<p>The Kepler set of a sequence <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1506_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="52" /> </InlineMediaObject> <EquationSource Format="TEX">\((a_n)_{n=0}^\infty\)</EquationSource> <EquationSource Format="MATHML"><math> <msubsup> <mrow> <mo stretchy="false">(</mo> <msub> <mi>a</mi> <mi>n</mi> </msub> <mo stretchy="false">)</mo> </mrow> <mrow> <mi>n</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>∞</mi> </msubsup> </math></EquationSource> </InlineEquation> is the closure of the set of consecutive ratios <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1506_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="134" /> </InlineMediaObject> <EquationSource Format="TEX">\(\{a_{n+1}/a_{n} : n\geq 0\}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">{</mo> <msub> <mi>a</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo stretchy="false">/</mo> <msub> <mi>a</mi> <mi>n</mi> </msub> <mo>:</mo> <mi>n</mi> <mo>≥</mo> <mn>0</mn> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation>. Following several studies, dealing with Kepler sets of recurrence sequences of order 2, we study here the case of recurrences of any order.</p>

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Kepler sets of linear recurrence sequences

  • D. Berend,
  • R. Kumar

摘要

The Kepler set of a sequence \((a_n)_{n=0}^\infty\) ( a n ) n = 0 is the closure of the set of consecutive ratios \(\{a_{n+1}/a_{n} : n\geq 0\}\) { a n + 1 / a n : n 0 } . Following several studies, dealing with Kepler sets of recurrence sequences of order 2, we study here the case of recurrences of any order.