# Zero Genesis Game 18｜Multiplicative Pair Amplifier、Direct–Reciprocal Selberg Duality 與 Zero-Selective Almost-Prime Signal **Series:** SECV / Zero Genesis Experimental Playground **Experiment:** Zero Genesis Game 18 **Version:** v0.1 **Author:** Neo.K **AI Collaboration:** Aletheia (GPT-5.6 Sol) **Status:** Exploratory / Multiplicative-Convolution and Pole-Filter Study **Important:** This document does **not** claim a proof of RH. Classical results on generalized von Mangoldt functions and their RH-equivalent twisted sums already cover part of the theorem-level territory considered here. The new terminology below is organizational unless explicitly stated otherwise. --- ## 0. Motivation Game 17 identified the RH-hard forcing mode as a Mellin / log-time resonance: $$ \boxed{ e^{(\beta-1/2)t}e^{i\gamma t} } $$ coming from a zero $$ \rho=\beta+i\gamma. $$ It also observed that the raw multiplicative pair transform $$ \left( -\frac{\zeta'}{\zeta} \right)^2 $$ retains a universal double pole at every simple zero, while Selberg's classical combination $$ \frac{\zeta''}{\zeta} $$ cancels that leading simple-zero signature. Game 18 asks: > Can multiplicative self-convolution amplify RH violations in a useful way, and is there an arithmetic differential filter that removes the equilibrium pole while preserving zero sensitivity? The answer has three levels: 1. raw multiplicative convolution amplifies pole order but not the critical exponent; 2. the generalized convolution hierarchy is information-complete but correlation-expensive; 3. using the reciprocal zeta function produces a natural **zero-selective dual Selberg hierarchy** that reverses the pole selectivity of the classical Selberg hierarchy. The $k=2$ reciprocal filter yields a particularly simple signed prime/semiprime signal with no $s=1$ pole and a double pole at every simple nontrivial zero. --- # 1. Logarithmic derivative Define $$ \boxed{ h(s) = -\frac{\zeta'(s)}{\zeta(s)}. } $$ For $\Re s>1$, $$ \boxed{ h(s) = \sum_{n\ge1} \frac{\Lambda(n)}{n^s}. } $$ At the pole $s=1$, $$ \boxed{ h(s) = \frac1{s-1} -\gamma + O(s-1), } $$ where $\gamma$ is Euler's constant. At a zero $\rho$ of multiplicity $m$, $$ \boxed{ h(s) = -\frac{m}{s-\rho} + c_\rho + O(s-\rho). } $$ --- # 2. Raw multiplicative pair Define the Dirichlet convolution $$ \boxed{ b_2 = \Lambda*\Lambda. } $$ Then $$ \boxed{ h(s)^2 = \sum_{n\ge1} \frac{b_2(n)}{n^s}. } $$ All coefficients satisfy $$ \boxed{ b_2(n)\ge0. } $$ Define the summatory multiplicative pair field $$ \boxed{ C_2(x) = \sum_{n\le x}b_2(n) = \sum_{uv\le x} \Lambda(u)\Lambda(v). } $$ --- # 3. Equilibrium double pole Near $$ s=1, $$ $$ h(s)^2 = \frac1{(s-1)^2} - \frac{2\gamma}{s-1} + O(1). $$ Perron's kernel contributes an additional factor $1/s$. Therefore the complete $s=1$ principal part generates the equilibrium $$ \boxed{ M_2(x) = x\log x - (1+2\gamma)x. } $$ This matches known RH-conditional twisted estimates for $\Lambda*\Lambda$. --- # 4. Zero double pole At a multiplicity-$m$ zero, $$ h(s)^2 = \frac{m^2}{(s-\rho)^2} - \frac{2mc_\rho}{s-\rho} + O(1). $$ Thus the leading Perron residue contributes $$ \boxed{ \frac{m^2}{\rho} x^\rho\log x } $$ plus lower-order $x^\rho$ terms. After critical normalization: $$ \boxed{ x^{-1/2} [ C_2(x)-M_2(x) ] } $$ contains $$ \boxed{ x^{\beta-1/2} e^{i\gamma\log x} [ A_\rho\log x+B_\rho ]. } $$ --- # 5. Pole-order amplification does not move the critical exponent The raw pair transform turns a simple pole into a double pole. But the growth exponent remains $$ \boxed{ \beta-\frac12. } $$ Thus: - pole order changes; - logarithmic visibility changes; - spectral abscissa does not. This is the first **Pole-Order Amplification / Criticality Conservation** phenomenon. --- # 6. RH criterion from the multiplicative pair residual Set $$ x=e^t. $$ Define $$ \boxed{ R_2(t) = e^{-t/2} [ C_2(e^t)-M_2(e^t) ]. } $$ Its centered Laplace transform is obtained from $$ \frac{h(s)^2}{s} $$ after subtracting the complete $s=1$ principal part. Under RH, known estimates for $k=2$ von Mangoldt convolution give polynomial growth in $t$. Conversely, if $R_2$ is a tempered distribution supported on $[0,\infty)$, its Laplace transform is holomorphic for $\Re z>0$, which excludes every zero with $\Re\rho>1/2$. Functional symmetry then gives RH. Hence, with the standard distributional/Perron conventions, $$ \boxed{ \mathrm{RH} \iff R_2 \text{ is tempered}. } $$ --- # 7. Known $k=2$ RH estimate Banks–Sinha obtain, under RH, the twisted estimate $$ \boxed{ \sum_{n\le x} (\Lambda*\Lambda)(n)n^{-iy} = \frac{x^{1-iy}(\log x-2\gamma)}{1-iy} - \frac{x^{1-iy}}{(1-iy)^2} + O\!\left( x^{1/2} \{\log(x+|y|)\}^{5} \right). } $$ For $y=0$: $$ \boxed{ C_2(x) = x\log x - (1+2\gamma)x + O(x^{1/2}\log^5x). } $$ So the raw pair field is already a known RH-sensitive generalized-von-Mangoldt object. --- # 8. General $k$-fold multiplicative convolution Define $$ \boxed{ b_k = \underbrace{ \Lambda*\cdots*\Lambda }_{k\text{ copies}}. } $$ Then $$ \boxed{ \sum_{n\ge1} \frac{b_k(n)}{n^s} = h(s)^k. } $$ Let $$ \boxed{ C_k(x) = \sum_{n\le x}b_k(n). } $$ --- # 9. Equilibrium hierarchy At $s=1$, $$ h(s)^k $$ has a pole of order $k$. Thus its deterministic main term is $$ \boxed{ M_k(x) = xP_{k-1}(\log x), } $$ where $$ P_{k-1} $$ is a polynomial of degree $k-1$ determined by the Laurent expansion of $h$ at $1$. --- # 10. Zero hierarchy At a multiplicity-$m$ zero: $$ \boxed{ h(s)^k \sim \frac{ (-m)^k }{ (s-\rho)^k }. } $$ Therefore the leading zero contribution to the summatory field is $$ \boxed{ \frac{ (-m)^k }{ \rho\,(k-1)! } x^\rho (\log x)^{k-1}. } $$ After critical normalization: $$ \boxed{ x^{-1/2} [ C_k(x)-M_k(x) ] \sim x^{\beta-1/2} (\log x)^{k-1} e^{i\gamma\log x}. } $$ --- # 11. Critical exponent is $k$-independent For every fixed finite $k$, $$ \boxed{ \Theta_k = \sup_\rho\Re\rho. } $$ Thus $$ \boxed{ \text{tensor order} \uparrow \quad\Rightarrow\quad \text{pole order} \uparrow, } $$ but $$ \boxed{ \text{critical exponent unchanged}. } $$ This is the general Pole-Order Amplification / Criticality Conservation principle. --- # 12. Banks–Sinha hierarchy Banks–Sinha study the same $k$-fold convolution hierarchy and prove RH-conditional estimates of the shape $$ \boxed{ \sum_{n\le x} b_k(n)n^{-iy} = \text{main residue} + O\!\left( x^{1/2} \{\log(x+|y|)\}^{2k+1} \right) } $$ for fixed $k$. They also prove converses of RH strength under corresponding $x^{1/2+\varepsilon}$-type uniform estimates. Thus higher convolution order increases logarithmic amplification but does not change the RH critical exponent. --- # 13. Arithmetic complexity cost The arithmetic object $b_k$ is already a $k$-body multiplicative convolution. If one then studies a quadratic energy such as $$ \int |C_k-M_k|^2 w(x)\,dx, $$ the squared arithmetic expression couples two $k$-body fields. Thus the direct arithmetic proof burden can rise to $2k$-body correlations. This motivates $$ \boxed{ \textbf{Amplification–Correlation Complexity Tradeoff}. } $$ > Spectral pole amplification by higher multiplicative convolution is purchased by higher arithmetic correlation complexity. There is no free amplifier. --- # 14. Quadratic completeness Although $b_2$ is a two-body statistic, analytically it retains all one-body information. Since $$ h(s)^2 $$ is known and $$ h(s)>0 $$ for real $s>1$, the meromorphic square root is fixed uniquely by $$ \boxed{ h(s) = +\sqrt{h(s)^2} \quad (s>1). } $$ Analytic continuation recovers $h$. Then $$ h = -\frac{d}{ds}\log\zeta $$ and the normalization $$ \zeta(s)\to1 \qquad (\Re s\to+\infty) $$ recover $\zeta$. So the multiplicative pair lift is analytically information-complete. --- # 15. Quadratic Completeness Principle We define $$ \boxed{ \textbf{Quadratic Completeness}. } $$ > The multiplicative pair field does not create new zeta information; it is an invertible nonlinear lift of the prime logarithmic derivative once the real-half-plane branch is fixed. Its value is conditioning / pole-order amplification, not theorem strength. --- # 16. Pair transform as a polynomial lift of Game 12 Game 12 used $$ \boxed{ D(s) = \frac{h(s)}s - \frac1{s-1}. } $$ Hence $$ \boxed{ h(s) = sD(s) + \frac{s}{s-1}. } $$ Therefore $$ \boxed{ \frac{h(s)^2}{s} = sD(s)^2 + \frac{2s}{s-1}D(s) + \frac{s}{(s-1)^2}. } $$ So after equilibrium subtraction, the raw pair residual is exactly a linear-plus-quadratic functional of the Game 12 transfer function. No new spectral support is created. --- # 17. Spectral Support Conservation under Polynomial Lift More generally, any finite polynomial lift in $h$ changes pole orders and coefficients but not their locations, unless special cancellations are deliberately introduced. This motivates $$ \boxed{ \textbf{Spectral Support Conservation under Polynomial Lift}. } $$ The raw $h^k$ hierarchy is therefore a pole-order amplifier, not a new support generator. --- # 18. Selberg's second-order filter The classical Selberg symmetry formula uses the generalized von Mangoldt function $$ \boxed{ \Lambda_2(n) = \Lambda(n)\log n + (\Lambda*\Lambda)(n). } $$ Its Dirichlet series is $$ \boxed{ \frac{\zeta''(s)}{\zeta(s)} = h(s)^2-h'(s). } $$ The summatory formula is $$ \boxed{ \sum_{n\le x}\Lambda_2(n) = 2x\log x + O(x). } $$ This is the core second-order object in Selberg's elementary proof of the prime number theorem. --- # 19. Residue calculus for differential filters Suppose locally $$ \boxed{ h(s) \sim \frac{r}{s-s_0}. } $$ Then $$ \boxed{ h'(s) \sim -\frac{r}{(s-s_0)^2}. } $$ Therefore $$ \boxed{ h^2-h' \sim \frac{ r(r+1) }{ (s-s_0)^2 }, } $$ while $$ \boxed{ h^2+h' \sim \frac{ r(r-1) }{ (s-s_0)^2 }. } $$ These two signs implement opposite residue filters. --- # 20. Selberg pole selectivity For $h=-\zeta'/\zeta$: ### At the zeta pole $s=1$ $$ r=+1. $$ Therefore $$ \boxed{ r(r+1)=2. } $$ Selberg preserves and doubles the equilibrium double pole. ### At a simple zeta zero $$ r=-1. $$ Then $$ \boxed{ r(r+1)=0. } $$ The universal double-pole signature disappears. For multiplicity $m$: $$ r=-m, $$ so the remaining double-pole coefficient is $$ \boxed{ m(m-1). } $$ --- # 21. Selberg Pole-Cancellation Principle We define $$ \boxed{ \textbf{Selberg Pole-Cancellation Principle}. } $$ > The second-order differential combination underlying Selberg's symmetry formula selectively preserves the equilibrium pole but cancels the universal double-pole signature of simple zeros. This helps explain why the object is exceptionally effective for PNT-scale control while not being maximally sensitive to zero localization. --- # 22. Anti-Selberg sign Now reverse the derivative sign: $$ \boxed{ h^2+h'. } $$ At the equilibrium pole: $$ r=+1, $$ so $$ \boxed{ r(r-1)=0. } $$ At a simple zero: $$ r=-1, $$ so $$ \boxed{ r(r-1)=2. } $$ At a multiplicity-$m$ zero: $$ \boxed{ r(r-1) = m(m+1). } $$ Thus the second-order pole selectivity is exactly reversed. --- # 23. Reciprocal-zeta origin Let $$ \boxed{ F(s) = \frac1{\zeta(s)}. } $$ Then $$ \boxed{ \frac{F'}F = h. } $$ By the logarithmic-derivative identity, $$ \boxed{ \frac{F''}F = h^2+h'. } $$ So the Anti-Selberg combination is not ad hoc. It is the direct second derivative ratio of the reciprocal zeta function. --- # 24. Arithmetic coefficients of the reciprocal filter Since $$ \frac1{\zeta(s)} = \sum_{n\ge1} \frac{\mu(n)}{n^s}, $$ we have $$ \boxed{ \left( \frac1{\zeta} \right)''(s) = \sum_{n\ge1} \frac{ \mu(n)(\log n)^2 }{ n^s }. } $$ Multiplying by $\zeta(s)$ gives $$ \boxed{ \frac{(1/\zeta)''}{1/\zeta} = \sum_{n\ge1} \frac{ \omega_2(n) }{ n^s}, } $$ where $$ \boxed{ \omega_2(n) = \sum_{d\mid n} \mu(d)(\log d)^2. } $$ Also $$ \boxed{ \omega_2 = \Lambda*\Lambda - \Lambda\log. } $$ --- # 25. Support of $\omega_2$ If $$ n=p^a, $$ then $$ \boxed{ \omega_2(n) = -(\log p)^2. } $$ If $$ n=p^aq^b, \qquad p\ne q, $$ then $$ \boxed{ \omega_2(n) = 2\log p\log q. } $$ If $n$ has at least three distinct prime factors, $$ \boxed{ \omega_2(n)=0. } $$ Thus the reciprocal second-order filter is supported only on prime powers and integers with exactly two distinct prime factors. --- # 26. Residual simple pole at $s=1$ Although the double pole at $s=1$ is canceled in $$ F''/F, $$ a simple pole remains. Use the Laurent/Taylor expansion $$ \frac1{\zeta(s)} = (s-1) - \gamma(s-1)^2 + O((s-1)^3). $$ Therefore $$ \boxed{ \frac{F''}F = -\frac{2\gamma}{s-1} + O(1). } $$ --- # 27. Full equilibrium annihilation Since $$ \frac{F'}F=h = \frac1{s-1} -\gamma + O(s-1), $$ define $$ \boxed{ \mathcal J_2(s) = \frac{F''(s)}{F(s)} + 2\gamma \frac{F'(s)}{F(s)}. } $$ Equivalently, $$ \boxed{ \mathcal J_2 = h^2+h'+2\gamma h. } $$ Then $$ \boxed{ \mathcal J_2 } $$ is holomorphic at $$ s=1. $$ --- # 28. Zero poles survive the normalization At a multiplicity-$m$ zero $\rho$ of $\zeta$, $F=1/\zeta$ has a pole of order $m$. Thus $$ \boxed{ \frac{F''}F \sim \frac{ m(m+1) }{ (s-\rho)^2 }. } $$ The added term $$ 2\gamma F'/F $$ has only a simple pole and does not affect the leading double pole. Therefore $$ \boxed{ \mathcal J_2(s) \sim \frac{ m(m+1) }{ (s-\rho)^2 } + O\!\left( \frac1{s-\rho} \right). } $$ --- # 29. Zero-selective arithmetic coefficient The Dirichlet coefficient of $\mathcal J_2$ is $$ \boxed{ j_2(n) = \omega_2(n) + 2\gamma\Lambda(n). } $$ Explicitly: ### prime powers $$ \boxed{ j_2(p^a) = -(\log p)^2 + 2\gamma\log p; } $$ ### two distinct prime factors $$ \boxed{ j_2(p^aq^b) = 2\log p\log q; } $$ ### at least three distinct prime factors $$ \boxed{ j_2(n)=0. } $$ This is a signed almost-prime signal. --- # 30. Zero-Selective Almost-Prime Signal We call $$ \boxed{ j_2 } $$ the: $$ \boxed{ \textbf{Zero-Selective Almost-Prime Signal} } $$ in this exploratory terminology. It has three notable features: 1. no $s=1$ pole; 2. a double pole at every simple nontrivial zero; 3. arithmetic support on prime powers and two-prime-factor integers. --- # 31. RH criterion from $j_2$ Define $$ \boxed{ J_2(x) = \sum_{n\le x} j_2(n). } $$ If RH holds, known critical estimates for $\Lambda*\Lambda$, together with the von Koch bound and partial summation for $\Lambda(n)\log n$, give a bound of the form $$ \boxed{ J_2(x) = O( x^{1/2}\log^A x ) } $$ for some fixed $A$; the $k=2$ generalized-von-Mangoldt estimate permits $A=5$ up to harmless lower-order contributions. Conversely, if $$ \boxed{ J_2(x) = O_\varepsilon( x^{1/2+\varepsilon} ) \quad \forall\varepsilon>0, } $$ then Abel/Perron continuation makes the Dirichlet series $\mathcal J_2(s)$ holomorphic in every half-plane $$ \Re s>\frac12+\varepsilon. $$ A zero with $\Re\rho>1/2$ would produce a double pole there, contradiction. Functional symmetry gives RH. Thus, with the standard technical conventions, $$ \boxed{ \mathrm{RH} \iff J_2(x) = O_\varepsilon( x^{1/2+\varepsilon} ) \quad \forall\varepsilon>0. } $$ This is a derived criterion; no novelty claim is made. --- # 32. Direct–Reciprocal Selberg Duality The $k=2$ picture generalizes. For a local model $$ G(s) \sim (s-s_0)^r, $$ we have $$ \boxed{ \frac{G^{(k)}}G \sim r^{\underline{k}} (s-s_0)^{-k}, } $$ where $$ r^{\underline{k}} = r(r-1)\cdots(r-k+1). $$ --- # 33. Direct hierarchy: $G=\zeta$ At $s=1$, $\zeta$ has exponent $$ r=-1. $$ Hence $$ \boxed{ (-1)^{\underline{k}} = (-1)^kk! \ne0. } $$ The equilibrium pole is preserved at full order. At a simple zero, $$ r=+1, $$ and for $$ k\ge2, $$ $$ \boxed{ 1^{\underline{k}}=0. } $$ So the highest-order simple-zero pole is suppressed. This is the structural pole geometry of the generalized Selberg hierarchy $$ \boxed{ \zeta^{(k)}/\zeta. } $$ --- # 34. Reciprocal hierarchy: $G=1/\zeta$ At $s=1$, $1/\zeta$ has exponent $$ r=+1. $$ Thus for $k\ge2$, $$ \boxed{ 1^{\underline{k}}=0. } $$ The equilibrium high-order pole is suppressed. At a multiplicity-$m$ zeta zero, $1/\zeta$ has exponent $$ r=-m. $$ Hence $$ \boxed{ (-m)^{\underline{k}} = (-1)^k m^{\overline{k}}, } $$ where $$ m^{\overline{k}} = m(m+1)\cdots(m+k-1). $$ Thus the zero pole is amplified at full order. --- # 35. Direct–Reciprocal Selberg Duality Principle We define: $$ \boxed{ \textbf{Direct–Reciprocal Selberg Duality} } $$ > The generalized derivative ratios of $\zeta$ are naturally equilibrium-selective and suppress high-order simple-zero poles, whereas the generalized derivative ratios of $1/\zeta$ reverse this selectivity: they suppress the high-order equilibrium pole and amplify zeta-zero poles. This is a structural duality. The underlying derivative-ratio identities are classical. --- # 36. General reciprocal arithmetic hierarchy For $$ F=1/\zeta, $$ define $$ \boxed{ \frac{F^{(k)}}F = \sum_{n\ge1} \frac{ \omega_k(n) }{ n^s}. } $$ Then $$ \boxed{ \omega_k(n) = (-1)^k \sum_{d\mid n} \mu(d)(\log d)^k. } $$ This may be expressed as a finite binomial combination of the standard generalized von Mangoldt functions. Also, $$ \boxed{ \omega_k(n)=0 } $$ if $n$ has more than $k$ distinct prime factors. Thus the reciprocal hierarchy is an almost-prime hierarchy. --- # 37. Full $s=1$ annihilation for general $k$ Write $$ \boxed{ F(s) = a_1(s-1) + a_2(s-1)^2 +\cdots, \qquad a_1\ne0. } $$ For $k\ge2$, $$ F^{(k)}/F $$ has at most a simple pole at $s=1$. Its residue is $$ \boxed{ c_k = \frac{k!a_k}{a_1}. } $$ Define $$ \boxed{ \mathcal J_k(s) = \frac{F^{(k)}(s)}{F(s)} - c_k \frac{F'(s)}{F(s)}. } $$ Then $$ \boxed{ \mathcal J_k } $$ is holomorphic at $s=1$. At a multiplicity-$m$ nontrivial zero, $$ \boxed{ \mathcal J_k(s) \sim (-1)^k \frac{ m^{\overline{k}} }{ (s-\rho)^k } +\text{lower pole orders}. } $$ This is a background-annihilated zero-pole amplifier hierarchy. --- # 38. Background-annihilated reciprocal hierarchy The general construction $$ \boxed{ \mathcal J_k } $$ has: - no equilibrium pole at $s=1$; - $k$th-order poles at nontrivial zeta zeros; - arithmetic coefficients supported on integers with at most $k$ distinct prime factors, plus the first-order correction used to annihilate the equilibrium pole. This is potentially a useful conditioning hierarchy. It does not change the RH critical exponent. --- # 39. Positivity–Background–Zero-Selectivity Triangle The second-order case displays a no-free-lunch geometry. ## Raw amplifier $$ \boxed{ h^2 } $$ has: - nonnegative coefficients $\Lambda*\Lambda$; - equilibrium double pole; - zero double poles. ## Selberg filter $$ \boxed{ h^2-h' } $$ has: - nonnegative generalized von Mangoldt coefficients; - strong equilibrium double pole; - simple-zero double pole canceled. ## Reciprocal zero-selective filter $$ \boxed{ h^2+h'+2\gamma h } $$ has: - equilibrium pole removed; - simple-zero double pole preserved and amplified; - signed arithmetic coefficients. --- # 40. Positivity–Background–Zero-Selectivity Principle We define: $$ \boxed{ \textbf{Positivity–Background–Zero-Selectivity Triangle} } $$ > In the natural second-order arithmetic filters considered here, maximizing zero selectivity while annihilating the deterministic equilibrium background forces one to abandon coefficient positivity and expose a signed global cancellation problem. Selberg chooses arithmetic positivity and tractability. The reciprocal filter chooses zero selectivity and background annihilation. Neither choice solves RH for free. --- # 41. Why the reciprocal filter is not automatically easier The summatory function $$ J_2(x) $$ has no large $x\log x$ or $x$ equilibrium term. This improves conditioning. But its coefficients are signed. So proving critical cancellation becomes a direct cancellation problem between: - negatively weighted prime-power states; - positively weighted two-prime-factor states; - the $2\gamma\Lambda$ counterterm. The difficulty has moved from background subtraction into arithmetic sign cancellation. --- # 42. Prime-versus-semiprime critical balance Explicitly, $$ \boxed{ J_2(x) } $$ is the weighted balance $$ \boxed{ 2 \sum_{\substack{ p^aq^b\le x\\ p Why does this zero-selective signed almost-prime signal never organize itself into a supercritical Mellin mode? That is the current fixed point. --- # References 1. Tao, T. *A Banach algebra proof of the prime number theorem* (2014), including the Selberg symmetry formula $$ \Lambda_2 = \Lambda\log + \Lambda*\Lambda. $$ 2. Banks, W. D., & Sinha, S. Work on the Riemann hypothesis via generalized von Mangoldt functions; Sinha's 2024 lecture notes give the $k$-fold convolution and generalized-von-Mangoldt twisted criteria. 3. Classical identity $$ \sum_n \Lambda^{*k}(n)n^{-s} = \left( -\frac{\zeta'}{\zeta}(s) \right)^k. $$ 4. Classical generalized von Mangoldt identity $$ \sum_n \Lambda_k(n)n^{-s} = (-1)^k \frac{\zeta^{(k)}(s)}{\zeta(s)}. $$ 5. Classical reciprocal-zeta Dirichlet series $$ \frac1{\zeta(s)} = \sum_n\mu(n)n^{-s} $$ and termwise differentiation in $\Re s>1$. 6. Standard Perron / Mellin residue calculus. 7. Standard Stieltjes expansion of $\zeta(s)$ at $s=1$.