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vf_d147aaea613dd946

Erdős Problem #572

Canonical assertion

declared status 'open'. Formalized: no. Show that for k3k\geq 3ex(n;C2k)n1+1k.\mathrm{ex}(n;C_{2k})\gg n^{1+\frac{1}{k}}. Current best: It is easy to see that ex(n;C2k+1)=n2/4\mathrm{ex}(n;C_{2k+1})=\lfloor n^2/4\rfloor for any k1k\geq 1 (and n>2k+1n>2k+1) (since no bipartite graph contains an odd cycle). Lazebnik, Ustimenko, and Woldar [LUW95] have shown that, for arbitrary k3k\geq 3,ex(n;C2k)n1+23k3+ν,\mathrm{ex}(n;C_{2k})\gg n^{1+\frac{2}{3k-3+\nu}},where ν=0\nu=0 if kk is odd and ν=1\nu=1 if kk is even. Prize: no. Tags: graph theory, turan number.

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erdos_deep:572
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Jun 16, 2026, 12:00 AM
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Exact record identityFinding ID, frontier identity, and pinned Git source
vf_d147aaea613dd946
vfr_0a25edabc16db143
ce8ba7d934c848408e0d91caca39e938698e3fc7
03f7371b496485f761f91961027fd48198dc7e93
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ce8ba7d934c848408e0d91caca39e938698e3fc7
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03f7371b496485f761f91961027fd48198dc7e93
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2026-07-20T19:20:20-04:00
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