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Apéry 常数

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Apéry 常数定义为

zeta(3)=1.2020569...,
(1)

(OEIS A002117) 其中 zeta(z) 是 Riemann zeta 函数。Apéry (1979) 证明了 zeta(3) 是 无理数，尽管尚不清楚它是否是 超越数。Sorokin (1994) 和 Nesterenko (1996) 随后构建了 zeta(3) 无理性的独立证明 (Hata 2000)。Apéry 的证明涉及使用连分数

6/(zeta(3))=5+K_(n=1)^infty(-n^6)/(17[n^3+(n+1)^3]-12(2n+1))
(2)

(Raayoni 2021, Elimelech et al. 2023)。

zeta(3) 自然出现在许多物理问题中，包括使用量子电动力学计算的电子回旋磁比的二阶和三阶项中。

下表总结了在计算 zeta(3) 的无理测度的上界方面的进展。这里，mu_4 的精确值由下式给出

mu_4=1+(4ln(sqrt(2)+1)+3)/(4ln(sqrt(2)+1)-3)
(3)
approx 13.4178202
(4)

(Hata 2000)。

mu_n上界参考
15.513891Rhin 和 Viola (2001)
28.830284Hata (1990)
312.74359Dvornicich 和 Viola (1987)
413.41782Apéry (1979), Sorokin (1994), Nesterenko (1996), Prévost (1996)

Beukers (1979) 使用以下形式的三重积分再现了 Apéry 对 zeta(3) 的有理逼近

int_0^1int_0^1int_0^1(L_n(x)L_n(y))/(1-(1-xy)u)dxdydu,
(5)

其中 L_n(x) 是 Legendre 多项式。Beukers 的积分由下式给出

zeta(3)=-1/2int_0^1int_0^1(ln(xy))/((1-xy))dxdy,
(6)

这是一个被称为 Hadjicostas 公式的特殊情况的结果。

该积分与 zeta(3) 密切相关，使用了以下奇特的恒等式

int_0^1int_0^1int_0^1(x^ry^s)/(1-(1-xy)u)dxdydu={2zeta(3)-sum_(l=1)^(r)2/(l^3) for r=s; sum_(l=min(r,s)+1)^(max(r,s))1/(|r-s|l^2) for r!=s
(7)
={2zeta(3)-2H_r^((3)) for r=s; (psi_1(1+min(r,s))-psi_1(1+max(r,s)))/(|r-s|) for r!=s,
(8)

其中 H_r^((n)) 是广义调和数，psi_k(x) 是 polygamma 函数 (Hata 2000)。

与 zeta(3) 相关的和包括

zeta(3)=5/2sum_(n=1)^(infty)((-1)^(n-1))/(n^3(2n; n))
(9)
=5/2sum_(k=1)^(infty)((-1)^(k+1)(k!)^2)/((2k)!k^3)
(10)

(Apéry 使用的)，相关的和

zeta(3)=2/3(ln2)^3+4sum_(k=1)^infty((-1)^(k+1))/(k^32^k(2k; k))
(11)

由 G. Huvent 于 2002 年首次证明 (Gourevitch)，并由 B. Cloitre 重新发现 (私人通讯，2004 年 10 月 8 日)，以及

sum_(k=0)^(infty)1/((2k+1)^3)=7/8zeta(3)
(12)
=lambda(3)
(13)
sum_(k=0)^(infty)1/((3k+1)^3)=(2pi^3)/(81sqrt(3))+(13)/(27)zeta(3)
(14)
sum_(k=0)^(infty)1/((4k+1)^3)=(pi^3)/(64)+7/(16)zeta(3)
(15)
sum_(k=0)^(infty)1/((6k+1)^3)=(pi^3)/(36sqrt(3))+(91)/(216)zeta(3),
(16)

其中 lambda(z) 是 Dirichlet lambda 函数。以上方程是 Ramanujan (Berndt 1985) 的一般结果的特例。

Apéry 常数由无限族 BBP 型公式给出，形式为

zeta(3)=4/3sum_(k=0)^(infty)((-1)^k)/((1+k)^3)
(17)
=4/3sum_(k=0)^(infty)(-1)^k[1/((3k+1)^3)-1/((3k+2)^3)+1/((3k+3)^3)]
(18)
=3/2sum_(k=0)^(infty)(-1)^k[1/((3k+1)^3)-1/((3k+2)^3)-2/((3k+3)^3)]
(19)
=4/3sum_(k=0)^(infty)(-1)^k[1/((5k+1)^3)-1/((5k+2)^3)+1/((5k+3)^3)-1/((5k+4)^3)+1/((5k+5)^3)]
(20)
=1/(15)sum_(k=0)^(infty)(-1)^k[(21)/((5k+1)^3)-(21)/((5k+2)^3)+(21)/((5k+3)^3)-(21)/((5k+4)^3)-(104)/((5k+5)^3)]
(21)
=4/3sum_(k=0)^(infty)(-1)^k[1/((7k+1)^3)-1/((7k+2)^3)+1/((7k+3)^3)-1/((7k+4)^3)+1/((7k+5)^3)-1/((7k+6)^3)+1/((7k+7)^3)]
(22)
=1/(30)sum_(k=0)^(infty)(-1)^k[(41)/((7k+1)^3)-(41)/((7k+2)^3)+(41)/((7k+3)^3)-(41)/((7k+4)^3)+(41)/((7k+5)^3)-(41)/((7k+6)^3)+(302)/((7k+7)^3)]
(23)

(E. W. Weisstein, 2006 年 2 月 25 日)，以及两个惊人的特殊和

zeta(3)=1/(672)sum_(k=0)^(infty)1/(4096^k)[(2048)/((24k+1)^3)-(11264)/((24k+2)^3)-(1024)/((24k+3)^3)+(11776)/((24k+4)^3)-(512)/((24k+5)^3)+(4096)/((24k+6)^3)+(256)/((24k+7)^3)+(3456)/((24k+8)^3)+(128)/((24k+9)^3)-(704)/((24k+10)^3)-(64)/((24k+11)^3)-(128)/((24k+12)^3)-(32)/((24k+13)^3)-(176)/((24k+14)^3)+(16)/((24k+15)^3)+(216)/((24k+16)^3)+8/((24k+17)^3)+(64)/((24k+18)^3)-4/((24k+19)^3)+(46)/((24k+20)^3)-2/((24k+21)^3)-(11)/((24k+22)^3)+1/((24k+23)^3)]
(24)
=9/(224)sum_(k=0)^(infty)1/(4096^k)[(1024)/((24k+2)^3)-(3072)/((24k+3)^3)+(512)/((24k+4)^3)+(1024)/((24k+6)^3)+(1152)/((24k+8)^3)+(384)/((24k+9)^3)+(64)/((24k+10)^3)+(128)/((24k+12)^3)+(16)/((24k+14)^3)+(48)/((24k+15)^3)+(72)/((24k+16)^3)+(16)/((24k+18)^3)+2/((24k+20)^3)-6/((24k+21)^3)+1/((24k+22)^3)].
(25)

确定这种类型的和在 Bailey et al. (2007, p. 225; 印刷错误已更正) 中作为一个练习给出。

一个美丽的 双重级数 用于 zeta(3) 由下式给出

zeta(3)=1/3sum_(i=1)^inftysum_(j=1)^infty((i-1)!(j-1)!)/((i+j)!)H_(i+j),
(26)

其中 H_n 是 调和数 (O. Oloa, 私人通讯，2005 年 12 月 30 日)。

Apéry 的证明依赖于证明和

a(n)=sum_(k=0)^n(n; k)^2(n+k; k)^2,
(27)

其中 (n; k) 是二项式系数，满足递推关系

n^3a_n-(34n^3-51n^2+27n-5)a_(n-1)+(n-1)^3a_(n-2)=0
(28)

(van der Poorten 1979, Zeilberger 1991)。特征多项式 x^2-34x+1 有根 (1+/-sqrt(2))^4，所以

lim_(n->infty)(a_(n+1))/(a_n)=(1+sqrt(2))^4
(29)

是无理数，并且 a_n 不能满足两项递推式 (Jin 和 Dickinson 2000)。

Apéry 常数也由下式给出

zeta(3)=8sum_(n=1)^infty(S_(n,2))/(n!n),
(30)

其中 S_(n,m) 是第一类 Stirling 数。这可以重写为

zeta(3)=1/2sum_(n=1)^(infty)1/(n^2)(1+1/2+...+1/n)
(31)
=1/2sum_(n=1)^(infty)(H_n)/(n^2),
(32)

其中 H_n 是第 n 个 调和数 (Castellanos 1988)。

Amdeberhan (1996) 使用 Wilf-Zeilberger 对 (F,G)，其中

F(n,k)=((-1)^kk!^2(sn-k-1)!)/((sn+k+1)!(k+1)),
(33)

s=1 得到

zeta(3)=5/2sum_(n=1)^infty(-1)^(n-1)1/((2n; n)n^3).
(34)

对于 s=2,

zeta(3)=1/4sum_(n=1)^infty(-1)^(n-1)(56n^2-32n+5)/((2n-1)^2)1/((3n; n)(2n; n)n^3)
(35)

(Boros 和 Moll 2004, p. 236; Amdeberhan 1996)，对于 s=3,

zeta(3)=sum_(n=0)^infty((-1)^n)/(72(4n; n)(3n; n))(5265n^4+13878n^3+13761n^2+6120n+1040)/((4n+1)(4n+3)(n+1)(3n+1)^2(3n+2)^2)
(36)

(Amdeberhan 1996)。G(n,k) 对应于 s=1 和 2 为

G(n,k)=(2(-1)^kk!^2(n-k)!)/((n+k+1)!(n+1)^2)
(37)

和

G(n,k)=((-1)^kk!^2(2n-k)!(3+4n)(4n^2+6n+k+3))/(2(2n+k+2)!(n+1)^2(2n+1)^2).
(38)

Amdeberhan 和 Zeilberger (1997) 使用了 Wilf-Zeilberger 对 恒等式，其中

F(n,k)=(-1)^k(n!^6(2n-k-1)!(k!)^3)/(2(n+k+1)!^2((2n)!)^3),
(39)

s=1, 和 t=1, 得到快速收敛级数

zeta(3)=sum_(n=0)^infty(-1)^n((n!)^(10)(205n^2+250n+77))/(64((2n+1)!)^5),
(40)

该级数用于计算 (3) 到小数点后 100 万位。Campbell (2022) 使用 WZ 方法获得

zeta(3)=-2/7-1/(448)sum_(n=1)^infty((-2^(12))^n(7168n^5-1664n^4-1328n^3+212n^2+49n-9))/(n^4(2n-1)(3n+1)(4n+1)(2n; n)(3n; n)(4n; 2n)^3).
(41)

zeta(3) 的积分 包括

zeta(3)=1/2int_0^infty(t^2)/(e^t-1)dt
(42)
=8/7[1/4pi^2ln2+2int_0^(pi/2)xln(sinx)dx].
(43)

Gosper (1990) 给出了

zeta(3)=1/4sum_(k=1)^infty(30k-11)/((2k-1)k^3(2k; k)^2).
(44)

一个涉及 Apéry 常数的连分数是

6/(zeta(3))=5-(1^6)/(117-)(2^6)/(535-)...(n^6)/(34n^3+51n^2+27n+5-)...
(45)

(Apéry 1979, Le Lionnais 1983)。

zeta(3) 与 Glaisher-Kinkelin 常数 A 和 polygamma 函数 psi_n(z) 相关，关系式为

zeta(3)=2/3pi^2[12psi_(-4)(1)-6lnA-ln(2pi)].
(46)

Gosper (1996) 将 zeta(3) 表示为矩阵乘积

lim_(N->infty)product_(n=1)^NM_n=[0 zeta(3); 0 1],
(47)

其中

M_n=[((n+1)^4)/(4096(n+5/4)^2(n+7/4)^2) (24570n^4+64161n^3+62152n^2+26427n+4154)/(31104(n+1/3)(n+1/2)(n+2/3)); 0 1]
(48)

每个项给出 12 位精度。前几项是

M_1=[1/(19600) (2077)/(1728); 0 1]
(49)
M_2=[1/(9801) (7561)/(4320); 0 1]
(50)
M_3=[9/(67600) (50501)/(20160); 0 1],
(51)

这给出

zeta(3) approx (423203577229)/(352066176000)=1.20205690315732....
(52)

给定三个随机选择的整数，没有公因子可以同时整除它们的概率是

[zeta(3)]^(-1) approx 1.20206^(-1) approx 0.831907.
(53)

另请参阅

Apéry 常数近似, Apéry 常数连分数, Apéry 常数数字, Hadjicostas 公式, 普朗克辐射函数, Riemann Zeta 函数, Riemann Zeta 函数 zeta(2), 三对数函数, Wilf-Zeilberger 对

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参考文献

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Apéry 常数

引用为

Weisstein, Eric W. "Apéry 常数。" 来自 MathWorld——Wolfram Web 资源。 https://mathworld.net.cn/AperysConstant.html

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