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<p>Seymour conjectured that in any oriented graph, there is a vertex having a second out-degree greater than or equal to its first out-degree. A similar conjecture posed by Sullivan says that any oriented graph contains a vertex having a second out-degree greater than or equal to its in-degree. Such a vertex is called a Sullivan-1 vertex. In this paper, we prove Sullivan’s conjecture for oriented graphs of order <i>n</i> with minimum degree at least <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="373_2025_2908_Article_IEq1.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="41" /> </InlineMediaObject> <EquationSource Format="TEX">\(n-3\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>-</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation>. Besides, we study the number of Sullivan-1 vertices in some oriented graphs.</p>

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A Note on Sullivan’s Second Neighborhood Conjecture

  • 期刊论文

  • 发表日期：

    2025年3月10日

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  • Ayman El Zein

摘要

Seymour conjectured that in any oriented graph, there is a vertex having a second out-degree greater than or equal to its first out-degree. A similar conjecture posed by Sullivan says that any oriented graph contains a vertex having a second out-degree greater than or equal to its in-degree. Such a vertex is called a Sullivan-1 vertex. In this paper, we prove Sullivan’s conjecture for oriented graphs of order n with minimum degree at least \(n-3\) n - 3 . Besides, we study the number of Sullivan-1 vertices in some oriented graphs.