<p>Let <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1216_Article_IEq1.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="88" /> </InlineMediaObject> <EquationSource Format="TEX">\(\overline{NT}(r, m, n)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mover> <mrow> <mi mathvariant="italic">NT</mi> </mrow> <mo>¯</mo> </mover> <mrow> <mo stretchy="false">(</mo> <mi>r</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> count the total number of parts in overpartitions of <i>n</i> with rank congruent to <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1216_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="81" /> </InlineMediaObject> <EquationSource Format="TEX">\(r \pmod {m}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>r</mi> <mspace width="4.44443pt" /> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <mi>m</mi> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>. Recently, Dan, Liu, and Yao proved the following Andrews–Beck type congruence for overpartitions using theta functions: for all <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1216_Article_IEq3.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="43" /> </InlineMediaObject> <EquationSource Format="TEX">\(n \ge 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation><Equation ID="Equ8"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1216_Article_Equ8.gif" Format="GIF" Height="51" Rendition="HTML" Resolution="72" Type="Linedraw" Width="545" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} \sum _{r = 1}^{3} r \overline{NT}(r, 4, n) = {\left\{ \begin{array}{ll} 2 \pmod 4, &amp; \text {if }\,\, n = 2k^{2} \,\, \text { or }\,\, n = 4k^{2} \,\,\text { for some}\,\, k \in \mathbb {Z^{+}}, \\ 0 \pmod 4, &amp; \text { otherwise.} \end{array}\right. } \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <munderover> <mo>∑</mo> <mrow> <mi>r</mi> <mo>=</mo> <mn>1</mn> </mrow> <mn>3</mn> </munderover> <mi>r</mi> <mover> <mrow> <mi mathvariant="italic">NT</mi> </mrow> <mo>¯</mo> </mover> <mrow> <mo stretchy="false">(</mo> <mi>r</mi> <mo>,</mo> <mn>4</mn> <mo>,</mo> <mi>n</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <mn>2</mn> <mspace width="10.0pt" /> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <mn>4</mn> <mo stretchy="false">)</mo> <mo>,</mo> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mtext>if</mtext> <mspace width="0.333333em" /> <mspace width="0.166667em" /> <mspace width="0.166667em" /> <mi>n</mi> <mo>=</mo> <mn>2</mn> <msup> <mi>k</mi> <mn>2</mn> </msup> <mspace width="0.166667em" /> <mspace width="0.166667em" /> <mspace width="0.333333em" /> <mtext>or</mtext> <mspace width="0.333333em" /> <mspace width="0.166667em" /> <mspace width="0.166667em" /> <mi>n</mi> <mo>=</mo> <mn>4</mn> <msup> <mi>k</mi> <mn>2</mn> </msup> <mspace width="0.166667em" /> <mspace width="0.166667em" /> <mspace width="0.333333em" /> <mtext>for some</mtext> <mspace width="0.166667em" /> <mspace width="0.166667em" /> <mi>k</mi> <mo>∈</mo> <msup> <mi mathvariant="double-struck">Z</mi> <mo>+</mo> </msup> <mo>,</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mn>0</mn> <mspace width="10.0pt" /> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <mn>4</mn> <mo stretchy="false">)</mo> <mo>,</mo> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mspace width="0.333333em" /> <mtext>otherwise.</mtext> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>They posed the problem of finding a combinatorial proof of this congruence. In this paper, we provide such a combinatorial proof.</p>

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A combinatorial proof of Andrews–Beck type congruences for overpartitions

  • Suparno Ghoshal,
  • Arijit Jana

摘要

Let \(\overline{NT}(r, m, n)\) NT ¯ ( r , m , n ) count the total number of parts in overpartitions of n with rank congruent to \(r \pmod {m}\) r ( mod m ) . Recently, Dan, Liu, and Yao proved the following Andrews–Beck type congruence for overpartitions using theta functions: for all \(n \ge 0\) n 0 \(\begin{aligned} \sum _{r = 1}^{3} r \overline{NT}(r, 4, n) = {\left\{ \begin{array}{ll} 2 \pmod 4, & \text {if }\,\, n = 2k^{2} \,\, \text { or }\,\, n = 4k^{2} \,\,\text { for some}\,\, k \in \mathbb {Z^{+}}, \\ 0 \pmod 4, & \text { otherwise.} \end{array}\right. } \end{aligned}\) r = 1 3 r NT ¯ ( r , 4 , n ) = 2 ( mod 4 ) , if n = 2 k 2 or n = 4 k 2 for some k Z + , 0 ( mod 4 ) , otherwise. They posed the problem of finding a combinatorial proof of this congruence. In this paper, we provide such a combinatorial proof.