Let n, b,  and c be integers with <InlineEquation ID="IEq1"> <EquationSource Format="TEX">$n\ge 0$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>. The generalized central trinomial coefficient <InlineEquation ID="IEq2"> <EquationSource Format="TEX">$T_n(b,c)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>T</mi> <mi>n</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi>b</mi> <mo>,</mo> <mi>c</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> is defined as the coefficient of <InlineEquation ID="IEq3"> <EquationSource Format="TEX">$x^n$</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mi>x</mi> <mi>n</mi> </msup> </math></EquationSource> </InlineEquation> in the expansion of <InlineEquation ID="IEq4"> <EquationSource Format="TEX">$(x^2+bx+c)^n$</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mrow> <mo stretchy="false">(</mo> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <mi>b</mi> <mi>x</mi> <mo>+</mo> <mi>c</mi> <mo stretchy="false">)</mo> </mrow> <mi>n</mi> </msup> </math></EquationSource> </InlineEquation>. Let <InlineEquation ID="IEq5"> <EquationSource Format="TEX">$p\ge 5$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>≥</mo> <mn>5</mn> </mrow> </math></EquationSource> </InlineEquation> be a prime. In this paper, by means of several combinatorial identities, we establish new congruences of the form <Equation ID="Equ29"> <EquationSource Format="TEX">$ \sum _{k=0}^{p-1}(2k+1)^a \epsilon ^k H_k^j \frac{T_k(b,c)^2}{d^k} $</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <msup> <mrow> <mo stretchy="false">(</mo> <mn>2</mn> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mi>a</mi> </msup> <msup> <mi>ϵ</mi> <mi>k</mi> </msup> <msubsup> <mi>H</mi> <mi>k</mi> <mi>j</mi> </msubsup> <mfrac> <mrow> <msub> <mi>T</mi> <mi>k</mi> </msub> <msup> <mrow> <mo stretchy="false">(</mo> <mi>b</mi> <mo>,</mo> <mi>c</mi> <mo stretchy="false">)</mo> </mrow> <mn>2</mn> </msup> </mrow> <msup> <mi>d</mi> <mi>k</mi> </msup> </mfrac> </mrow> </math></EquationSource> </Equation>modulo <InlineEquation ID="IEq6"> <EquationSource Format="TEX">$p^3$</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mi>p</mi> <mn>3</mn> </msup> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq7"> <EquationSource Format="TEX">$p^2$</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mi>p</mi> <mn>2</mn> </msup> </math></EquationSource> </InlineEquation>, where <InlineEquation ID="IEq8"> <EquationSource Format="TEX">$a=1,3$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>a</mi> <mo>=</mo> <mn>1</mn> <mo>,</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq9"> <EquationSource Format="TEX">$j=1,2$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> <mo>,</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>, and <InlineEquation ID="IEq10"> <EquationSource Format="TEX">$\epsilon \in \{-1,1\}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>ϵ</mi> <mo>∈</mo> <mo stretchy="false">{</mo> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mn>1</mn> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation>. Here <InlineEquation ID="IEq11"> <EquationSource Format="TEX">$H_n$</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>H</mi> <mi>n</mi> </msub> </math></EquationSource> </InlineEquation> denotes the nth harmonic number and <InlineEquation ID="IEq12"> <EquationSource Format="TEX">$d:=b^2-4c$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>d</mi> <mo>:</mo> <mo>=</mo> <msup> <mi>b</mi> <mn>2</mn> </msup> <mo>-</mo> <mn>4</mn> <mi>c</mi> </mrow> </math></EquationSource> </InlineEquation> satisfies <InlineEquation ID="IEq13"> <EquationSource Format="TEX">$p\not \mid d$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>∤</mo> <mi>d</mi> </mrow> </math></EquationSource> </InlineEquation>. As an illustration, in the case <InlineEquation ID="IEq14"> <EquationSource Format="TEX">$b=c=1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>b</mi> <mo>=</mo> <mi>c</mi> <mo>=</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, we obtain <Equation ID="Equ30"> <EquationSource Format="TEX">$ \sum _{k=0}^{p-1}(2k+1)H_k^2(-1)^k\frac{T_k^2}{3^k} \equiv -\frac{p}{2}\left( 3-\frac{3}{2}q_p(3)+\left( \frac{p}{3}\right) \right) \pmod {p^2}, $</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <mrow> <mo stretchy="false">(</mo> <mn>2</mn> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <msubsup> <mi>H</mi> <mi>k</mi> <mn>2</mn> </msubsup> <msup> <mrow> <mo stretchy="false">(</mo> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mi>k</mi> </msup> <mfrac> <msubsup> <mi>T</mi> <mi>k</mi> <mn>2</mn> </msubsup> <msup> <mn>3</mn> <mi>k</mi> </msup> </mfrac> <mo>≡</mo> <mo>-</mo> <mfrac> <mi>p</mi> <mn>2</mn> </mfrac> <mfenced close=")" open="("> <mn>3</mn> <mo>-</mo> <mfrac> <mn>3</mn> <mn>2</mn> </mfrac> <msub> <mi>q</mi> <mi>p</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mn>3</mn> <mo stretchy="false">)</mo> </mrow> <mo>+</mo> <mfenced close=")" open="("> <mfrac> <mi>p</mi> <mn>3</mn> </mfrac> </mfenced> </mfenced> <mspace width="10.0pt" /> <mrow> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <msup> <mi>p</mi> <mn>2</mn> </msup> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> </mrow> </math></EquationSource> </Equation>where <InlineEquation ID="IEq15"> <EquationSource Format="TEX">$T_n$</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>T</mi> <mi>n</mi> </msub> </math></EquationSource> </InlineEquation> denotes the central trinomial coefficient, <InlineEquation ID="IEq16"> <EquationSource Format="TEX">$\left( \frac{\cdot }{\cdot }\right) $</EquationSource> <EquationSource Format="MATHML"><math> <mfenced close=")" open="("> <mfrac> <mo>·</mo> <mo>·</mo> </mfrac> </mfenced> </math></EquationSource> </InlineEquation> is the Legendre symbol, and <InlineEquation ID="IEq17"> <EquationSource Format="TEX">$q_p(3)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>q</mi> <mi>p</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mn>3</mn> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> is the Fermat quotient.

On some congruences related to generalized central trinomial coefficients and harmonic numbers

期刊论文
发表日期：

2026年3月2日

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Yassine Otmani,
Hacène Belbachir

摘要

Let n, b, and c be integers with $n\ge 0$ $n \geq 0$ . The generalized central trinomial coefficient $T_n(b,c)$ $T_{n} (b, c)$ is defined as the coefficient of $x^n$ $x^{n}$ in the expansion of $(x^2+bx+c)^n$ ${(x^{2} + b x + c)}^{n}$ . Let $p\ge 5$ $p \geq 5$ be a prime. In this paper, by means of several combinatorial identities, we establish new congruences of the form $ \sum _{k=0}^{p-1}(2k+1)^a \epsilon ^k H_k^j \frac{T_k(b,c)^2}{d^k} $ $\sum_{k = 0}^{p - 1} {(2 k + 1)}^{a} ϵ^{k} H_{k}^{j} \frac{T_{k} {(b, c)}^{2}}{d^{k}}$ modulo $p^3$ $p^{3}$ and $p^2$ $p^{2}$ , where $a=1,3$ $a = 1, 3$ , $j=1,2$ $j = 1, 2$ , and $\epsilon \in \{-1,1\}$ $ϵ \in {- 1, 1}$ . Here $H_n$ $H_{n}$ denotes the nth harmonic number and $d:=b^2-4c$ $d : = b^{2} - 4 c$ satisfies $p\not \mid d$ $p ∤ d$ . As an illustration, in the case $b=c=1$ $b = c = 1$ , we obtain $ \sum _{k=0}^{p-1}(2k+1)H_k^2(-1)^k\frac{T_k^2}{3^k} \equiv -\frac{p}{2}\left( 3-\frac{3}{2}q_p(3)+\left( \frac{p}{3}\right) \right) \pmod {p^2}, $ $\sum_{k = 0}^{p - 1} (2 k + 1) H_{k}^{2} {(- 1)}^{k} \frac{T_{k}^{2}}{3^{k}} \equiv - \frac{p}{2} (3 - \frac{3}{2} q_{p} (3) + (\frac{p}{3})) (mod p^{2}),$ where $T_n$ $T_{n}$ denotes the central trinomial coefficient, $\left( \frac{\cdot }{\cdot }\right) $ $(\frac{\cdot}{\cdot})$ is the Legendre symbol, and $q_p(3)$ $q_{p} (3)$ is the Fermat quotient.