<p>Let <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq10.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\( k \geq 2 \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation> be an integer. One of the generalization of the classical Fibonacci sequence is defined by the recurrence relation<InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq11.gif" Format="GIF" Height="26" Rendition="HTML" Resolution="72" Type="Linedraw" Width="182" /> </InlineMediaObject> <EquationSource Format="TEX">\( F_{n}^{(k)}=F_{n-1}^{(k)} + \cdots + F_{n-k}^{(k)}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msubsup> <mi>F</mi> <mrow> <mi>n</mi> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo stretchy="false">)</mo> </mrow> </msubsup> <mo>=</mo> <msubsup> <mi>F</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo stretchy="false">)</mo> </mrow> </msubsup> <mo>+</mo> <mo>⋯</mo> <mo>+</mo> <msubsup> <mi>F</mi> <mrow> <mi>n</mi> <mo>-</mo> <mi>k</mi> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo stretchy="false">)</mo> </mrow> </msubsup> </mrow> </math></EquationSource> </InlineEquation> for all <InlineEquation ID="IEq12"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq12.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="43" /> </InlineMediaObject> <EquationSource Format="TEX">\( n \geq 2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation> with the initial values <InlineEquation ID="IEq13"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq13.gif" Format="GIF" Height="24" Rendition="HTML" Resolution="72" Type="Linedraw" Width="61" /> </InlineMediaObject> <EquationSource Format="TEX">\( F_{i}^{(k)}=0 \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msubsup> <mi>F</mi> <mrow> <mi>i</mi> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo stretchy="false">)</mo> </mrow> </msubsup> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation> for <InlineEquation ID="IEq14"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq14.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="113" /> </InlineMediaObject> <EquationSource Format="TEX">\( i=2-k, \ldots, 0 \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>i</mi> <mo>=</mo> <mn>2</mn> <mo>-</mo> <mi>k</mi> <mo>,</mo> <mo>…</mo> <mo>,</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq15"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq15.gif" Format="GIF" Height="24" Rendition="HTML" Resolution="72" Type="Linedraw" Width="61" /> </InlineMediaObject> <EquationSource Format="TEX">\( F_{1}^{(k)}=1\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msubsup> <mi>F</mi> <mrow> <mn>1</mn> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo stretchy="false">)</mo> </mrow> </msubsup> <mo>=</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation> <InlineEquation ID="IEq16"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq16.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="35" /> </InlineMediaObject> <EquationSource Format="TEX">\(. F_{n}^{(k)} \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo>.</mo> <msubsup> <mi>F</mi> <mrow> <mi>n</mi> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo stretchy="false">)</mo> </mrow> </msubsup> </mrow> </math></EquationSource> </InlineEquation> is an order <InlineEquation ID="IEq17"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq17.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="13" /> </InlineMediaObject> <EquationSource Format="TEX">\( k \)</EquationSource> <EquationSource Format="MATHML"><math> <mi>k</mi> </math></EquationSource> </InlineEquation> generalization of the Fibonacci sequence and it is called <InlineEquation ID="IEq18"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq2.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="13" /> </InlineMediaObject> <EquationSource Format="TEX">\( k\)</EquationSource> <EquationSource Format="MATHML"><math> <mi>k</mi> </math></EquationSource> </InlineEquation>-generalizedFibonacci sequence or shortly <InlineEquation ID="IEq19"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq2.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="13" /> </InlineMediaObject> <EquationSource Format="TEX">\( k\)</EquationSource> <EquationSource Format="MATHML"><math> <mi>k</mi> </math></EquationSource> </InlineEquation>-Fibonacci sequence. Banks and Luca [7], among other things, determined all Fibonacci numbers which are concatenations of two Fibonacci numbers. In this paper, we consider the analogue of this problem in more general manner by taking into account the concatenations of two terms of the same sequence in base <InlineEquation ID="IEq20"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq20.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="40" /> </InlineMediaObject> <EquationSource Format="TEX">\(b \geq 2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>b</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>. First, we show that there exists only finitely many such concatenations for each <InlineEquation ID="IEq21"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq10.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\( k \geq 2 \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq22"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq22.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="40" /> </InlineMediaObject> <EquationSource Format="TEX">\( b \geq 2 \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>b</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>. Next, we completely determine all these concatenations for all <InlineEquation ID="IEq23"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq23.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\( k \geq 2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq24"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10474_2025_1517_Article_IEq24.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="78" /> </InlineMediaObject> <EquationSource Format="TEX">\( 2 \leq b \leq 10 \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>2</mn> <mo>≤</mo> <mi>b</mi> <mo>≤</mo> <mn>10</mn> </mrow> </math></EquationSource> </InlineEquation>.</p>

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On \( b\)-concatenations of two \( k\)-generalized Fibonacci numbers

  • M. Alan,
  • A. Altassan

摘要

Let \( k \geq 2 \) k 2 be an integer. One of the generalization of the classical Fibonacci sequence is defined by the recurrence relation \( F_{n}^{(k)}=F_{n-1}^{(k)} + \cdots + F_{n-k}^{(k)}\) F n ( k ) = F n - 1 ( k ) + + F n - k ( k ) for all \( n \geq 2\) n 2 with the initial values \( F_{i}^{(k)}=0 \) F i ( k ) = 0 for \( i=2-k, \ldots, 0 \) i = 2 - k , , 0 and \( F_{1}^{(k)}=1\) F 1 ( k ) = 1 \(. F_{n}^{(k)} \) . F n ( k ) is an order \( k \) k generalization of the Fibonacci sequence and it is called \( k\) k -generalizedFibonacci sequence or shortly \( k\) k -Fibonacci sequence. Banks and Luca [7], among other things, determined all Fibonacci numbers which are concatenations of two Fibonacci numbers. In this paper, we consider the analogue of this problem in more general manner by taking into account the concatenations of two terms of the same sequence in base \(b \geq 2\) b 2 . First, we show that there exists only finitely many such concatenations for each \( k \geq 2 \) k 2 and \( b \geq 2 \) b 2 . Next, we completely determine all these concatenations for all \( k \geq 2\) k 2 and \( 2 \leq b \leq 10 \) 2 b 10 .