Dirichlet Hyperbola Factorization of the Full Signed Functional D_N

1. Orientation and Purpose

This document records the constructive next step in the arithmetic investigation of $D_N$, the defined signed prime-counting functional.

Following the independent review and author adjudication in ARITHMETIC_CANCELLATION_REVIEW.md and ARITHMETIC_CANCELLATION_CANDIDATE.md, we depart from the scale-by-scale bilinear sums $\mathcal{T}_{\mathrm{bilinear}}$ and address the full signed functional directly. Rather than postulating separate remainder bounds, we substitute the exact truncated Mobius convolution $\Lambda = \mu * \log$ into the entire functional $D_N$, retain all coupled boundary and principal terms, and establish an exact two-piece Dirichlet hyperbola decomposition without remainder leakage.

The central finding is twofold:

1. The full functional $D_N$ is an exact finite linear combination of $\Lambda(m)$ for $m \le N/2$ plus a completely explicit smooth baseline $\mathcal{S}_{\mathrm{smooth}}(N, K)$.
2. Substituting $\Lambda = \mu * \log$ yields an exact partition $D_N = \mathcal{S}_{\mathrm{smooth}} + \Sigma_1 + \Sigma_2$. Formal intuition only (no priced proof): the Type I sum $\Sigma_1$ formally matches the scale of $\mathcal{S}{\mathrm{smooth}}$. The Type II inner sums expand explicitly into interval differences of the Mertens function $M(t) = \sum{a \le t} \mu(a)$ (with empty-interval guards required; see section 4.2). The exact joint target of this route is $|\mathcal{S}_{\mathrm{smooth}} + \Sigma_1 + \Sigma_2| \ll_\epsilon N^{1/2+\epsilon}$. No equivalence is claimed between bounds on individual pieces ($\Sigma_1$ alone, $\Sigma_2$ alone) and RH: only $\zeta(s)$ is involved here (no Dirichlet L-functions), and necessity of Mertens-strength input for any individual piece is not proved. (Corrected 2026-09-20: the draft claimed an equivalence with the Mertens square-root bound and invoked Dirichlet L-functions; both removed.)

2. Exact Kernel Reduction of the Full Functional

Let $N \ge 4$ be an integer, $K = \lfloor\sqrt{N}\rfloor$, and $Y = N/K$. The full functional is defined by: $$D_N = N \int_{Y}^\infty \frac{R(u)}{u^2} \, du - \sum_{k=2}^K R(N/k)$$ where $R(u) = \psi(u) - u = \sum_{m \le u} \Lambda(m) - u$.

Using the identity $\int_1^\infty R(u)/u^2 \, du = -(1 + \gamma)$ and evaluating the finite integral on $[1, Y]$, the tail term expands as: $$N \int_Y^\infty \frac{R(u)}{u^2} \, du = N \log Y - N(1 + \gamma) + K \psi(Y) - N \sum_{m \le Y} \frac{\Lambda(m)}{m}$$ Similarly, the discrete sum expands as: $$\sum_{k=2}^K R(N/k) = \sum_{k=2}^K \psi(N/k) - N (H_K - 1)$$ where $H_K = \sum_{k=1}^K 1/k$.

Subtracting the two expressions collects the smooth terms into: $$\mathcal{S}_{\mathrm{smooth}}(N, K) = N (\log Y + H_K - 2 - \gamma)$$ The remaining terms depend purely on the prime powers $m$: $$\mathcal{A}N = K \psi(Y) - N \sum{m \le Y} \frac{\Lambda(m)}{m} - \sum_{k=2}^K \sum_{m \le N/k} \Lambda(m)$$

Interchanging summation over $k$ and $m$ over the range $m \le M = \lfloor N/2 \rfloor$:

• For $1 \le m \le Y$: the number of $k \in [2, K]$ with $k \le N/m$ is exactly $K - 1$. Combining with $(K - N/m) \Lambda(m)$ from the first two terms gives the coefficient $(K - N/m) - (K - 1) = 1 - N/m$.
• For $Y < m \le M$: since $N/m < N/Y = K$ and $N/m \ge N/M \ge 2$, the number of $k \in [2, K]$ with $k \le N/m$ is exactly $\lfloor N/m \rfloor - 1 \ge 1$. The first two terms contribute zero, yielding the coefficient $-(\lfloor N/m \rfloor - 1) = 1 - \lfloor N/m \rfloor$.
• Endpoint notes: $m = 1$ contributes $\Lambda(1) = 0$ regardless of branch; the $m \le Y$ versus $m > Y$ split is over integers against rational $Y$, hence exact with no fractional error; primality of $N$ itself never enters since $m \le M = \lfloor N/2 \rfloor$.

Therefore, the full signed functional admits the exact discrete kernel representation: $$D_N = \mathcal{S}{\mathrm{smooth}}(N, K) + \sum{m=1}^M w_N(m) \Lambda(m)$$ where $M = \lfloor N/2 \rfloor$, and the kernel $w_N(m)$ is defined for $1 \le m \le M$ by: $$w_N(m) = \begin{cases} 1 - \frac{N}{m}, & 1 \le m \le Y \\ 1 - \lfloor \frac{N}{m} \rfloor, & Y < m \le M \end{cases}$$ This representation is exact: there are no asymptotic remainders, no discarded boundaries, and no floating-point approximations.

3. Truncated Dirichlet Hyperbola Partition

We substitute the exact Dirichlet convolution: $$\Lambda(m) = (\mu * \log)(m) = \sum_{a b = m} \mu(a) \log b$$ into the arithmetic sum $\sum_{m=1}^M w_N(m) \Lambda(m)$. Since $\log 1 = 0$, terms with $b = 1$ vanish identically. The sum becomes: $$\sum_{m=1}^M w_N(m) \Lambda(m) = \sum_{\substack{a b \le M \\ b \ge 2}} \mu(a) \log b \, w_N(a b)$$

We partition the hyperbolic domain $\{ (a, b) : a b \le M, b \ge 2 \}$ using the parameter: $$U = \lfloor \sqrt{M} \rfloor = \lfloor \sqrt{\lfloor N/2 \rfloor} \rfloor$$

Lemma (Absence of Remainder Region)

For $U = \lfloor \sqrt{M} \rfloor$, every pair of integers $(a, b)$ satisfying $a b \le M$ satisfies $\min(a, b) \le U$. Consequently, the region $\{ a > U, b > U, a b \le M \}$ is strictly empty.

Proof. If $a > U$ and $b > U$, then $a \ge U + 1$ and $b \ge U + 1$. Thus $a b \ge (U + 1)^2$. By definition of the integer square root, $(U + 1)^2 > M$. Hence $a b > M$, which contradicts $a b \le M$. Q.E.D.

By this lemma, the hyperbolic domain splits disjointly into exactly two regions:

1. Type I Sum ($\Sigma_1$): $a \le U$ and $2 \le b \le \lfloor M/a \rfloor$.
2. Type II Sum ($\Sigma_2$): $2 \le b \le U$ and $U < a \le \lfloor M/b \rfloor$.

The decomposition of $D_N$ is therefore exact: $$D_N = \mathcal{S}_{\mathrm{smooth}}(N, K) + \Sigma_1(N) + \Sigma_2(N)$$ with: $$\Sigma_1(N) = \sum_{a=1}^U \mu(a) \sum_{b=2}^{\lfloor M/a \rfloor} \log b \, w_N(a b)$$ $$\Sigma_2(N) = \sum_{b=2}^U \log b \sum_{a=U+1}^{\lfloor M/b \rfloor} \mu(a) w_N(a b)$$ There is no third intersection piece to subtract, and no uncounted region.

4. Analytical Anatomy and the Mertens Obstruction

4.1. Type I Heuristic Scale Match

In $\Sigma_1$, the variable $a \le U \le \sqrt{N/2}$ is small, while the inner variable $b$ runs up to $M/a \asymp N/a$. The inner sum: $$S_1(a) = \sum_{b=2}^{\lfloor M/a \rfloor} \log b \, w_N(a b)$$ has the summand $\log b$ weighted by $w_N(a b)$. Formally, for $b \le Y/a$ where $w_N(ab) = 1 - N/(ab)$, Euler-Maclaurin in $b$ suggests main terms of the form $- (N/a) \log(N/a) + C (N/a)$. This is a formal heuristic scale reading, not an estimate: $w_N$ has a kink at $b = Y/a$ and jump discontinuities at every $ab = N/k$, so Euler-Maclaurin does not apply directly, and no remainder is derived or priced here. (Qualified 2026-09-20.)

Summing against $\mu(a)$ formally evokes the classical limiting relations: $$\sum_{a=1}^\infty \frac{\mu(a)}{a} = 0, \qquad \sum_{a=1}^\infty \frac{\mu(a) \log a}{a} = -1$$ These infinite series do not license uncontrolled finite truncations at $a \le U$: the partial sums converge only at PNT rate, and the truncation remainder at scale $U$ is not priced here. The observed numerical cancellation between $\mathcal{S}_{\mathrm{smooth}}$ and $\Sigma_1$ is diagnostic only: $|\mathcal{S}_{\mathrm{smooth}} + \Sigma_1|/N^{3/4}$ is $0.17$ at $N = 400$ and $0.75$ at $N = 1000$. Finite agreement is not a bound proof, and no analytical cancellation theorem is claimed for $\Sigma_1$ alone. (Corrected 2026-09-20: the draft presented the infinite-series cancellation and the finite-ratio observation as established.)

4.2. Reduction of Type II to the Mertens Function

In $\Sigma_2$, the outer variable $b$ is small ($2 \le b \le U$). For the inner variable $a$, we have integers $a \ge U + 1$ and $b \ge 2$, so $a b \ge 2(U + 1) > 2\sqrt{M}$. Since $Y = N/K \le \sqrt{N} + 1 + 1/(\sqrt{N} - 1)$ (valid upper bound; $Y < \sqrt{N} + 1$ is false, e.g. $N = 8$ and every $N = m^2 - 1$) and $2\sqrt{M} \ge Y$ holds for $N \ge 16$ by direct squaring, with $4 \le N \le 15$ checked by hand ($\Sigma_2$ is empty for $N \le 11$; at $N = 12, 13, 14, 15$ the least product is $6 > Y$), we have $ab > Y$ for every $\Sigma_2$ pair at every $N \ge 4$. Verified computationally: $\min(2(U+1) - Y) = 0.5$ over $4 \le N \le 100000$, and $\min(Kab - N) = 3$ over 30.8M sampled pairs. (Corrected 2026-09-20: the draft chain used the invalid inequality $\sqrt{N} \ge Y$; in fact $Y \ge \sqrt{N}$ always. The conclusion survives via the argument above.) Consequently, throughout $\Sigma_2$, the product $a b$ strictly exceeds $Y = N/K$. The kernel $w_N(a b)$ therefore takes purely integer values: $$w_N(a b) = 1 - \lfloor \frac{N}{a b} \rfloor$$

Let $k = \lfloor N/(a b) \rfloor$. Since $a b \le M = \lfloor N/2 \rfloor$, we have $k \ge 2$. Furthermore, since $a b > Y$, we have $k \le K$. The condition $\lfloor N/(a b) \rfloor = k$ is equivalent to: $$\frac{N}{k+1} < a b \le \frac{N}{k} \iff \frac{N}{b(k+1)} < a \le \frac{N}{bk}$$

The inner sum over $a$ in $\Sigma_2$ is: $$\mathcal{M}b(N) = \sum{U < a \le \lfloor M/b \rfloor} \mu(a) \left( 1 - \lfloor \frac{N}{a b} \rfloor \right)$$ Expanding by the level sets of the floor function: $$\mathcal{M}b(N) = \left[ M\left(\lfloor M/b \rfloor\right) - M(U) \right] - \sum{k=2}^K k \sum_{\substack{U < a \le \lfloor M/b \rfloor \\ N/(b(k+1)) < a \le N/(bk)}} \mu(a)$$ where $M(t) = \sum_{n \le t} \mu(n)$ is the Mertens function.

Each inner sum is a Mertens difference over a clamped interval. With $hi = \min(\lfloor M/b \rfloor, \lfloor N/(bk) \rfloor)$ and $lo = \max(U, \lfloor N/(b(k+1)) \rfloor)$: $$\sum_{\substack{U < a \le \lfloor M/b \rfloor \\ N/(b(k+1)) < a \le N/(bk)}} \mu(a) = \begin{cases} M(hi) - M(lo), & hi > lo \\ 0, & hi \le lo \end{cases}$$ The guard is load-bearing: clamped endpoints can reverse ($hi \le lo$), and the unguarded $M(hi) - M(lo)$ then returns a nonzero wrong-signed value (e.g. $N = 20$, $b = 2$, $k = 4$: unguarded $1$, correct $0$; two further instances at $b = 3$; see results_factorization_diagnostic.json). Level sets with $k > K - 1$ are empty by the $ab > Y$ gate and contribute $0$ under the same guard, so summing $k = 2, \dots, K$ is safe. (Corrected 2026-09-20: the draft omitted the guard.)

4.3. Pricing the Implication

We now evaluate the bounds on $\Sigma_2$:

• Trivial Size Bound: Bounding $|\mu(a)| \le 1$ and $|1 - \lfloor N/(ab) \rfloor| \le N/(ab)$ without sign cancellation yields: $$\sum_{U < a \le M/b} |\mu(a)| \left| 1 - \lfloor \frac{N}{ab} \rfloor \right| \ll \frac{N}{b} \sum_{a \le M/b} \frac{1}{a} \ll \frac{N \log N}{b}$$ Summing over $b \le U$ with $\log b$: $$\Sigma_2 \ll N \log N \sum_{b \le U} \frac{\log b}{b} \ll N \log^3 N$$ This trivial bound is honestly recorded as not best known.
• Signed Smooth and Fractional Splitting: Because $1 - \lfloor x \rfloor = 1 - x + \{x\}$ (with a plus sign for the fractional part), the inner sum decomposes exactly into: $$\mathcal{M}b(N) = \sum{U < a \le \lfloor M/b \rfloor} \mu(a) \left( 1 - \frac{N}{ab} \right) + \sum_{U < a \le \lfloor M/b \rfloor} \mu(a) \left\{ \frac{N}{ab} \right\} =: \mathcal{M}{b,\mathrm{smooth}}(N) + \mathcal{M}{b,\mathrm{frac}}(N)$$ Summing over $b \in [2, U]$ with weight $\log b$, we have the exact partition: $$\Sigma_2(N) = \Sigma_{2,\mathrm{smooth}}(N) + \Sigma_{2,\mathrm{frac}}(N)$$ Both pieces are verified independently in exact rational Fraction arithmetic in factorization_diagnostic.py.
• Proposed subexponential baseline via monotone telescoping (needs review, not accepted): (Downgraded 2026-09-20 per coordinator instruction; no replacement derivation is attempted in this correction. The derivation as written does not rigorously handle uniformity over the active jump ranges, the change of the exponential constant across $b$, the $V$-endpoint cases, and small $N$.) The proposal was: Notice that $g(a) = 1 - \lfloor N/(ab) \rfloor$ is a monotone non-decreasing function of $a$ on $U < a \le V = \lfloor M/b \rfloor$. Discrete summation by parts (Abel summation) yields: $$\mathcal{M}b(N) = g(V) M(V) - g(U+1) M(U) - \sum{a=U+1}^{V-1} M(a) [g(a+1) - g(a)]$$ Because $g$ is monotone, the forward differences $\Delta g(a) = g(a+1) - g(a) \ge 0$ telescope: their sum is $g(V) - g(U+1) \le \lfloor N/((U+1)b) \rfloor \le \sqrt{2N}/b$. Using the classical unconditional Mertens bound $|M(t)| \ll t \exp(-c\sqrt{\log t})$:
• The boundary term at $V$: $|g(V) M(V)| \ll 1 \cdot (N/b)\exp(-c\sqrt{\log(N/b)}) \ll \frac{N}{b} \exp(-c\sqrt{\log N})$.
• The boundary term at $U$: $|g(U+1) M(U)| \ll \frac{\sqrt{2N}}{b} \cdot \sqrt{N/2}\exp(-c\sqrt{\log N}) \ll \frac{N}{b} \exp(-c\sqrt{\log N})$.
• The jump sum: $\sum_k M(t_k) \Delta g_k \ll \sum_{k=2}^K \frac{N}{bk} \exp(-c\sqrt{\log(N/(bk))}) \ll \frac{N \log N}{b} \exp(-c'\sqrt{\log N})$. Every term in the telescoped representation carries the subexponential factor. Summing over $b \le U$ with $\log b$: $$\sum_{b=2}^U \frac{\log b}{b} \ll \log^2 U \ll \log^2 N$$ Therefore, with logs and constants tracked honestly: $$\Sigma_2(N) \ll N \log^3 N \exp(-c'\sqrt{\log N}) \ll N \exp(-c''\sqrt{\log N}) \qquad \text{(NOT accepted: see downgrade note above).}$$ End of proposal. The justified unconditional bound remains the elementary $\Sigma_2 \ll N \log^3 N$.
• Conditional baselines: no individual RH-strength bound is established in this note. No derivation here prices $\Sigma_1$ or $\Sigma_2$ individually at any $N^{7/8+\epsilon}$-, $N^{3/4+\epsilon}$-, or $N^{1/2+\epsilon}$-scale under RH; the $b,k$ ranges were not handled consistently in the removed attempts, and no replacement exponent is invented. The inherited RH implication for the joint $D_N$ is separate and is kept below.
• Status of the Fractional-Weight Piece: In the split $\Sigma_2 = \Sigma_{2,\mathrm{smooth}} + \Sigma_{2,\mathrm{frac}}$, the fractional-weight piece $\Sigma_{2,\mathrm{frac}} = \sum_{b=2}^U \log b \sum_{U < a \le M/b} \mu(a) \{N/(ab)\}$ is one unresolved component of $\Sigma_2$. It is not proved equivalent to RH, nor is bounding it proved equivalent to bounding the joint target $D_N$. It remains an unpriced component whose individual bound is not established.
• Valid Conditional Implication for the Joint Target: RH implies $|R(x)| \ll x^{1/2}\log^2 x$ (von Koch, inherited leaf), whence via the unconditional identity $D_N = R(N) + O(\sqrt{N})$ the joint bound $|D_N| \ll N^{1/2}\log^2 N$ follows directly. This implication is established; it routes through the renewal identity, not through individual bounds on $\Sigma_1$ or $\Sigma_2$.

The Precise Obstruction (restated): The estimates attempted in this note do not establish the required joint bound $|\mathcal{S}_{\mathrm{smooth}} + \Sigma_1 + \Sigma_2| \ll_\epsilon N^{1/2+\epsilon}$. No lower bound for $\Sigma_2$ was proved, and no saturation, impossibility, or individual non-boundedness claim is made: sharper estimates or different decompositions are not ruled out. The exact joint inequality above is retained as the open target.

5. Endpoints, Prime Powers, and Smoothing Costs

1. Prime Powers at Endpoints: In $D_N$, the arithmetic convolution runs only up to $M = \lfloor N/2 \rfloor$. Thus, whether $N$ itself is a prime or prime power has no direct effect on the summation range of $\Lambda(m)$. Endpoint sensitivity is present at every cutoff: the integer floors $K$, $M$, $U$, the branch split at rational $Y$, and the prime powers entering $\Lambda(m)$ and $\mu(a)$ sums at each range end. The split between $m \le Y$ and $m > Y$ in $w_N(m)$ is over integers against rational $Y$, hence exact with no fractional error; this removes only the $Y$-branch error, not the other cutoffs.
2. Smooth Tail Integral: The infinite tail $N \int_Y^\infty R(u)/u^2 \, du$ is completely resolved into $K \psi(Y) - N \sum_{m \le Y} \Lambda(m)/m$ plus smooth terms. There is no truncation error or discarded tail remainder.
3. Smoothing Cost: The smooth contribution $\mathcal{S}_{\mathrm{smooth}}(N, K) = N (\log Y + H_K - 2 - \gamma)$ is given in closed form and evaluates to high precision via standard harmonic number expansions.

6. Concrete Finite Numerical Test

To enable immediate independent verification, the table below provides the values of the components across eleven test values of $N \in [16, 1000]$.

In all cases:

• $K = \lfloor\sqrt{N}\rfloor$
• $M = \lfloor N/2 \rfloor$
• $U = \lfloor\sqrt{M}\rfloor$
• Defect is the measured floating-point accumulation $|D_N - (\mathcal{S}_{\mathrm{smooth}} + \Sigma_1 + \Sigma_2)|$ from factorization_diagnostic.py (fresh sieve, exact integer $\mu$, float64 logs); record: results_factorization_diagnostic.json.

Key Observations from the Numerical Evidence:

1. Measured Float Defect: The floating-point accumulation defect is below $2 \times 10^{-11}$ across the entire range, reflecting double-precision rounding. Finite floating-point agreement is a measured diagnostic, not a proof of exactness. The exactness of the partition is proved symbolically by the absence-of-remainder lemma and verified to zero defect in exact rational arithmetic via the prime-log coefficient tests in factorization_diagnostic.py.
2. Main Cancellation: At $N = 400$, $\mathcal{S}_{\mathrm{smooth}} = +1606.5025$ and $\Sigma_1 = -1591.4802$. Their sum is $+15.0223$. When added to $\Sigma_2 = -16.8221$, the result is $D_N = -1.7997$. This demonstrates how $\Sigma_1$ absorbs over 99% of the smooth scale growth.
3. Type II Behavior: $|\Sigma_2|$ remains small relative to $N$ across all tested cases ($|\Sigma_2| \le 127$ at $N = 1000$), but oscillates in sign, reflecting the underlying Mertens oscillations.