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Erdős problem 36

This example calculates the value of M1M 1. The set is {1,2}\{1, 2\}, so the only partition is A={1},B={2}A = \{1\}, B = \{2\} (or vice versa). The possible differences are 1−2=−11 - 2 = -1 and 2−1=12 - 1 = 1. The Overlap for k=−1k=-1 is 1 (if A={1},B={2}A=\{1\}, B=\{2\}) and for k=1k=1 also 1 (if A={2},B={1}A=\{2\}, B=\{1\} ). The MaxOverlap is 11, since the Overlap is 00 for other kk. Thus, M1=1M 1 = 1.

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The source's own material, retained exactly.

18 retained statements36c6d0fef71a

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FormalConjectures/ErdosProblems/

36.lean

Retained formal statement11 of 18

A lower bound of 4−15\sqrt{4 - \sqrt{15}}.

FormalConjectures/ErdosProblems/36.leanErdos36.minimum_overlap.variants.lower.haugland_19961 lineExact file
(4 - 15 ^ (1 / 2)) ^ (1 / 2) < Filter.liminf Erdos36.MinOverlapQuotient Filter.atTop
SolvedStatement only, no proof