> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification. # Reverse Goldbach Structural Closure Program 03 ## M6/Wheel Obstruction Geometry and Counterexample Coverage Cost ### Reverse Goldbach Structural Closure Program 03 ### M6/Wheel Obstruction Geometry and Counterexample Coverage Cost **Series name:** Reverse Goldbach Structural Closure Program **English name:** Reverse Goldbach Structural Closure Program **Abbreviation:** RG-SCP **Article:** 03 / planned 08–10 **Authors:** Neo.K; Aletheia (GPT-5.6 Sol) **Date:** 2026-09-28 **Version:** v0.1 **Nature:** Structural analysis draft / obstruction geometry draft / counterexample cost model **Status:** Not a proof of the Goldbach conjecture; this article studies the structural coverage cost that a hypothetical Goldbach failure must pay under M6 and primorial wheels --- ## Abstract The second article of the Reverse Goldbach Structural Closure Program established: $$ F_G(N)=1 $$ if and only if there exists a total proper obstruction map: $$ \Phi_N: \mathbb P_N^{-} \to \mathbb P $$ such that all reduced prime inputs: $$ p\in\mathbb P_N^{-} $$ satisfy: $$ \Phi_N(p)\mid(N-p), $$ and: $$ \Phi_N(p)\le\sqrt{N-p}. $$ If the canonical least-factor obstruction is chosen: $$ \Phi_N^{\min}(p) = \operatorname{lpf}(N-p), $$ then a hypothetical Goldbach counterexample $N$ induces a complete prime-indexed obstruction partition. This article further studies: > Do the residue structure and small-prime-divisor structure of different $N$ change the cost of this obstruction partition? The first layer starts with: $$ M_6 $$ as displayed above. For: $$ N\bmod6\in\{0,2,4\}, $$ prime inputs: $$ p\bmod6\in\{1,5\} $$ form different easy/hard geometries. where: - The easy branch can be directly obstructed by $3$; - The complement of the hard branch already avoids $2$ and $3$, so if it is still composite, its least prime factor is at least $5$. This article then generalizes this concept to the primorial wheel: $$ M_k = \prod_{i=1}^{k}p_i. $$ Define the wheel-hard set: $$ \mathbb H_{M_k}(N) = \left\{ p\in\mathbb P_N^{-}: \gcd(N-p,M_k)=1 \right\}. $$ If $N$ is a Goldbach failure, then: $$ p\in\mathbb H_{M_k}(N) \Longrightarrow \Phi_N^{\min}(p)>p_k. $$ Another central result of this article is **divisor self-suppression**: If: $$ r\mid N, $$ then $r$ almost loses its ability to serve as an obstruction color, and can only obstruct: $$ p=r. $$ Thus, if a hypothetical counterexample itself has many small prime factors, it simultaneously: 1. gains residue structure in certain quotients; 2. yet loses the corresponding small-prime obstruction channels; 3. pushes the remaining obstruction burden toward larger prime factors. On this basis, this article defines: - hard-branch ratio; - wheel residual pressure; - obstruction depth; - suppressed-prime cost; - effective obstruction alphabet; - normalized coverage cost; - prime-weighted obstruction cost; - structural compression ratio. The central question advances from: $$ \text{Can coverage be achieved?} $$ to: $$ \boxed{ \text{Can a hypothetical counterexample achieve coverage at low cost, with finite sources, and consistently across scales?} } $$ --- # 1. Review of the Previous Article’s Results Let: $$ \mathbb P_N^{-} = \left\{ p\in\mathbb P: 2\le p\le\frac N2 \right\}. $$ If: $$ F_G(N)=1, $$ then for each: $$ p\in\mathbb P_N^{-} $$ we have: $$ N-p $$ is composite. Thus, the canonical least-factor obstruction map is: $$ \Phi_N^{\min}(p) = \operatorname{lpf}(N-p). $$ and satisfies: $$ \Phi_N^{\min}(p) \le \sqrt{N-p}. $$ Thus, the complete data of a hypothetical failure state can be written as: $$ \mathfrak O_N = \left( N, \mathbb P_N^{-}, \Phi_N^{\min}, \mathcal L_N, D_G(N), \Theta_N, \mathcal D(N), \Pi_N \right). $$ This article focuses on: $$ \Phi_N^{\min} $$ how it is constrained by modular geometry and divisor structure. --- # 2. Why Start with M6 for every: $$ p>3 $$ prime, we have: $$ p\equiv\pm1\pmod6. $$ Therefore: $$ M_6 $$ is not a complete description of primality, but is the first natural layer that compresses: - parity exclusion; - divisibility-by-$3$ exclusion; - prime candidate branching; into a very small quotient. For Goldbach failure, this quotient is not merely a candidate filter, but directly determines: $$ N-p $$ whether it can be obstructed by $3$. Thus: $$ M_6 $$ is the first nontrivial layer of obstruction geometry. --- # 3. M6 residue state of $N$ Because: $$ N $$ is even, so: $$ N\bmod6 \in \{0,2,4\}. $$ whereas for: $$ p>3, $$ Only: $$ p\bmod6 \in \{1,5\}. $$ Thus, we only need to analyze: $$ 3\times2 $$ residue interactions. --- # 4. M6 complement table Consider: $$ N-p\bmod6. $$ This gives: | $N\bmod6$ | $p\bmod6=1$ | $p\bmod6=5$ | |---|---:|---:| | $0$ | $5$ | $1$ | | $2$ | $1$ | $3$ | | $4$ | $3$ | $5$ | If the complement is: $$ 3\pmod6, $$ and greater than $3$, it must be divisible by $3$ and is therefore composite. If the complement is: $$ 1 \quad\text{or}\quad 5 \pmod6, $$ then it avoids $2$ and $3$. Thus, one can define: - **easy branch**: obstruction is provided directly by $3$; - **hard branch**: a prime factor of at least $5$ must be used. --- # 5. M6 hard branch Define: $$ \mathbb H_6(N) = \left\{ p\in\mathbb P_N^{-}: p>3, \quad \gcd(N-p,6)=1 \right\}. $$ If: $$ F_G(N)=1, $$ Then: $$ p\in\mathbb H_6(N) \Longrightarrow \Phi_N^{\min}(p)\ge5. $$ This is not an approximation, but an exact consequence. --- # 6. M6 easy branch Define: $$ \mathbb E_6(N) = \left\{ p\in\mathbb P_N^{-}: p>3, \quad 3\mid(N-p) \right\}. $$ In a failure state, if: $$ N-p>3, $$ Then: $$ \Phi_N^{\min}(p)=3. $$ Thus: $$ \mathbb E_6(N) $$ is a low-cost obstruction sector. --- # 7. Three M6 Geometries ## 7.1 $N\equiv0\pmod6$ for every: $$ p>3, $$ we have: $$ N-p\equiv\pm1\pmod6. $$ Therefore: $$ \mathbb E_6(N)=\varnothing. $$ whereas: $$ \mathbb H_6(N) = \mathbb P_N^{-}\setminus\{2,3\} $$ ignoring very small boundary cases. Thus, a hypothetical failure with $N\equiv0\pmod6$ must, for almost the entire prime domain, use: $$ r\ge5 $$ obstructions. --- ## 7.2 $N\equiv2\pmod6$ For: $$ p\equiv5\pmod6 $$ prime inputs, $$ 3\mid(N-p). $$ Thus, this branch forms the easy branch. For: $$ p\equiv1\pmod6, $$ Then: $$ \gcd(N-p,6)=1. $$ Thus, it forms the hard branch. --- ## 7.3 $N\equiv4\pmod6$ The situation is reversed. For: $$ p\equiv1\pmod6 $$ forms the easy branch, whereas: $$ p\equiv5\pmod6 $$ forms the hard branch. --- # 8. M6 hard-branch ratio Define: $$ \eta_6(N) = \frac{ |\mathbb H_6(N)| }{ |\mathbb P_N^{-}| }. $$ At large scales, ignoring small-prime boundaries, intuitively: - If $N\equiv0\pmod6$, $\eta_6(N)$ is close to $1$; - If $N\equiv2$ or $4\pmod6$, $\eta_6(N)$ corresponds approximately to one of the two prime residue branches. This article does not, however, treat this density intuition as a theorem. What is genuinely provable is the branch membership rule, not its exact asymptotic proportion. --- # 9. The First Coverage-Cost Ordering Thus, a structural ordering can be proposed at the M6 layer: $$ N\equiv0\pmod6 $$ a hypothetical failure of this form puts more pressure on small-factor obstruction than: $$ N\equiv2 \quad\text{or}\quad 4 \pmod6 $$ a failure of this form. It must, however, be stated precisely: > Here, “greater pressure” means only lower availability of the $r=3$ obstruction channel; it does not mean that we have proved such $N$ are less likely to be Goldbach counterexamples. Thus, we define only: $$ \boxed{ \text{relative obstruction burden} } $$ and do not define: $$ \text{probability of being counterexample}. $$ --- # 10. From M6 to General Primorial Wheels Let: $$ p_1=2, \quad p_2=3, \quad p_3=5, \ldots $$ and define: $$ M_k = \prod_{i=1}^{k}p_i. $$ For example: $$ M_1=2, $$ $$ M_2=6, $$ $$ M_3=30, $$ $$ M_4=210, $$ $$ M_5=2310. $$ At each layer, examine whether the complement passes the small-prime divisibility screen. --- # 11. Wheel-covered set Define: $$ \mathbb E_{M_k}(N) = \left\{ p\in\mathbb P_N^{-}: \gcd(N-p,M_k)>1 \right\}. $$ Every complement in this set is obstructed by at least: $$ p_1,\ldots,p_k $$ one of these small primes. --- # 12. Wheel-hard set Define: $$ \mathbb H_{M_k}(N) = \left\{ p\in\mathbb P_N^{-}: \gcd(N-p,M_k)=1 \right\}. $$ Then: $$ \mathbb P_N^{-} = \mathbb E_{M_k}(N) \bigsqcup \mathbb H_{M_k}(N). $$ If: $$ F_G(N)=1, $$ then for: $$ p\in\mathbb H_{M_k}(N) $$ we have: $$ \boxed{ \Phi_N^{\min}(p)>p_k. } $$ --- # 13. Hard-branch nesting Because: $$ M_k\mid M_{k+1}, $$ Thus: $$ \gcd(N-p,M_{k+1})=1 $$ necessarily implies: $$ \gcd(N-p,M_k)=1. $$ Therefore: $$ \boxed{ \mathbb H_{M_{k+1}}(N) \subseteq \mathbb H_{M_k}(N). } $$ This provides a natural residual filtration: $$ \mathbb H_2(N) \supseteq \mathbb H_6(N) \supseteq \mathbb H_{30}(N) \supseteq \mathbb H_{210}(N) \supseteq \cdots. $$ --- # 14. Wheel residual ratio Define: $$ \eta_{M_k}(N) = \frac{ |\mathbb H_{M_k}(N)| }{ |\mathbb P_N^{-}| }. $$ we have: $$ \eta_{M_{k+1}}(N) \le \eta_{M_k}(N). $$ This quantity describes: > After the first $k$ primes have served as obstruction channels, how many prime inputs still have complements that must rely on larger prime factors to be obstructed. --- # 15. Wheel residual pressure Knowing only the survivor count is not enough. If: $$ p\in\mathbb H_{M_k}(N), $$ Then: $$ \Phi_N^{\min}(p)\ge p_{k+1}. $$ Thus, define the simplest lower-bound pressure: $$ \Pi_{M_k}(N) = |\mathbb H_{M_k}(N)| \cdot p_{k+1}. $$ It is not an exact factor cost, but supplies a minimum-burden proxy: $$ \boxed{ \text{survivor count} \times \text{minimum next obstruction scale}. } $$ --- # 16. Prime-weighted obstruction cost More directly, one can define, for the canonical obstruction map: $$ C_{\mathrm{obs}}(N) = \sum_{p\in\mathbb P_N^{-}} \log \Phi_N^{\min}(p). $$ One can also define: $$ C_{\mathrm{obs}}^{(\alpha)}(N) = \sum_{p\in\mathbb P_N^{-}} \left( \Phi_N^{\min}(p) \right)^\alpha, $$ where: $$ \alpha>0. $$ The first emphasizes the multiplicative information scale; the second emphasizes penalizing large obstruction factors. --- # 17. Normalized obstruction cost To compare $N$ of different sizes, define: $$ \overline C_{\mathrm{obs}}(N) = \frac{ 1 }{ |\mathbb P_N^{-}| } \sum_{p\in\mathbb P_N^{-}} \log \Phi_N^{\min}(p). $$ If: $$ F_G(N)=1, $$ then this is the average least-factor obstruction information cost that each prime input must pay. --- # 18. Reinterpreting Obstruction Depth The previous article defined: $$ D_G(N) = \max_{p\in\mathbb P_N^{-}} \Phi_N^{\min}(p). $$ Now one can regard: $$ D_G(N) $$ as the maximum obstruction scale of a hypothetical failure state. If: $$ D_G(N) $$ is large, this means at least one prime complement must reach a very deep factor scale before it can be proved composite. If: $$ D_G(N) $$ is small, this means all complements are captured by a shallow factor system. --- # 19. Obstruction alphabet Define the canonical obstruction alphabet: $$ \Sigma_N = \left\{ r\in\mathbb P: \exists p\in\mathbb P_N^{-}, \quad \Phi_N^{\min}(p)=r \right\}. $$ Its size: $$ A_G(N) = |\Sigma_N| $$ is called the obstruction alphabet size. This describes: > How many different least-prime-factor types a hypothetical failure state actually uses to achieve coverage. --- # 20. Effective obstruction alphabet Not every small prime is available for a given $N$. If: $$ r\mid N, $$ then by self-suppression: $$ r\mid(N-p) \Longrightarrow p=r. $$ Thus, $r$ almost loses its coverage ability for other prime inputs. Define the available obstruction alphabet up to depth $y$: $$ \Sigma_{\mathrm{avail}}(N;y) = \left\{ r\in\mathbb P: r\le y, \quad r\nmid N \right\}. $$ If the self-cell must be retained: $$ p=r, $$ it can be handled separately as a singleton exception. --- # 21. Divisor self-suppression Let: $$ \mathcal D(N) = \left\{ r\in\mathbb P: r\mid N \right\}. $$ then for: $$ p\notin\mathcal D(N), $$ we have: $$ \Phi_N^{\min}(p)\notin\mathcal D(N). $$ Thus, the canonical coloring of a hypothetical failure must avoid: $$ \mathcal D(N) $$ the colors in this set, except for a possible self-index cell. This forms: $$ \boxed{ \text{forbidden obstruction colors induced by }N. } $$ --- # 22. Suppression count Define the suppression count up to scale $y$: $$ S_N(y) = \left| \mathcal D(N) \cap \{r\in\mathbb P:r\le y\} \right|. $$ along with the normalized suppression ratio: $$ \sigma_N(y) = \frac{ S_N(y) }{ \pi(y) }. $$ where: $$ \pi(y) $$ is the number of primes not exceeding $y$. If: $$ \sigma_N(y) $$ is high, this means $N$ itself absorbs many small primes, preventing those primes from being widely used as obstruction channels. --- # 23. Smoothness paradox This produces a phenomenon worth further study: If $N$ has many small prime factors, intuitively its structure in modular space is more regular. At the same time, however, these small prime factors lose obstruction power through self-suppression. Thus, for hypothetical Goldbach failure, small-prime smoothness may produce a tension: $$ \boxed{ \text{more modular structure} \quad\Longleftrightarrow\quad \text{fewer reusable small obstruction colors}. } $$ This is not a contradiction theorem. It is a structural pressure that needs to be quantified. --- # 24. Suppressed Wheel For the primorial: $$ M_k, $$ Define: $$ M_k^{\mathrm{eff}}(N) = \prod_{\substack{ 1\le i\le k\\ p_i\nmid N }} p_i. $$ This is not a traditional wheel, but an effective wheel for obstruction availability. Because: $$ p_i\mid N $$ when this holds, $$ p_i $$ is suppressed for complement obstruction on most prime inputs. Thus, one can subsequently compare: $$ M_k $$ and: $$ M_k^{\mathrm{eff}}(N). $$ --- # 25. Effective hard branch Define: $$ \mathbb H_{M_k}^{\mathrm{eff}}(N) = \left\{ p\in\mathbb P_N^{-}: \gcd(N-p,M_k^{\mathrm{eff}}(N))=1 \right\}. $$ This set represents: > The prime inputs still not captured by the currently available small obstruction colors after excluding the small primes suppressed by $N$ itself. Compared with the traditional: $$ \mathbb H_{M_k}(N) $$ it is closer to the counterexample’s actual obstruction burden. --- # 26. Effective residual ratio Define: $$ \eta_{M_k}^{\mathrm{eff}}(N) = \frac{ |\mathbb H_{M_k}^{\mathrm{eff}}(N)| }{ |\mathbb P_N^{-}| }. $$ Because: $$ M_k^{\mathrm{eff}}(N) $$ removes suppressed primes, we generally have: $$ \mathbb H_{M_k}(N) \subseteq \mathbb H_{M_k}^{\mathrm{eff}}(N). $$ Therefore: $$ \eta_{M_k}(N) \le \eta_{M_k}^{\mathrm{eff}}(N). $$ This means: > A traditional wheel may overestimate the actual coverage ability of certain small primes for hypothetical failure. --- # 27. Obstruction availability matrix One can define the matrix: $$ A_{p,r}^{(N)} = \begin{cases} 1, & r\mid(N-p),\\ 0, & \text{otherwise}. \end{cases} $$ where the rows are: $$ p\in\mathbb P_N^{-}, $$ and the columns are: $$ r\in\mathcal R_N. $$ Goldbach failure is equivalent to: > Every row has at least one proper-factor column equal to $1$. That is: $$ \forall p, \quad \sum_r A_{p,r}^{(N)} \ge1. $$ --- # 28. Canonical coloring from the matrix If the least prime factor is used, each row selects its leftmost available obstruction: $$ \Phi_N^{\min}(p) = \min \left\{ r: A_{p,r}^{(N)}=1 \right\}. $$ Thus: $$ \Phi_N^{\min} $$ is in fact: $$ A^{(N)} $$ the canonical row selector of this matrix. This allows the obstruction cover to be turned into a matrix / graph-theoretic problem. --- # 29. Bipartite obstruction graph Define the bipartite graph: $$ G_N^{\mathrm{obs}} = ( P_N, R_N, E_N ), $$ where: $$ P_N = \mathbb P_N^{-}, $$ $$ R_N = \mathcal R_N, $$ and: $$ (p,r)\in E_N \Longleftrightarrow r\mid(N-p). $$ Goldbach failure is equivalent to: $$ \deg(p)\ge1 $$ for every: $$ p\in P_N $$ holds. --- # 30. Coverage as left-totality Therefore: $$ F_G(N)=1 $$ can be rewritten as: $$ \boxed{ G_N^{\mathrm{obs}} \text{ is total on the left prime vertex set.} } $$ That is, there is no isolated prime vertex. Conversely, if one finds: $$ p\in P_N $$ satisfying: $$ \deg(p)=0, $$ Then: $$ N-p $$ has no proper prime factor, Thus: $$ N-p $$ must be prime, immediately yielding: $$ N=p+(N-p). $$ --- # 31. Graph deficiency Define: $$ \delta_N = \left| \left\{ p\in P_N: \deg(p)=0 \right\} \right|. $$ Then: $$ R_G(N)>0 \Longleftrightarrow \delta_N>0. $$ whereas: $$ F_G(N)=1 \Longleftrightarrow \delta_N=0. $$ Thus, the Goldbach representation count and obstruction graph deficiency have a direct dual relationship. --- # 32. Truncated obstruction graph Retaining only: $$ r\le p_k $$ obstruction primes gives: $$ G_{N,k}^{\mathrm{obs}}. $$ Its isolated left vertices correspond exactly to: $$ \mathbb H_{M_k}(N) $$ the elements of this set after appropriate boundary handling. Thus, the wheel-hard branch can be reinterpreted as: $$ \boxed{ \text{isolated left vertices of a truncated obstruction graph}. } $$ --- # 33. Graph-depth filtration we have: $$ G_{N,1}^{\mathrm{obs}} \subseteq G_{N,2}^{\mathrm{obs}} \subseteq \cdots $$ As $k$ increases, more obstruction-prime columns are added. The isolated left set shrinks monotonically: $$ I_{N,1} \supseteq I_{N,2} \supseteq \cdots. $$ This agrees exactly with: $$ \mathbb H_{M_k}(N) $$ the nesting of this set. --- # 34. Coverage-cost curve Thus, one can define a discrete curve: $$ k \mapsto \eta_{M_k}(N). $$ A prime cutoff can also be used: $$ y \mapsto \eta_y(N). $$ For hypothetical failure, one must eventually have: $$ \eta_y(N)=0 $$ when: $$ y\ge D_G(N). $$ What is truly worth comparing is: > At what scale, and how quickly, does $\eta_y(N)$ fall to $0$? --- # 35. Structural compression ratio If new prime factors are merely added one by one, source complexity grows with: $$ \pi(y) $$ this quantity. Define a conceptual compression ratio: $$ \kappa_N(y) = \frac{ -\log \left( \max\{ \eta_y(N),\varepsilon \} \right) }{ \operatorname{SourceComplexity}(y) } $$ where: $$ \varepsilon>0 $$ merely prevents logarithmic divergence in numerical experiments. This quantity is not a theorem metric, but studies: > How much the hard residual is compressed for each additional unit of source complexity. If: $$ \kappa_N(y) $$ approaches extremely low values over the long term, this indicates poor proof efficiency of wheel refinement. --- # 36. Source Cost and Target Cost It is necessary to distinguish: $$ C_{\mathrm{source}} $$ and: $$ C_{\mathrm{target}}. $$ The former comprises: - modulus size; - number of prime divisors tested; - graph columns; - certificate size. The latter comprises: - hard branch size; - ambiguity; - residual quantifier burden. An ideal new method is not merely: $$ C_{\mathrm{source}} \uparrow $$ in exchange for: $$ C_{\mathrm{target}} \downarrow. $$ but seeks: $$ \boxed{ \text{small source increase} \to \text{large target collapse}. } $$ --- # 37. Exhaustive-refinement warning If coverage ultimately requires: $$ p_1,p_2,\ldots,p_{\pi(\sqrt N)} $$ all of these to enter, then the method is essentially close to: $$ \text{factor exhaustion}. $$ Although this is effective for a fixed $N$, it does not constitute a new Goldbach closure mechanism. Thus, this series must distinguish: $$ \boxed{ \text{verification completeness} } $$ and: $$ \boxed{ \text{structural compression}. } $$ --- # 38. The Actual Role of M6 M6 should not be exaggerated into: $$ \text{Goldbach proof engine}. $$ Its actual value is: 1. the first nontrivial quotient; 2. the first easy/hard obstruction splitter; 3. the first fixed-point interface; 4. the first layer that clearly exposes the interaction between self-suppression and residue burden; 5. a minimal model for subsequent wheel geometry. Thus: $$ \boxed{ M_6 = \text{minimal obstruction geometry laboratory}. } $$ --- # 39. Connection with M6 Fixed Points If the earlier fixed-point representation: $$ \operatorname{Fix}(T_6) $$ faithfully represents primes on the target domain, then one can define the fixed-point hard branch: $$ \mathbb H_6^{\mathrm{fix}}(N) = \left\{ p\in\operatorname{Fix}(T_6): p\le\frac N2, \quad \gcd(N-p,6)=1 \right\}. $$ Goldbach failure requires: $$ \forall p\in \mathbb H_6^{\mathrm{fix}}(N), $$ there exists: $$ r\ge5 $$ such that: $$ r\mid(N-p). $$ This genuinely connects the fixed-point fiber and obstruction graph. --- # 40. Fixed-point coverage burden Define: $$ B_6^{\mathrm{fix}}(N) = \left| \mathbb H_6^{\mathrm{fix}}(N) \right|. $$ It represents: > How many prime states hypothetical failure must additionally handle with prime factors of at least $5$ in the most basic M6 fixed-point sector. One can subsequently also define: $$ B_{M_k}^{\mathrm{fix}}(N). $$ --- # 41. Counterexample cost vector A single scalar cannot completely describe the obstruction burden. Thus, define: $$ \mathbf C_G(N) = \left( \eta_6(N), \eta_{30}(N), \eta_{210}(N), D_G(N), A_G(N), \overline C_{\mathrm{obs}}(N), \sigma_N(y), B_6^{\mathrm{fix}}(N) \right). $$ This is the multi-axis cost signature of a hypothetical counterexample. --- # 42. Why Not Directly Sum Them into a Total Score For now, this series does not define: $$ \operatorname{TotalScore}(N). $$ because the cost components have different mathematical meanings: - hard ratio is cardinality; - depth is factor scale; - alphabet size is diversity; - suppression is available-channel loss; - fixed-point burden is structural subdomain size. Arbitrary weighting would create an artificial metric. Thus, the vector structure is retained first; whether a natural invariant exists will be studied later. --- # 43. Additional Pressure from the Minimal Counterexample If: $$ N_* = \min\mathcal C_G, $$ then, beyond obstruction cost, there is: $$ \forall M There may be a constrained trade-off among the residue structure, small-prime-divisor structure, wheel-hard residual, and obstruction depth of a hypothetical Goldbach counterexample. Formally, one can seek an inequality of the form: $$ \mathcal P_G \left( \eta_{M_k}(N), \sigma_N(p_k), D_G(N), A_G(N) \right) \ge0. $$ At present: $$ \mathcal P_G $$ has not yet been discovered. Thus, this is a research objective, not a theorem. --- # 48. Self-Suppression Pressure Hypothesis Another, more specific direction is: If: $$ S_N(y) $$ is large, meaning that $N$ has many small prime factors not exceeding $y$, then the effective obstruction alphabet: $$ \Sigma_{\mathrm{avail}}(N;y) $$ becomes smaller. This may therefore force: $$ \eta_y^{\mathrm{eff}}(N) $$ or: $$ D_G(N) $$ to increase. One can seek a form: $$ S_N(y)\uparrow \Longrightarrow \text{effective residual pressure}\uparrow. $$ This still requires a rigorous proof or counterexample testing. --- # 49. Primorial-$N$ Pressure Tests It is particularly worth testing when: $$ N $$ itself is close to or equal to a primorial multiple: $$ N=cM_k. $$ In this case: $$ p_i\mid N $$ for every: $$ i\le k. $$ Thus, many of the first $k$ small primes are self-suppressed. This forms a class of extreme test objects: $$ \boxed{ \text{highly smooth }N \text{ vs. } \text{deep obstruction burden}. } $$ If Goldbach pairs instead always turn out to be easy to find for such $N$, this may reveal an important direction. Finite experiments, however, cannot directly be promoted to a theorem. --- # 50. Prime-$N/2$ boundary issue If: $$ N=2p $$ and $p$ itself is prime, then: $$ N=p+p $$ immediately forms a Goldbach representation. Thus, a hypothetical failure $N$ must satisfy: $$ \frac N2 \notin\mathbb P. $$ Thus: $$ \boxed{ F_G(N)=1 \Longrightarrow N/2 \text{ composite}. } $$ This is a simple necessary condition that should nevertheless be included in the state. --- # 51. Half-point obstruction For: $$ p=N/2, $$ it appears only when $N/2$ is prime in: $$ \mathbb P_N^{-}. $$ and once it appears, $$ N-p=p $$ is also prime. Thus, the reduced domain of a failure state automatically cannot contain a midpoint prime. This can be denoted by: $$ K_{\mathrm{mid}}(N): \quad N/2\notin\mathbb P. $$ --- # 52. Odd-complement sector for every: $$ p>2, $$ since $N$ is even, $$ N-p $$ is odd. Thus, in the canonical obstruction map: $$ \Phi_N^{\min}(p)\neq2 $$ for every: $$ p>2. $$ Thus, large-scale obstruction geometry actually begins with: $$ 3 $$ as displayed above. For the hard geometry of $N\equiv0\pmod6$, it begins more directly with: $$ 5 $$ as displayed above. --- # 53. $3$-suppression when $3\mid N$ If: $$ 3\mid N, $$ then by divisor self-suppression: $$ 3\mid(N-p) $$ can occur only at: $$ p=3 $$ this input. Therefore: $$ N\equiv0\pmod6 $$ is precisely the case in which: $$ r=2 $$ and: $$ r=3 $$ both almost lose general obstruction power. This gives: $$ \boxed{ N\equiv0\pmod6 } $$ as the maximal low-prime suppression branch in M6 obstruction geometry. --- # 54. Generalized suppression modulo $M_k$ If: $$ M_k\mid N, $$ then for each: $$ r\mid M_k, \quad r\in\mathbb P, $$ we have: $$ r\mid(N-p) \Longrightarrow p=r. $$ Thus, for: $$ p>p_k, $$ every: $$ r\le p_k $$ cannot obstruct. Therefore: $$ \boxed{ M_k\mid N, \quad p>p_k \Longrightarrow \Phi_N^{\min}(p)>p_k } $$ holds under the failure assumption. --- # 55. Primorial suppression theorem ## Theorem RG-SCP-03.1 If: $$ M_k\mid N, $$ and: $$ F_G(N)=1, $$ then for every: $$ p\in\mathbb P_N^{-} $$ satisfying: $$ p>p_k, $$ we have: $$ \Phi_N^{\min}(p)>p_k. $$ --- ## Proof If: $$ \Phi_N^{\min}(p)\le p_k, $$ Then: $$ \Phi_N^{\min}(p) $$ is: $$ M_k $$ a prime factor of this primorial. From: $$ M_k\mid N, $$ we obtain: $$ \Phi_N^{\min}(p)\mid N. $$ And because: $$ \Phi_N^{\min}(p)\mid(N-p), $$ Hence: $$ \Phi_N^{\min}(p)\mid p. $$ Because $p$ is prime: $$ p=\Phi_N^{\min}(p)\le p_k, $$ and: $$ p>p_k $$ a contradiction. Hence: $$ \Phi_N^{\min}(p)>p_k. $$ This completes the proof. --- # 56. The Importance of This Theorem This is a more general version than merely: $$ N\bmod6 $$ this residue. It means: > If a hypothetical failure itself is divisible by the primorial $M_k$, then the complements of all prime inputs greater than $p_k$ must avoid the first $k$ prime factors. Therefore: $$ M_k\mid N $$ creates a large-scale hard sector. --- # 57. Primorial-hard prime sector Define: $$ \mathbb P_{N,k}^{>} = \left\{ p\in\mathbb P_N^{-}: p>p_k \right\}. $$ If: $$ M_k\mid N $$ and: $$ F_G(N)=1, $$ Then: $$ \mathbb P_{N,k}^{>} \subseteq \mathbb H_{M_k}(N). $$ Thus: $$ \boxed{ M_k\mid N \Longrightarrow \eta_{M_k}(N) \text{ is forced to remain in a high-residual state} } $$ at least for most prime inputs. --- # 58. Primorial-pressure route This immediately provides a new research route: Study: $$ N=cM_k $$ even integers of this form. If one can prove: > For every sufficiently large $k$, whenever any $N$ is divisible by $M_k$, there must exist some: $$ p>p_k $$ such that: $$ N-p $$ is prime, then an entire family of highly smooth counterexamples can be excluded. This has not yet been proved. But obstruction geometry has already supplied the precise target of this route. --- # 59. Coverage cost lower bound target For: $$ M_k\mid N, $$ hypothetical failure requires: $$ \forall p\in\mathbb P_{N,k}^{>}, \quad \Phi_N^{\min}(p)>p_k. $$ Therefore: $$ C_{\mathrm{obs}}(N) \ge |\mathbb P_{N,k}^{>}| \log p_{k+1}. $$ This is a direct lower bound. More precisely: $$ \boxed{ C_{\mathrm{obs}}(N) \ge \sum_{p\in\mathbb P_{N,k}^{>}} \log p_{k+1} = |\mathbb P_{N,k}^{>}| \log p_{k+1}. } $$ --- # 60. Limitations of the Cost Lower Bound This lower bound does not directly yield a contradiction. Because obstruction cost may be large yet still finite. Thus: $$ C_{\mathrm{obs}}(N)\uparrow $$ cannot directly imply: $$ F_G(N)=0. $$ Its actual uses are: 1. comparing failure-state families; 2. finding possible extremal structures; 3. seeking conflicts with other upper bounds; 4. forming a future contradiction route. --- # 61. Upper-bound hunting One form that might genuinely produce a proof is: first prove: $$ F_G(N)=1 \Longrightarrow C_{\mathrm{obs}}(N) \ge L(N), $$ then prove from an independent structure: $$ C_{\mathrm{obs}}(N) \le U(N), $$ if ultimately: $$ L(N)>U(N), $$ there is a contradiction. That is: $$ \boxed{ \text{lower obstruction burden} > \text{maximum admissible obstruction capacity}. } $$ This is the first explicit appearance of a potential contradiction architecture in this series. --- # 62. Capacity notion Thus, define the conceptual obstruction capacity: $$ \operatorname{Cap}_N(y) $$ representing the obstruction coverage capacity legitimately available under a source cutoff $y$. We hope to find, in the future: $$ \operatorname{Demand}_N(y) $$ and: $$ \operatorname{Cap}_N(y). $$ If: $$ \operatorname{Demand}_N(y) > \operatorname{Cap}_N(y), $$ then that failure state is inadmissible. These two quantities do not yet have mature definitions. --- # 63. Demand / Capacity program In the future, obstruction cover can be turned into: $$ \boxed{ \text{Demand} \le \text{Capacity} } $$ a necessary condition. Here, demand may derive from: - hard branch size; - fixed-point fiber size; - prime density; - self-suppression; - least-factor lower bounds. Capacity may derive from: - residue class occupancy; - sieve-type upper bounds; - overlap limits; - graph degree limits; - independence constraints; - arithmetic progression restrictions. Only if nonoverlapping upper and lower bounds can be established might genuine closure be formed. --- # 64. Why This Is More Conservative than “Directly Applying Sieve Methods” This series is not saying: > Sieve methods cannot do it, so we will not use sieve methods. Instead, it first rewrites the problem as: $$ \text{obstruction demand/capacity}. $$ Any existing sieve theorem that can provide: $$ \operatorname{Cap}_N(y) $$ an upper bound can be directly inserted into this framework. Thus, RG-SCP is a meta-structure accommodating existing methods, not a replacement for all existing number theory. --- # 65. Relationship to the Parity Problem The well-known parity limitation of sieve methods tells us: Certain sieve information has difficulty distinguishing structures with different prime-factor parities. RG-SCP does not claim here to have overcome the parity problem. Instead, obstruction geometry should actively record: $$ \text{which constraints remain confined to sieve-visible sources} $$ and: $$ \text{which new sources genuinely increase target discrimination}. $$ This is exactly the role of the Prime-Surface Insufficiency gate. --- # 66. Source provenance graph Every obstruction constraint should record its source. For example: ```text K1: parity K2: mod 3 K3: mod 5 K4: divisor self-suppression K5: M6 fixed-point K6: prime density estimate K7: certificate theorem ``` and establish: $$ G_{\mathrm{prov}}. $$ If two constraints are merely: $$ N-p\bmod30 $$ different readouts of this quantity, they cannot be presented as two independent proof sources. --- # 67. New-source gate A new invariant: $$ I(N) $$ is worth promotion only when it satisfies at least one of the following: 1. splits fibers that old sources cannot split; 2. supplies coupling across $p$; 3. supplies cross-wheel consistency; 4. reduces the quantifier residual; 5. gives a nontrivial capacity bound; 6. supports a descending transform; 7. can generate a finite certificate. --- # 68. Serialization extension This article extends the obstruction state to: ```yaml rg_scp_obstruction_geometry: version: N: reduced_prime_domain: residue_state: mod_6: mod_30: mod_210: divisor_set: suppressed_obstruction_primes: canonical_obstruction_map: m6: easy_branch: hard_branch: hard_ratio: wheels: - modulus: cutoff_prime: covered_count: hard_count: hard_ratio: effective_hard_count: obstruction_alphabet: obstruction_depth: average_log_cost: weighted_costs: graph: left_vertices: right_vertices: edge_rule: isolated_left_vertices: fixed_point: enabled: hard_branch: source_provenance: proof_constraints: heuristic_metrics: certificates: checkpoint_hash: ``` --- # 69. Pre-Experimental Principles If subsequent work begins computing the obstruction signatures of many: $$ N $$ of these values, it should preregister: - $N$ range; - wheel depths; - prime generator; - factor verifier; - canonical lpf implementation; - boundary handling; - metrics; - stopping rule. Avoid changing the metric only after examining the data. --- # 70. Positive control Choose a value known to have a Goldbach representation: $$ N. $$ Then the obstruction graph must have at least one isolated left vertex: $$ p $$ such that: $$ N-p $$ is prime. If the program instead produces a total proper obstruction map, the implementation is faulty. --- # 71. Negative control for truncated wheel For a fixed: $$ M_k, $$ there should be many: $$ p\in\mathbb H_{M_k}(N) $$ surviving for general $N$. If a small wheel claims almost always to achieve complete coverage, suspect first: - oracle leakage; - factorization leakage; - domain bug; - boundary bug. --- # 72. Midpoint control If: $$ N/2 $$ is prime, Then: $$ N $$ must have a representation: $$ N = \frac N2 + \frac N2. $$ Thus, any hypothetical failure generator must first exclude: $$ N/2\in\mathbb P. $$ --- # 73. Divisor-suppression control If: $$ r\mid N, $$ then, in the experiment: $$ \mathcal O_r(N) $$ apart from: $$ p=r $$ it must contain no other prime input. If there occurs: $$ p\neq r $$ and: $$ r\mid N, \quad r\mid(N-p), $$ the program must be wrong, because this would imply: $$ r\mid p. $$ --- # 74. M6 control If: $$ N\equiv0\pmod6, $$ then all: $$ p>3 $$ complements must avoid $2$ and $3$. Thus, any: $$ p>3 $$ labeled as: $$ \Phi_N^{\min}(p)=3 $$ indicates implementation failure. --- # 75. Proved Structures and Unproved Hypotheses of This Article ## Proved / Directly Provable 1. M6 easy/hard branch classification; 2. wheel-hard nesting; 3. divisor self-suppression; 4. primorial suppression theorem; 5. obstruction-cost lower bound under primorial divisibility; 6. graph left-total characterization; 7. midpoint necessary condition. ## Not Yet Proved 1. high suppression necessarily leads to contradiction; 2. obstruction cost has a universal upper bound; 3. primorial multiples are less likely to be counterexamples; 4. hard-branch ratio has a sufficiently strong lower bound; 5. capacity is always smaller than demand; 6. the fixed-point layer can complete global closure. --- # 76. Central Theorem RG-SCP-03.2 The most important exact structural result of this article can be compressed into: If: $$ M_k\mid N, $$ and: $$ F_G(N)=1, $$ then for every: $$ p\in\mathbb P, \quad p_kp_k. $$ That is: $$ \boxed{ M_k\mid N \land F_G(N)=1 \Longrightarrow \mathbb P_{N,k}^{>} \subseteq \mathbb H_{M_k}(N). } $$ This directly turns the primorial divisibility of $N$ itself into a large-scale hard-obstruction obligation. --- # 77. Research Significance The point of this result is not: > Primorial multiples have difficulty being Goldbach counterexamples. This cannot currently be claimed. The actual conclusion is: > If a primorial multiple really is a Goldbach counterexample, it cannot use the first $k$ small primes to obstruct almost all prime inputs greater than $p_k$. Thus, the complements of all these prime inputs must: $$ \boxed{ \text{avoid the first }k\text{ primes, yet still all be composite}. } $$ This is a very strong synchronized structural requirement. --- # 78. From “Many Composites” to “Simultaneously Avoiding Small Factors” Composites in general are not unusual. What is truly special is: $$ \forall p>p_k, $$ many: $$ N-p $$ simultaneously satisfy: $$ \gcd(N-p,M_k)=1 $$ yet are all composite. This can be regarded as: $$ \boxed{ \text{small-factor-avoiding composite wall}. } $$ If Goldbach failure exists in: $$ M_k\mid N $$ this branch, it must build such a wall. --- # 79. Small-Factor-Avoiding Composite Wall Define: $$ \mathcal W_k(N) = \left\{ N-p: p\in\mathbb P, \quad p_k Can there exist a large set generated by the prime-complement map, each of whose elements avoids the first $k$ prime factors, yet all of whose elements remain composite? This question is more structured than merely “every complement is composite.” It may become: - sieve upper bound; - density contradiction; - graph capacity; - fixed-point transport; - spectral analysis; a shared interface for these. --- # 81. Interface with Prime Approximation / Skew For: $$ \mathcal W_k(N) $$ one can use the heuristic channel to analyze: - composite density; - least-factor distribution; - deviation from expected rough-number behavior; - prime-gap skew; - approximation residual. But these can only generate hypotheses: $$ \mathcal H \to \text{candidate invariant}. $$ cannot directly imply: $$ \mathcal W_k(N) $$ is impossible. --- # 82. Possible Interface with Rough Numbers The set: $$ \left\{ m: \gcd(m,M_k)=1 \right\} $$ can be understood as the rough sector avoiding the first $k$ prime factors. Goldbach failure on primorial-divisible $N$ requires: $$ \mathcal W_k(N) $$ to fall entirely within the rough composite sector. Thus, later interfaces may be established with: - rough number density; - Buchstab-type structure; - sieve remainder; - almost-prime decomposition; these subjects. This article does not directly introduce the relevant theorems here, to avoid unverified cross-layer inferences. --- # 83. The Natural Direction for the Next Article This article has completed: $$ \text{obstruction cover} \to \text{wheel geometry} \to \text{coverage cost} \to \text{suppression pressure}. $$ The most natural next article formally returns to the existing M6 fixed-point line: # RG-SCP 04 ## Fixed-Point Fibers and Goldbach Counterexample Exclusion ### Fixed-Point Fibers and Goldbach Counterexample Exclusion It will centrally study: $$ \mathcal F_N \cap \left( \operatorname{Fix}(T_6) \times \operatorname{Fix}(T_6) \right) $$ how it couples with: $$ \mathbb H_6(N) $$ and: $$ \Phi_N^{\min} $$ these structures. The goal is not to claim that the fixed-point form has proved Goldbach, but to ask: > Must a Goldbach failure create a global hole in the fixed-point fiber that can be tracked, transported, or perhaps even descended? --- # 84. Conclusion The second article proved: $$ \text{Goldbach failure} \Longleftrightarrow \text{complete proper obstruction assignment}. $$ The third article further points out: > The cost of this assignment is not uniform, but is strongly constrained by the residue and prime-divisor structure of $N$ itself. At the M6 layer: $$ N\equiv0\pmod6 $$ makes: $$ 2 $$ and: $$ 3 $$ both almost lose general obstruction power. At the general primorial layer: If: $$ M_k\mid N, $$ then for every: $$ p>p_k, $$ a hypothetical failure must satisfy: $$ \operatorname{lpf}(N-p)>p_k. $$ Therefore: $$ N $$ the more it absorbs small primes, the more it suppresses the corresponding small-prime obstruction channels. This produces a central tension: $$ \boxed{ \text{arithmetic smoothness of }N \quad\leftrightarrow\quad \text{obstruction-channel scarcity}. } $$ Thus, if a Goldbach counterexample exists, it must not only establish a complete obstruction cover; it must also continually push the entire prime-complement domain toward deeper composite structure as it gradually loses its own small-prime-factor coverage ability. This is what this article calls: $$ \boxed{ \text{counterexample coverage cost}. } $$ The next key that might genuinely form Goldbach closure is not merely to make cost larger, but to find: $$ \boxed{ \text{obstruction demand} > \text{admissible obstruction capacity} } $$ this structural contradiction. --- # 85. Shortest Version If only one sentence is retained: > **A Goldbach counterexample not only needs to make all prime complements composite; it must also continually complete global coverage using deeper prime factors while $N$ itself gradually closes off small-prime-factor obstruction channels.** Its central exact statement is: $$ \boxed{ M_k\mid N \land F_G(N)=1 \Longrightarrow \forall p\in\mathbb P, \quad p_kp_k. } $$ The next question is: $$ \boxed{ \text{Can this kind of small-factor-avoiding composite wall remain legitimately present in a fixed-point fiber over the long term?} } $$