<p>We utilize the Wilf-Zeilberger (WZ) method to establish congruences related to truncated Ramanujan-type series. By constructing hypergeometric terms <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1193_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="93" /> </InlineMediaObject> <EquationSource Format="TEX">\(f(k, a, b, \ldots )\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>f</mi> <mo stretchy="false">(</mo> <mi>k</mi> <mo>,</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mo>…</mo> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> with Gosper-summable differences and selecting appropriate parameters, we derive several congruences modulo <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1193_Article_IEq2.gif" Format="GIF" Height="12" Rendition="HTML" Resolution="72" Type="Linedraw" Width="13" /> </InlineMediaObject> <EquationSource Format="TEX">\(p\)</EquationSource> <EquationSource Format="MATHML"><math> <mi>p</mi> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1193_Article_IEq3.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(p^2\)</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mi>p</mi> <mn>2</mn> </msup> </math></EquationSource> </InlineEquation> for primes <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1193_Article_IEq4.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\(p &gt; 2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>&gt;</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>. For instance, we prove that for any prime <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1193_Article_IEq4.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\(p &gt; 2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>&gt;</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>, <Equation ID="Equ26"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1193_Article_Equ26.gif" Format="GIF" Height="52" Rendition="HTML" Resolution="72" Type="Linedraw" Width="305" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} \sum _{n=0}^{p-1} \frac{10n+3}{2^{3n}}\left( {\begin{array}{c}3n\\ n\end{array}}\right) \left( {\begin{array}{c}2n\\ n\end{array}}\right) ^2 \equiv 0 \pmod {p}, \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <munderover> <mo>∑</mo> <mrow> <mi>n</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <mfrac> <mrow> <mn>10</mn> <mi>n</mi> <mo>+</mo> <mn>3</mn> </mrow> <msup> <mn>2</mn> <mrow> <mn>3</mn> <mi>n</mi> </mrow> </msup> </mfrac> <mfenced close=")" open="("> <mrow> <mtable> <mtr> <mtd> <mrow> <mn>3</mn> <mi>n</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <mi>n</mi> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> <msup> <mfenced close=")" open="("> <mrow> <mtable> <mtr> <mtd> <mrow> <mn>2</mn> <mi>n</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <mi>n</mi> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> <mn>2</mn> </msup> <mo>≡</mo> <mn>0</mn> <mspace width="10.0pt" /> <mrow> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <mi>p</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>and <Equation ID="Equ27"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="11139_2025_1193_Article_Equ27.gif" Format="GIF" Height="52" Rendition="HTML" Resolution="72" Type="Linedraw" Width="352" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} \sum _{n=0}^{p-1} \frac{(-1)^n(20n^2+8n+1)}{2^{12n}}\left( {\begin{array}{c}2n\\ n\end{array}}\right) ^5 \equiv 0 \pmod {p^2}. \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <munderover> <mo>∑</mo> <mrow> <mi>n</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo stretchy="false">(</mo> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mi>n</mi> </msup> <mrow> <mo stretchy="false">(</mo> <mn>20</mn> <msup> <mi>n</mi> <mn>2</mn> </msup> <mo>+</mo> <mn>8</mn> <mi>n</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> </mrow> <msup> <mn>2</mn> <mrow> <mn>12</mn> <mi>n</mi> </mrow> </msup> </mfrac> <msup> <mfenced close=")" open="("> <mrow> <mtable> <mtr> <mtd> <mrow> <mn>2</mn> <mi>n</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <mi>n</mi> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> <mn>5</mn> </msup> <mo>≡</mo> <mn>0</mn> <mspace width="10.0pt" /> <mrow> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <msup> <mi>p</mi> <mn>2</mn> </msup> <mo stretchy="false">)</mo> </mrow> <mo>.</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>These results partially confirm conjectures by Sun and provide some novel congruences.</p>

错误:搜索内容不能为空,请输入英文关键词
错误:关键词超出字数限制,请精简
高级检索

Finding congruences with the WZ method

  • Li-Quan Feng,
  • Qing-Hu Hou

摘要

We utilize the Wilf-Zeilberger (WZ) method to establish congruences related to truncated Ramanujan-type series. By constructing hypergeometric terms \(f(k, a, b, \ldots )\) f ( k , a , b , ) with Gosper-summable differences and selecting appropriate parameters, we derive several congruences modulo \(p\) p and \(p^2\) p 2 for primes \(p > 2\) p > 2 . For instance, we prove that for any prime \(p > 2\) p > 2 , \(\begin{aligned} \sum _{n=0}^{p-1} \frac{10n+3}{2^{3n}}\left( {\begin{array}{c}3n\\ n\end{array}}\right) \left( {\begin{array}{c}2n\\ n\end{array}}\right) ^2 \equiv 0 \pmod {p}, \end{aligned}\) n = 0 p - 1 10 n + 3 2 3 n 3 n n 2 n n 2 0 ( mod p ) , and \(\begin{aligned} \sum _{n=0}^{p-1} \frac{(-1)^n(20n^2+8n+1)}{2^{12n}}\left( {\begin{array}{c}2n\\ n\end{array}}\right) ^5 \equiv 0 \pmod {p^2}. \end{aligned}\) n = 0 p - 1 ( - 1 ) n ( 20 n 2 + 8 n + 1 ) 2 12 n 2 n n 5 0 ( mod p 2 ) . These results partially confirm conjectures by Sun and provide some novel congruences.