<p>Let <i>n</i> be a fixed non-zero integer and let <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq1.gif" Format="GIF" Height="13" Rendition="HTML" Resolution="72" Type="Linedraw" Width="47" /> </InlineMediaObject> <EquationSource Format="TEX">\(m&gt;1\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mo>&gt;</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation> be a positive integer. We say a set <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="110" /> </InlineMediaObject> <EquationSource Format="TEX">\(\{a_1, a_2,..., a_m\}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">{</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msub> <mi>a</mi> <mi>m</mi> </msub> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> of positive integers is a <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq3.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="131" /> </InlineMediaObject> <EquationSource Format="TEX">\(D(n)-m-\textrm{tuple}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>D</mi> <mo stretchy="false">(</mo> <mi>n</mi> <mo stretchy="false">)</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mtext>tuple</mtext> </mrow> </math></EquationSource> </InlineEquation> if <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq4.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="62" /> </InlineMediaObject> <EquationSource Format="TEX">\(a_ia_j+n\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>a</mi> <mi>i</mi> </msub> <msub> <mi>a</mi> <mi>j</mi> </msub> <mo>+</mo> <mi>n</mi> </mrow> </math></EquationSource> </InlineEquation> is a perfect square <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq5.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="141" /> </InlineMediaObject> <EquationSource Format="TEX">\(\forall i,j\in \{1, 2,..., m\}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo>∀</mo> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>∈</mo> <mo stretchy="false">{</mo> <mn>1</mn> <mo>,</mo> <mn>2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <mi>m</mi> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> with <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq6.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="39" /> </InlineMediaObject> <EquationSource Format="TEX">\(i\not =j\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>i</mi> <mo>≠</mo> <mi>j</mi> </mrow> </math></EquationSource> </InlineEquation>. By counting solutions to the congruence <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq7.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="117" /> </InlineMediaObject> <EquationSource Format="TEX">\(x^2\equiv {n}\hspace{0.1cm}\pmod {b}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>≡</mo> <mi>n</mi> <mspace width="2.84544pt" /> <mspace width="4.44443pt" /> <mrow> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <mi>b</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> for <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq8.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="118" /> </InlineMediaObject> <EquationSource Format="TEX">\(b\in {\{1, 2,..., N\}}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>b</mi> <mo>∈</mo> <mrow> <mo stretchy="false">{</mo> <mn>1</mn> <mo>,</mo> <mn>2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <mi>N</mi> <mo stretchy="false">}</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>, we find the asymptotic behaviour of the number of <i>D</i>(<i>n</i>)-pairs and <i>D</i>(<i>n</i>)-triples with elements up to <i>N</i>. If <i>n</i> is a perfect square, these numbers grow as <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq9.gif" Format="GIF" Height="23" Rendition="HTML" Resolution="72" Type="Linedraw" Width="73" /> </InlineMediaObject> <EquationSource Format="TEX">\(\frac{6}{\pi ^2}N\log N\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfrac> <mn>6</mn> <msup> <mi>π</mi> <mn>2</mn> </msup> </mfrac> <mi>N</mi> <mo>log</mo> <mi>N</mi> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq10.gif" Format="GIF" Height="23" Rendition="HTML" Resolution="72" Type="Linedraw" Width="73" /> </InlineMediaObject> <EquationSource Format="TEX">\(\frac{3}{\pi ^2}N\log N\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfrac> <mn>3</mn> <msup> <mi>π</mi> <mn>2</mn> </msup> </mfrac> <mi>N</mi> <mo>log</mo> <mi>N</mi> </mrow> </math></EquationSource> </InlineEquation>, respectively. Otherwise, they grow as <i>C</i>(<i>n</i>)<i>N</i> and <InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1149_Article_IEq11.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="61" /> </InlineMediaObject> <EquationSource Format="TEX">\(\frac{1}{2}C(n)N\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mi>C</mi> <mrow> <mo stretchy="false">(</mo> <mi>n</mi> <mo stretchy="false">)</mo> </mrow> <mi>N</mi> </mrow> </math></EquationSource> </InlineEquation>, where <i>C</i>(<i>n</i>) is a constant depending only on <i>n</i> which we determine as a function of some <i>L</i>-series of Dirichlet characters.</p>

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On the asymptotics of D(n)-pairs and triples

  • Javier Badesa

摘要

Let n be a fixed non-zero integer and let \(m>1\) m > 1 be a positive integer. We say a set \(\{a_1, a_2,..., a_m\}\) { a 1 , a 2 , . . . , a m } of positive integers is a \(D(n)-m-\textrm{tuple}\) D ( n ) - m - tuple if \(a_ia_j+n\) a i a j + n is a perfect square \(\forall i,j\in \{1, 2,..., m\}\) i , j { 1 , 2 , . . . , m } with \(i\not =j\) i j . By counting solutions to the congruence \(x^2\equiv {n}\hspace{0.1cm}\pmod {b}\) x 2 n ( mod b ) for \(b\in {\{1, 2,..., N\}}\) b { 1 , 2 , . . . , N } , we find the asymptotic behaviour of the number of D(n)-pairs and D(n)-triples with elements up to N. If n is a perfect square, these numbers grow as \(\frac{6}{\pi ^2}N\log N\) 6 π 2 N log N and \(\frac{3}{\pi ^2}N\log N\) 3 π 2 N log N , respectively. Otherwise, they grow as C(n)N and \(\frac{1}{2}C(n)N\) 1 2 C ( n ) N , where C(n) is a constant depending only on n which we determine as a function of some L-series of Dirichlet characters.