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Openai/68ec50da-cf00-8005-b5f6-b683506e5853
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===== Using the constants above and the sharp Θ29(1)\Theta_{29}^{(1)}Θ29(1), the inequalities in the manuscript produce: ===== * Case 1 (A∗≠∅A^\ast\neq\varnothingA∗=∅, some a∈A∗a\in A^\asta∈A∗ with 3∣a3\mid a3∣a): ∣A∣/N≤0.0338826+o(1)<1/25|A|/N \le \textbf{0.0338826} + o(1) < 1/25∣A∣/N≤0.0338826+o(1)<1/25. * Case 2i (A∗≠∅A^\ast\neq\varnothingA∗=∅, none in A∗A^\astA∗ divisible by 333, but some of A70,A99,A38,A251A_{70},A_{99},A_{38},A_{251}A70,A99,A38,A251 divisible by 333): ∣A∣/N≤0.0330251+o(1)<1/25|A|/N \le \textbf{0.0330251} + o(1) < 1/25∣A∣/N≤0.0330251+o(1)<1/25. * Case 2ii (A∗≠∅A^\ast\neq\varnothingA∗=∅, and no 3∣3\mid3∣ any element of A∗∪A70∪A99∪A38∪A251A^\ast\cup A_{70}\cup A_{99}\cup A_{38}\cup A_{251}A∗∪A70∪A99∪A38∪A251): ∣A∣/N≤0.0305740+o(1)<1/25|A|/N \le \textbf{0.0305740} + o(1) < 1/25∣A∣/N≤0.0305740+o(1)<1/25. * Case 3i (A∗=∅A^\ast=\varnothingA∗=∅, both A38A_{38}A38 and A251A_{251}A251 nonempty): ∣A∣/N≤0.0366926+o(1)<1/25|A|/N \le \textbf{0.0366926} + o(1) < 1/25∣A∣/N≤0.0366926+o(1)<1/25. * Case 3ii (A∗=∅A^\ast=\varnothingA∗=∅, exactly one of A38,A251A_{38},A_{251}A38,A251 nonempty): ∣A∣/N≤0.0339793+o(1)<1/25|A|/N \le \textbf{0.0339793} + o(1) < 1/25∣A∣/N≤0.0339793+o(1)<1/25. * Case 4a (A∗=A38=A251=∅A^\ast=A_{38}=A_{251}=\varnothingA∗=A38=A251=∅, both A70,A99A_{70},A_{99}A70,A99 nonempty): ∣A∣/N≤0.0386192+o(1)<1/25|A|/N \le \textbf{0.0386192} + o(1) < 1/25∣A∣/N≤0.0386192+o(1)<1/25. * Case 4b (A∗=A38=A251=∅A^\ast=A_{38}=A_{251}=\varnothingA∗=A38=A251=∅, exactly one of A70,A99A_{70},A_{99}A70,A99 nonempty): ∣A∣/N≤0.0339793+o(1)<1/25|A|/N \le \textbf{0.0339793} + o(1) < 1/25∣A∣/N≤0.0339793+o(1)<1/25. * Case 5 (A⊆A7∪A18A\subseteq A_7\cup A_{18}A⊆A7∪A18): If only one of A7,A18A_7,A_{18}A7,A18 is nonempty, then ∣A∣≤N/25+o(N)|A|\le N/25+o(N)∣A∣≤N/25+o(N). If both are nonempty, the cross‑pairs cannot use p=5p=5p=5, and Lemma \ref{lem:sf-AP} plus the standard CRT covering implies one of the two classes is o(N)o(N)o(N); hence the extremal configuration is a single class n≡7(mod25)n\equiv7\pmod{25}n≡7(mod25) or 18(mod25)18\pmod{25}18(mod25), exactly as you claim.
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