<p>In 2020, Mao and Pan proved a nice result associated with Van Hamme’s (H.2) supercongruence. In the same year, Guo and Zudilin supplied the following generalization of Mao and Pan’s formula: for any odd prime <i>p</i>, <Equation ID="Equ14"> <EquationSource Format="TEX">\(\begin{aligned} \sum _{k=0}^{(p+1)/2} \frac{(-\frac{1}{2})_k^3}{k!^3} \equiv p\,\dfrac{(\frac{1}{4})_{(p-1)/2}}{(\frac{7}{4})_{(p-1)/2}} {\left\{ \begin{array}{ll} \hspace{-4.5pt}\pmod {p^3} &amp; \text {if }p\equiv 1\pmod 4,\\ \hspace{-4.5pt}\pmod {p^2} &amp; \text {if }p\equiv 3\pmod 4. \end{array}\right. } \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>p</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <msubsup> <mrow> <mo stretchy="false">(</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> <mi>k</mi> <mn>3</mn> </msubsup> <mrow> <mi>k</mi> <msup> <mo>!</mo> <mn>3</mn> </msup> </mrow> </mfrac> <mo>≡</mo> <mi>p</mi> <mspace width="0.166667em" /> <mstyle displaystyle="true" scriptlevel="0"> <mfrac> <msub> <mrow> <mo stretchy="false">(</mo> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>p</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </msub> <msub> <mrow> <mo stretchy="false">(</mo> <mfrac> <mn>7</mn> <mn>4</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>p</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </msub> </mfrac> </mstyle> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <mspace width="-4.5pt" /> <mspace width="10.0pt" /> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <msup> <mi>p</mi> <mn>3</mn> </msup> <mo stretchy="false">)</mo> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mtext>if</mtext> <mspace width="0.333333em" /> <mi>p</mi> <mo>≡</mo> <mn>1</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> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mspace width="-4.5pt" /> <mspace width="10.0pt" /> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <msup> <mi>p</mi> <mn>2</mn> </msup> <mo stretchy="false">)</mo> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mtext>if</mtext> <mspace width="0.333333em" /> <mi>p</mi> <mo>≡</mo> <mn>3</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> </mtr> </mtable> </mrow> </mfenced> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>With the help of the <i>q</i>-Dixon formula, the creative microscoping method, and the Chinese remainder theorem for coprime polynomials, we shall establish a <i>q</i>-supercongruence modulo the third power of a cyclotomic polynomial. When <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\(q\rightarrow 1\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>q</mi> <mo stretchy="false">→</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, it produces the coming generalization of Guo and Zudilin’s supercongruence in the <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(p\equiv 3\pmod 4\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>≡</mo> <mn>3</mn> <mspace width="4.44443pt" /> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <mn>4</mn> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> case: <Equation ID="Equ15"> <EquationSource Format="TEX">\(\begin{aligned} \sum _{k=0}^{(p+1)/2}\frac{(-\frac{1}{2})_k^3}{k!^3}\equiv \frac{(\frac{1}{4})_{(p-1)/2}}{(\frac{7}{4})_{(p-1)/2}} \bigg \{p-\frac{p^{3}}{4}H_{(p+1)/4}^{(2)}\bigg \}\pmod {p^3}. \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>p</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <msubsup> <mrow> <mo stretchy="false">(</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> <mi>k</mi> <mn>3</mn> </msubsup> <mrow> <mi>k</mi> <msup> <mo>!</mo> <mn>3</mn> </msup> </mrow> </mfrac> <mo>≡</mo> <mfrac> <msub> <mrow> <mo stretchy="false">(</mo> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>p</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </msub> <msub> <mrow> <mo stretchy="false">(</mo> <mfrac> <mn>7</mn> <mn>4</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>p</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </msub> </mfrac> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">{</mo> </mrow> <mi>p</mi> <mo>-</mo> <mfrac> <msup> <mi>p</mi> <mn>3</mn> </msup> <mn>4</mn> </mfrac> <msubsup> <mi>H</mi> <mrow> <mo stretchy="false">(</mo> <mi>p</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>4</mn> </mrow> <mrow> <mo stretchy="false">(</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> </msubsup> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">}</mo> </mrow> <mspace width="10.0pt" /> <mrow> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <msup> <mi>p</mi> <mn>3</mn> </msup> <mo stretchy="false">)</mo> </mrow> <mo>.</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation></p>

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A New Result Associated with Van Hamme’s (H.2) Supercongruence

  • Chuanan Wei,
  • Yuanbo Yu

摘要

In 2020, Mao and Pan proved a nice result associated with Van Hamme’s (H.2) supercongruence. In the same year, Guo and Zudilin supplied the following generalization of Mao and Pan’s formula: for any odd prime p, \(\begin{aligned} \sum _{k=0}^{(p+1)/2} \frac{(-\frac{1}{2})_k^3}{k!^3} \equiv p\,\dfrac{(\frac{1}{4})_{(p-1)/2}}{(\frac{7}{4})_{(p-1)/2}} {\left\{ \begin{array}{ll} \hspace{-4.5pt}\pmod {p^3} & \text {if }p\equiv 1\pmod 4,\\ \hspace{-4.5pt}\pmod {p^2} & \text {if }p\equiv 3\pmod 4. \end{array}\right. } \end{aligned}\) k = 0 ( p + 1 ) / 2 ( - 1 2 ) k 3 k ! 3 p ( 1 4 ) ( p - 1 ) / 2 ( 7 4 ) ( p - 1 ) / 2 ( mod p 3 ) if p 1 ( mod 4 ) , ( mod p 2 ) if p 3 ( mod 4 ) . With the help of the q-Dixon formula, the creative microscoping method, and the Chinese remainder theorem for coprime polynomials, we shall establish a q-supercongruence modulo the third power of a cyclotomic polynomial. When \(q\rightarrow 1\) q 1 , it produces the coming generalization of Guo and Zudilin’s supercongruence in the \(p\equiv 3\pmod 4\) p 3 ( mod 4 ) case: \(\begin{aligned} \sum _{k=0}^{(p+1)/2}\frac{(-\frac{1}{2})_k^3}{k!^3}\equiv \frac{(\frac{1}{4})_{(p-1)/2}}{(\frac{7}{4})_{(p-1)/2}} \bigg \{p-\frac{p^{3}}{4}H_{(p+1)/4}^{(2)}\bigg \}\pmod {p^3}. \end{aligned}\) k = 0 ( p + 1 ) / 2 ( - 1 2 ) k 3 k ! 3 ( 1 4 ) ( p - 1 ) / 2 ( 7 4 ) ( p - 1 ) / 2 { p - p 3 4 H ( p + 1 ) / 4 ( 2 ) } ( mod p 3 ) .