Let <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_657_Article_IEq4.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">$k\ge 2$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation> be a fixed integer. The k-generalized Lucas sequence <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_657_Article_IEq5.gif" Format="GIF" Height="24" Rendition="HTML" Resolution="72" Type="Linedraw" Width="70" /> </InlineMediaObject> <EquationSource Format="TEX">$\{L_{n}^{(k)}\}_{n\ge 0}$</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mrow> <mo stretchy="false">{</mo> <msubsup> <mi>L</mi> <mrow> <mi>n</mi> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo stretchy="false">)</mo> </mrow> </msubsup> <mo stretchy="false">}</mo> </mrow> <mrow> <mi>n</mi> <mo>≥</mo> <mn>0</mn> </mrow> </msub> </math></EquationSource> </InlineEquation> starts with the positive integer initial values k, 1, 3, <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_657_Article_IEq6.gif" Format="GIF" Height="4" Rendition="HTML" Resolution="72" Type="Linedraw" Width="22" /> </InlineMediaObject> <EquationSource Format="TEX">$\ldots $</EquationSource> <EquationSource Format="MATHML"><math> <mo>…</mo> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_657_Article_IEq7.gif" Format="GIF" Height="18" Rendition="HTML" Resolution="72" Type="Linedraw" Width="61" /> </InlineMediaObject> <EquationSource Format="TEX">$2^{k-1}-1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mn>2</mn> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, and each term afterward is the sum of the k consecutive preceding elements. In this paper, we find all solutions of the equation <Equation ID="Equ20"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_657_Article_Equ20.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="161" /> </MediaObject> <EquationSource Format="TEX">$\begin{aligned} L_n^{(k)}=(2^a-1)(2^b-1) \end{aligned}$</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <msubsup> <mi>L</mi> <mi>n</mi> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo stretchy="false">)</mo> </mrow> </msubsup> <mo>=</mo> <mrow> <mo stretchy="false">(</mo> <msup> <mn>2</mn> <mi>a</mi> </msup> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mrow> <mo stretchy="false">(</mo> <msup> <mn>2</mn> <mi>b</mi> </msup> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>in integers <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_657_Article_IEq8.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="90" /> </InlineMediaObject> <EquationSource Format="TEX">$n\ge 2,k\ge 2$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <mn>2</mn> <mo>,</mo> <mi>k</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_657_Article_IEq9.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="72" /> </InlineMediaObject> <EquationSource Format="TEX">$b\ge a\ge 0$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>b</mi> <mo>≥</mo> <mi>a</mi> <mo>≥</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>.

The Diophantine Equation $L_n^{(k)}=(2^a-1)(2^b-1)$

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发表日期：

2025年6月10日

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Nurettin Irmak,
Florian Luca

摘要

Let $k\ge 2$ $k \geq 2$ be a fixed integer. The k-generalized Lucas sequence $\{L_{n}^{(k)}\}_{n\ge 0}$ ${L_{n}^{(k)}}_{n \geq 0}$ starts with the positive integer initial values k, 1, 3, $\ldots $ $\dots$ , $2^{k-1}-1$ $2^{k - 1} - 1$ , and each term afterward is the sum of the k consecutive preceding elements. In this paper, we find all solutions of the equation $\begin{aligned} L_n^{(k)}=(2^a-1)(2^b-1) \end{aligned}$ $\begin{matrix} L_{n}^{(k)} = (2^{a} - 1) (2^{b} - 1) \end{matrix}$ in integers $n\ge 2,k\ge 2$ $n \geq 2, k \geq 2$ , $b\ge a\ge 0$ $b \geq a \geq 0$ .

The Diophantine Equation \(L_n^{(k)}=(2^a-1)(2^b-1)\)

摘要