The Tribonacci sequence <InlineEquation ID="IEq1"> <EquationSource Format="TEX">$\{T_{n}\}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">{</mo> <msub> <mi>T</mi> <mi>n</mi> </msub> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> is defined by the initial terms <InlineEquation ID="IEq2"> <EquationSource Format="TEX">$T_{0}=0,$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>T</mi> <mn>0</mn> </msub> <mo>=</mo> <mn>0</mn> <mo>,</mo> </mrow> </math></EquationSource> </InlineEquation> <InlineEquation ID="IEq3"> <EquationSource Format="TEX">$T_{1}=T_{2}=1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>=</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> <mo>=</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation> and by the recursion <InlineEquation ID="IEq4"> <EquationSource Format="TEX">$T_{n+3}=T_{n+2}+T_{n+1}+T_{n}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>T</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>3</mn> </mrow> </msub> <mo>=</mo> <msub> <mi>T</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>2</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>T</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>T</mi> <mi>n</mi> </msub> </mrow> </math></EquationSource> </InlineEquation> for <InlineEquation ID="IEq5"> <EquationSource Format="TEX">$n\ge 0.$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <mn>0</mn> <mo>.</mo> </mrow> </math></EquationSource> </InlineEquation> The Tribonacci–Lucas sequence <InlineEquation ID="IEq6"> <EquationSource Format="TEX">$\{S_{m}\}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">{</mo> <msub> <mi>S</mi> <mi>m</mi> </msub> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> satisfies the same recurrence relation as the Tribonacci sequence but with initial conditions <InlineEquation ID="IEq7"> <EquationSource Format="TEX">$S_{0}=S_{2}=3,$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>S</mi> <mn>0</mn> </msub> <mo>=</mo> <msub> <mi>S</mi> <mn>2</mn> </msub> <mo>=</mo> <mn>3</mn> <mo>,</mo> </mrow> </math></EquationSource> </InlineEquation> <InlineEquation ID="IEq8"> <EquationSource Format="TEX">$S_{1}=1.$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>S</mi> <mn>1</mn> </msub> <mo>=</mo> <mn>1</mn> <mo>.</mo> </mrow> </math></EquationSource> </InlineEquation> In this note we use Baker’s theory to solve the exponential diophantine equation <InlineEquation ID="IEq9"> <EquationSource Format="TEX">$T_{n}=\pm S_{m}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>T</mi> <mi>n</mi> </msub> <mo>=</mo> <mo>±</mo> <msub> <mi>S</mi> <mi>m</mi> </msub> </mrow> </math></EquationSource> </InlineEquation> in integers n, m. We show that <InlineEquation ID="IEq10"> <EquationSource Format="TEX">$\{T_{n}\}\cap \{\pm S_{m}\}=\{-271,-47,-3,\pm 1,5,7\}.$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mrow> <mo stretchy="false">{</mo> <msub> <mi>T</mi> <mi>n</mi> </msub> <mo stretchy="false">}</mo> </mrow> <mo>∩</mo> <mrow> <mo stretchy="false">{</mo> <mo>±</mo> <msub> <mi>S</mi> <mi>m</mi> </msub> <mo stretchy="false">}</mo> </mrow> <mo>=</mo> <mrow> <mo stretchy="false">{</mo> <mo>-</mo> <mn>271</mn> <mo>,</mo> <mo>-</mo> <mn>47</mn> <mo>,</mo> <mo>-</mo> <mn>3</mn> <mo>,</mo> <mo>±</mo> <mn>1</mn> <mo>,</mo> <mn>5</mn> <mo>,</mo> <mn>7</mn> <mo stretchy="false">}</mo> </mrow> <mo>.</mo> </mrow> </math></EquationSource> </InlineEquation>

Common terms of Tribonacci and Tribonacci–Lucas sequences

期刊论文
发表日期：

2025年8月14日

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Eric Fernando Bravo

摘要

The Tribonacci sequence $\{T_{n}\}$ ${T_{n}}$ is defined by the initial terms $T_{0}=0,$ $T_{0} = 0,$ $T_{1}=T_{2}=1$ $T_{1} = T_{2} = 1$ and by the recursion $T_{n+3}=T_{n+2}+T_{n+1}+T_{n}$ $T_{n + 3} = T_{n + 2} + T_{n + 1} + T_{n}$ for $n\ge 0.$ $n \geq 0 .$ The Tribonacci–Lucas sequence $\{S_{m}\}$ ${S_{m}}$ satisfies the same recurrence relation as the Tribonacci sequence but with initial conditions $S_{0}=S_{2}=3,$ $S_{0} = S_{2} = 3,$ $S_{1}=1.$ $S_{1} = 1 .$ In this note we use Baker’s theory to solve the exponential diophantine equation $T_{n}=\pm S_{m}$ $T_{n} = \pm S_{m}$ in integers n, m. We show that $\{T_{n}\}\cap \{\pm S_{m}\}=\{-271,-47,-3,\pm 1,5,7\}.$ ${T_{n}} \cap {\pm S_{m}} = {- 271, - 47, - 3, \pm 1, 5, 7} .$