Solving this quadratic inequality in∆yields ∆≤ 6n+5− √ 4n2 −28n−15 4 or∆≥ 6n+5+ √ 4n2 −28n−15 4

Taking the nearest integer|B|=4n−2∆ +2 gives an upper bound: X v∈V(G) d(v)≤f 2(4n−2∆ +2)=2∆ 2 −(6n+5)∆ +6n 2 +11n+3

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math.CO · 2026-04-13 · unverdicted · novelty 6.0

K_{n,n+1} is the unique graph on 2n+1 vertices with at least n^2 + n edges avoiding equal-degree vertices at path distance 5, for all n >= 11.

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Paths of length five with equal-degree endpoints math.CO · 2026-04-13 · unverdicted · none · ref 5
K_{n,n+1} is the unique graph on 2n+1 vertices with at least n^2 + n edges avoiding equal-degree vertices at path distance 5, for all n >= 11.