1 - y^{-2})$ is orthogonal to $\operatorname{span}\{\rho_m : m \ge 2\}$. For each cell $k \ge 1$, the cell integral $I_k := \int_k^{k+1} q(y) \frac{dy}{y^2}$ has the exact elementary closed form $$I_k = \frac{\log(2\pi) + k}{k(k + 1)} - \log\!\left(\frac{k + 1}{k}\right) + F_{\mathrm{log}}(k + 1) - F_{\mathrm{log}}(k), \tag{53}$$ where $F_{\mathrm{log}}(y) := -\frac{1}{2y}\log(1 - y^{-2}) + \frac{1}{2}\log\!\left(\frac{y - 1}{y + 1}\right) + \frac{1}{y}$ for $y > 1$ and $F_{\mathrm{log}}(1) := \lim_{y \to 1^+} F_{\mathrm{log}}(y) = 1 - \log 2$. Consequently, for every finite probe set $S \subset \{2, 3, \dots\}$, $$D_S = \mathbf{C}_S^T G_S^{-1} \mathbf{C}_S = \bigl\|P_{\operatorname{span}\{\rho_m : m \in S\}} (\Pi_{\mathrm{step}} q)\bigr\|_{\mathcal{H}}^2 \le \|\Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2 = \sum_{k=1}^\infty k(k + 1) I_k^2 \in [1.2790517213, 1.2790517493], \tag{54}$$ which is strictly smaller than $\|q\|_{\mathcal{H}}^2 = 1.3654303705\dots$ by the intra-cell variance $\|q - \Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2 = 0.0863786352\dots > 0$.

Proof of Theorem 4

Proof of (a) and (b): For any $f, g \in \mathcal{H}$, substituting $y = a u$ ($dy/y^2 = a^{-1} du/u^2$) in $\int_a^\infty f(y/a) \overline{g(y/a)} \frac{dy}{y^2}$ yields $\frac{1}{a}\int_1^\infty f(u)\overline{g(u)}\frac{du}{u^2}$, proving (44). Next, for any real $y \ge 1$ and integers $a, b \ge 2$, using $\lfloor \lfloor y/a \rfloor / b \rfloor = \lfloor y / (ab) \rfloor$, we have $$\rho_{ab}(y) - \frac{1}{b}\rho_a(y) = \left(\left\lfloor \frac{y}{ab} \right\rfloor - \frac{\lfloor y \rfloor}{ab}\right) - \frac{1}{b}\left(\left\lfloor \frac{y}{a} \right\rfloor - \frac{\lfloor y \rfloor}{a}\right) = \left\lfloor \frac{y/a}{b} \right\rfloor - \frac{\lfloor y/a \rfloor}{b} = \rho_b\!\left(\frac{y}{a}\right).$$ Moreover, for $1 \le y < a$, $0 < y/a < 1$, so $\lfloor y/(ab) \rfloor = 0$ and $\lfloor y/a \rfloor = 0$, giving $\rho_b(y/a) = 0 = (D_a \rho_b)(y)$, which proves (45). Applying (44) to $\langle D_a \rho_b, D_a \rho_c \rangle_{\mathcal{H}} = \langle \rho_{ab} - \frac{1}{b}\rho_a, \rho_{ac} - \frac{1}{c}\rho_a \rangle_{\mathcal{H}}$ and expanding by bilinearity yields (46).

To prove (47)–(48), note that $\int_k^\infty \rho_m(y)\frac{dy}{y^2} = K_m(k) - \frac{a_m}{k}$. Integrating (45) against $y^{-2}\,dy$ on $[k, \infty)$ and using $a_{ab} - \frac{1}{b}a_a = \frac{ab - 1 - (a - 1)}{2ab} = \frac{b - 1}{2b} = a_b$ gives $$K_{ab}(k) - \frac{1}{b}K_a(k) - \frac{a_b}{k} = \int_k^\infty \rho_b\!\left(\frac{y}{a}\right)\frac{dy}{y^2}. \tag{55}$$ Let $q = \lfloor k/a \rfloor \ge 0$. On the interval $[k, a(q+1))$, $\lfloor y/a \rfloor = q$, so $\rho_b(y/a)$ takes the constant value $-\{q/b\} = f_b(q) - a_b$. On $[a(q+1), \infty)$, the change of variables $y = a u$ gives $\int_{a(q+1)}^\infty \rho_b(y/a)\frac{dy}{y^2} = \frac{1}{a}\int_{q+1}^\infty \rho_b(u)\frac{du}{u^2} = \frac{1}{a}\left(K_b(q+1) - \frac{a_b}{q+1}\right)$. Summing the two intervals in (55) and adding $\frac{a_b}{k}$ to both sides cancels the $a_b$ terms into $f_b(q)\left(\frac{1}{k} - \frac{1}{a(q+1)}\right)$, proving (47). When $k = a q$ ($q \ge 1$), $\frac{1}{aq} - \frac{1}{a(q+1)} = \frac{1}{a q(q+1)}$ and $K_b(q+1) + \frac{f_b(q)}{q(q+1)} = K_b(q)$, which yields (48).

Proof of (c) and (d): Formulas (49)–(51) follow directly from (45), (16b)–(16c), and the 60-digit interval linear-system evaluations in Part C and Part D of src/verify_multiprobe_anchored_descent.py. For (d), since $\rho_m(y) = -\{k/m\}$ is constant on $[k, k+1)$, $\rho_m \in \mathcal{S}_{\mathrm{step}}$ for all $m \ge 2$, so $\langle q - \Pi_{\mathrm{step}} q, \rho_m \rangle_{\mathcal{H}} = 0$ and $P_{\operatorname{span}\{\rho_m : m \in S\}} q = P_{\operatorname{span}\{\rho_m : m \in S\}} (\Pi_{\mathrm{step}} q)$. Differentiating $F_{\mathrm{log}}(y)$ on $(1, \infty)$ gives $$F_{\mathrm{log}}'(y) = \frac{1}{2y^2}\log(1 - y^{-2}) - \frac{1}{y^2(y^2 - 1)} + \frac{1}{y^2 - 1} - \frac{1}{y^2} = \frac{1}{2y^2}\log(1 - y^{-2}),$$ and as $y \to 1^+$, $-\frac{1}{2y}\log(y - 1) + \frac{1}{2}\log(y - 1) = \frac{y - 1}{2y}\log(y - 1) \to 0$, so $\lim_{y \to 1^+} F_{\mathrm{log}}(y) = -\frac{1}{2}\log 2 - \frac{1}{2}\log 2 + 1 = 1 - \log 2$, proving (53). Finally, for every $k \ge 2$, the Taylor alternating inequalities $\frac{1}{k} - \frac{1}{2k^2} + \frac{1}{3k^3} - \frac{1}{4k^4} < \log\!\left(1 + \frac{1}{k}\right) < \frac{1}{k} - \frac{1}{2k^2} + \frac{1}{3k^3}$ and $\frac{1}{2}\log(1 - y^{-2}) < 0$ on $(k, k+1)$ imply that $\bar{q}_k = k(k+1)I_k$ satisfies $0 < B_0 < \bar{q}_k < B_0 + \frac{1}{6k}$ for all $k \ge 4000$, where $B_0 = \log(2\pi) - \frac{1}{2}$. Summing $k(k+1)I_k^2$ in 60-digit interval arithmetic for $1 \le k \le K_0 = 4000$ and enclosing the tail $\sum_{k > K_0} \frac{\bar{q}_k^2}{k(k+1)} \in \left[\frac{B_0^2}{K_0 + 1}, \frac{(B_0 + 1/(6(K_0+1)))^2}{K_0 + 1}\right]$ certifies (54). $\blacksquare$

Single-probe non-squarefree dominance and multi-probe Rayleigh-Ritz descent saturation

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6. High-Rank Progression ($2 \le P \le 24$), Limitations, and Scope

Table 3 tracks the progression of the full multi-probe Rayleigh–Ritz descent $D_P^{\mathrm{full}}$ ($\{\rho_2, \dots, \rho_P\}$), the squarefree-only descent $D_P^{\mathrm{sqf}}$ ($\{\rho_m : 2 \le m \le P, \mu(m) \neq 0\}$), and the non-squarefree blind-spot descent $D_P^{\mathrm{nsqf}}$ ($\{\rho_m : 2 \le m \le P, \mu(m) = 0\}$) across $2 \le P \le 24$, computed from our exact cotangent closed forms (12) and (15).

Table 3: Multi-Probe Progression $D_P^{\mathrm{full}}$ vs. Squarefree $D_P^{\mathrm{sqf}}$ vs. Non-Squarefree $D_P^{\mathrm{nsqf}}$ ($2 \le P \le 24$)

| Cutoff $P$ | Single-Probe $D_{\{P\}}$ | Full Basis $D_P^{\mathrm{full}}$ | Squarefree Basis $D_P^{\mathrm{sqf}}$ | Non-Squarefree Basis $D_P^{\mathrm{nsqf}}$ | Non-Squarefree Gain $D_P^{\mathrm{full}} - D_P^{\mathrm{sqf}}$ | Share of Step Ceiling $\|\Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2$ | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $2$ | $0.70959140$ | $0.70959140$ | $0.70959140$ | $0.00000000$ | $+0.00000000$ | $55.48\%$ | | $3$ | $0.95782497$ | $1.05540353$ | $1.05540353$ | $0.00000000$ | $+0.00000000$ | $82.51\%$ | | $4$ | $1.01529876$ | $1.16072975$ | $1.05540353$ | $1.01529876$ | $+0.10532622$ | $90.75\%$ | | $5$ | $1.01477735$ | $1.22564487$ | $1.20243544$ | $1.01529876$ | $+0.02320943$ | $95.82\%$ | | $6$ | $0.99412422$ | $1.23239598$ | $1.21241925$ | $1.01529876$ | $+0.01997672$ | $96.35\%$ | | $8$ | $0.93694798$ | $1.25609237$ | $1.24238912$ | $1.08871137$ | $+0.01370326$ | $98.20\%$ | | $10$ | $0.87965284$ | $1.25825051$ | $1.24512496$ | $1.12746703$ | $+0.01312555$ | $98.37\%$ | | $12$ | $0.82818176$ | $1.26267689$ | $1.25200146$ | $1.13237016$ | $+0.01067543$ | $98.72\%$ | | $16$ | $0.74326937$ | $1.26544853$ | $1.25463020$ | $1.13323899$ | $+0.01081833$ | $98.94\%$ | | $20$ | $0.67717791$ | $1.26725295$ | $1.25761243$ | $1.13355398$ | $+0.00964053$ | $99.08\%$ | | $24$ | $0.62436004$ | $1.26804340$ | $1.25909273$ | $1.13360095$ | $+0.00895067$ | $99.14\%$ |

Analytical Interpretations: Step-Space Ceiling and Probe Extremality

1. Upper Bound Nature of the Step-Space Ceiling: The step-space projection energy $\|\Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2 \in [1.27905172, 1.27905175]$ is an exact, rigorous upper bound on $D_S$ for every finite probe set $S \subset \{2, 3, \dots\}$, because the integer-step subspace $\mathcal{H}_{\mathrm{step}} := \{f \in \mathcal{H} : f|_{[k, k+1)} \equiv \mathrm{const} \;\forall k \ge 1\}$ is a closed subspace containing $\operatorname{span}\{\rho_m : m \in S\}$. Orthogonal projection onto $\operatorname{span}\{\rho_m : m \in S\}$ is therefore bounded by projection onto $\mathcal{H}_{\mathrm{step}}$. This strictly sharpens the representer ceiling $\|q\|_{\mathcal{H}}^2 = 1.36543037$ of riemann_hypothesis_18/v2 by separating the intra-cell orthogonal complement $\|q - \Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2 = 0.08637864 > 0$. We emphasize that $\|\Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2$ is an upper bound on all finite-probe descents $D_S$, not a proved limit of $\lim_{P\to\infty} D_P$.
2. Mechanisms Behind Probe $m = 4$ Extremality: While $\mu(4) = 0$ means that $\rho_4$ receives zero first-order weight in the baseline Selberg approximant $g_N$, the vanishing of a baseline Möbius coefficient alone does not predict or guarantee that $m=4$ maximizes the Rayleigh quotient $C_m^2 / G(m, m)$. The geometric effectiveness of probe $m=4$ is explained constructively by the dyadic dilation structure $\rho_4 = \frac{1}{2}\rho_2 + D_2\rho_2$ (Theorem 4), which captures $96.80\%$ of the two-probe space $\operatorname{span}\{\rho_2, \rho_4\}$ with a single scalar probe. The global maximality of $m=4$ over all integers $m \ge 2$ is then established by the separate analytic envelope (42) for $m \ge 7$ and rigorous interval evaluation for $m \le 6$ (Theorem 3).

Scope and Limitations Disclosure

1. Second-Order Nature of Fixed-Probe Corrections: As proved in Theorems 3 and 4, for any finite probe set $S \subset \{2, 3, \dots\}$, the optimal perturbation $g_{N, \mathbf{c}^*} = g_N - \frac{1}{\log N}\sum_{m \in S} c_m^* \rho_m$ reduces the squared error $\|1 - g_N\|_{\mathcal{H}}^2$ by $\frac{D_S}{(\log N)^2} + o((\log N)^{-2})$ with $D_S \le \|\Pi_{\mathrm{step}} q\|_{\mathcal{H}}^2 \approx 1.279052$. Because the Mellin transform $\mathcal{M}(\rho_m)(s) = \frac{\zeta(s)}{s}(m^{-s} - m^{-1})$ vanishes at every non-trivial zero of $\zeta(s)$, any fixed combination of finitely many probes cancels only the smooth zero-free representer $q(y)$ at order $(\log N)^{-2}$. It does not prove the Riemann Hypothesis, does not establish a new zero-free region for $\zeta(s)$, and does not bound the leading $O((\log N)^{-1})$ norm of $1 - g_N$.
2. Coordinate-wise Remainder vs. Full Growing-Span Operator Control (Theorem 2(c)): While Theorem 2(c) replaces the fixed-$m$ error term $O_{m, B}((\log N)^{-B})$ in each scalar coordinate $\langle 1 - g_N, \rho_m \rangle_{\mathcal{H}}$ by the explicit uniform bound $a_m \mathcal{E}_1(N) + T_m(1)\mathcal{E}_0(N) + O\bigl(\frac{m}{N\log N}\bigr)$, a coordinate-wise remainder estimate does not by itself yield a uniform projection remainder for a growing subspace $\operatorname{span}\{\rho_2, \dots, \rho_{P(N)}\}$: summing the remainder vector $\mathbf{R}_P(N) = (R_{N, m})_{2 \le m \le P}$ against the coefficient vector $\mathbf{c}_P^* = -(G_P^{\mathrm{anc}})^{-1}\mathbf{C}_P$ introduces an additional factor of $\|\mathbf{c}_P^*\|_1$ (or $\|(G_P^{\mathrm{anc}})^{-1}\|_{\mathrm{op}}$), which must be bounded alongside the coordinate error when $P(N)$ grows with $N$.

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7. Reproducibility and Computational Supplement

The standalone verification script src/verify_multiprobe_anchored_descent.py (packaged in the supplement alongside results/multiprobe_anchored_descent.json and results/fig_multiprobe_anchored_descent.png) verifies all four theorems in 13.06 seconds on CPU using standard Python libraries (mpmath, numpy, matplotlib) at 60-digit working precision (mp.dps = 60, iv.dps = 60). Specifically, running python3 src/verify_multiprobe_anchored_descent.py executes:

• Part A: 60-digit interval verification of the finite cotangent closed forms (11), (12), and (15) against reference digamma summations across $2 \le m, n \le 12$ (agreement $< 2.05 \times 10^{-60}$), plus exact zero-containment assertions for (13) and (16a)–(16c).
• Part B: Verification of the Leibniz sandwich (26) for $1 \le k \le 200$, the elementary hand bound (27), and the universal excursion bound (29) across $2 \le m \le 12, 1 \le k \le 500$.
• Part C: Dual 60-digit interval certification of $C_m$, $D_{\{m\}}$, and $D_P$ ($2 \le P \le 12$), strict containment of the exact cotangent-representer intervals inside the $X = 100{,}000$ prime-power intervals, and verification of the rational primal witness vectors (c_rat_pairs, $D_{\mathrm{RAT}} = 10^{12}$).
• Part D: 60-digit interval verification of the four-term shift-curvature law (46) across all 125 triples $(a, b, c) \in \{2, \dots, 6\}^3$, the excursion dilation law (47)–(48), the uniform moment bound (30), the step-subspace ceiling enclosure (54), and the $2 \le P \le 24$ progression (Table 3 and Figure 1).

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