Algorithms for Some Euler-Type Identities for Multiple Zeta Values

Abstract

Multiple zeta values are the numbers defined by the convergent series 𝜁(𝑠₁,𝑠₂,…,𝑠_𝑘)=∑_{𝑛1>𝑛2>⋯>𝑛𝑘>0}(1/𝑛^𝑠1₁𝑛^𝑠2₂⋯𝑛^𝑠𝑘_𝑘), where 𝑠₁, 𝑠₂, …, 𝑠_𝑘 are positive integers with 𝑠₁>1. For 𝑘≤𝑛, let 𝐸(2𝑛,𝑘) be the sum of all multiple zeta values with even arguments whose weight is 2𝑛 and whose depth is 𝑘. The well-known result 𝐸(2𝑛,2)=3𝜁(2𝑛)/4 was extended to 𝐸(2𝑛,3) and 𝐸(2𝑛,4) by Z. Shen and T. Cai. Applying the theory of symmetric functions, Hoffman gave an explicit generating function for the numbers 𝐸(2𝑛,𝑘) and then gave a direct formula for 𝐸(2𝑛,𝑘) for arbitrary 𝑘≤𝑛. In this paper we apply a technique introduced by Granville to present an algorithm to calculate 𝐸(2𝑛,𝑘) and prove that the direct formula can also be deduced from Eisenstein's double product.

1. Introduction

The multiple zeta sums, 𝜁𝑠₁,𝑠₂,…,𝑠_𝑚=_{𝑛1>𝑛2>⋯>𝑛𝑘>0}1𝑛^𝑠1₁𝑛^𝑠2₂⋯𝑛^𝑠𝑚_𝑚, (1) are also called Euler-Zagier sums, where 𝑠₁,𝑠₂,…,𝑠_𝑚 are positive integers with 𝑠₁≥2. Clearly, the Riemann zeta function 𝜁(𝑠), 𝜁(𝑠)=^∞_𝑛=11𝑛^𝑠,𝑠>1, (2) is the case 𝑚=1 in (1). The multiple zeta functions have attracted considerable interest in recent years.

For Riemann’s zeta function 𝜁(𝑠), Euler proved the following identity: _{𝑗1+𝑗2𝑗=𝑛1,𝑗2≥1}𝜁2𝑗₁𝜁2𝑗₂=2𝑛+12𝜁(2𝑛). (3) Recently, some identities similar to (3) have also been established. Given two positive integers 𝑛 and 𝑘 (suppose 𝑛≥𝑘), define a number 𝑍(𝑛,𝑘) by 𝑍(𝑛,𝑘)=_{𝑗1+𝑗2+⋯+𝑗𝑘𝑗=𝑛1,𝑗2,…,𝑗𝑘≥1}𝜁2𝑗₁𝜁2𝑗₂⋯𝜁2𝑗_𝑘. (4) Then, for 𝑘∈{3,4,…,9}, the value of 𝑍(𝑛,𝑘) is known [1–5].

Following [6], for 𝑘≤𝑛, let 𝐸(2𝑛,𝑘) be the sum of all multiple zeta values with even arguments whose weight is 2𝑛 and whose depth is 𝑘; that is, 𝐸(2𝑛,𝑘)=_{𝑗1+⋯+𝑗𝑘𝑗=𝑛1,𝑗2,…,𝑗𝑘≥1}𝜁2𝑗₁,2𝑗₂,…,2𝑗_𝑘. (5)

In [7], Gangl et al. proved the following identities: 𝐸()=32𝑛,24𝜁(2𝑛),for𝑛≥2,_{𝑛−1𝑟=1}𝜁()=12𝑟+1,2𝑛−2𝑟−14𝜁(2𝑛),for𝑛≥2. (6)(7)

Recently, using harmonic shuffle relations, Shen and Cai proved the following results in [8]: 𝐸()=52𝑛,38𝜁()−12𝑛4𝜁(2)𝜁()𝐸()=2𝑛−2,for𝑛≥3,2𝑛,435𝜁()−5642𝑛𝜁(2)𝜁()162𝑛−2,for𝑛≥4. (8)

In [6], applying the theory of symmetric functions, Hoffman established the generating function for the numbers 𝐸(2𝑛,𝑘). He proved that 1+_{𝑛≥𝑘≥1}𝐸(2𝑛,𝑘)𝑡^𝑛𝑠^𝑘=sin𝜋√1−𝑠√𝑡√1−𝑠sin𝜋√𝑡. (9) Based on this generating function, some formulas for 𝐸(2𝑛,𝑘) for arbitrary 𝑛≥𝑘 are given. For example, Hoffman obtained that 𝐸()=𝜁()2𝑛,𝑘2𝑛2^2(𝑘−1)𝑘−2𝑘−1_{[(𝑘−1)/2]𝑗=1}𝜁2𝑗𝜁2𝑛−2𝑗2^2𝑘−32𝑗+1𝐵_2𝑗𝑘,2𝑘−2𝑗−1 (10) where 𝐵_2𝑗 is the 2𝑗th Bernoulli number.

In this paper we use a technique introduced by Granville [9] to present an elementary recursion algorithm to calculate 𝐸(2𝑛,𝑘), we also give some direct formula for 𝐸(𝑛,𝑘) for arbitrary 𝑛≥𝑘. Our algorithm may be of some interest if we note that it is obtained through an elementary analytic method and that the statement of the algorithm is fairly simple.

2. Statements of the Theorems

Theorem 1. Let 𝑁 denote a positive integer. Let 𝑎₀^(𝑁),𝑎₁^(𝑁),𝑎₂^(𝑁),…, be a series of numbers defined by ^∞_{𝑟=1𝑟≠𝑁}𝑥1+𝑟²−𝑁²=𝑎₀^(𝑁)+𝑎₁^(𝑁)𝑥+⋯+𝑎^(𝑁)_𝑘−1𝑥^𝑘−1+⋯. (11) Then, for any two positive integers 𝑛 and 𝑘 with 𝑛≥𝑘, one has 𝐸(2𝑛,𝑘)=^∞_𝑁=1𝑎^(𝑁)_𝑘−1𝑁^{2𝑛−2𝑘+2}. (12)

Theorem 2. Given a positive integer 𝑁, we have ^∞_{𝑟=1𝑟≠𝑁}𝑥1+𝑟²−𝑁²=2𝑁²(−1)^𝑁+1sin𝜋𝑁²−𝑥𝜋𝑥𝑁².−𝑥 (13)

When 𝑘 is not large, we may use the following recursion algorithm to calculate 𝑎_𝑘^(𝑁) then use Theorem 1 to get the formula for 𝐸(2𝑛,𝑘).

Theorem 3. The coefficients 𝑎₀^(𝑁),𝑎₁^(𝑁),…,𝑎_𝑘^(𝑁),…, can be calculated recursively by the following formulas: 𝑎₀^(𝑁)=1;𝑘𝑎_𝑘^(𝑁)=−^𝑘−1_𝑗=0𝑏_𝑘−𝑗𝑎_𝑗^(𝑁),for𝑘≥1, (14) where 𝑏₁,𝑏₂,…, are the numbers defined by 𝑏_𝑗≜^∞_{𝑟=1𝑟≠𝑁}1𝑁²−𝑟^2𝑗,∀𝑗=1,2,3,…. (15)

In [6], Hoffman established an interesting result [6, Lemma 1.3] to obtain his formula (10) for 𝐸(2𝑛,𝑘). This lemma might be deduced from the theory of Bessel functions. Using the expressions for the Bessel functions of the first kind with a half integer index, we may deduce from the generating function (13) a direct formula for 𝑎_𝑘^(𝑁).

Theorem 4. For 𝑘≥1, one has 𝑎^(𝑁)_𝑘−1=𝜋^𝑘−1𝑘!(2𝑁)^𝑘−1sin𝑘𝜋2^[𝑘/2]_𝑗=0(−1)^𝑗𝑘+2𝑗!!!⋅12𝑗𝑘−2𝑗(2𝑁𝜋)^2𝑗−cos𝑘𝜋2×^{[(𝑘−1)/2]}_𝑗=0(−1)^𝑗𝑘+2𝑗+1!!!⋅12𝑗+1𝑘−2𝑗−1(2𝑁𝜋)^2𝑗+1,𝑎^(𝑁)_𝑘−1=12^2𝑘−2𝑘!^{[(𝑘−1)/2]}_𝑗=0(−1)^𝑗(2𝜋)^2𝑗2𝑘−1−2𝑗!!!⋅1𝑘−1−2𝑗2𝑗+1𝑁^{2𝑘−2−2𝑗}. (16)(17)

To deduce (17) from (16), we only need to write the expression of 𝑎^(𝑁)_𝑘−1, respectively, according to whether 𝑘 is odd or even, and use [(𝑘−1)/2]−𝑗 (if 𝑘 is odd) or [𝑘/2]−𝑗 (if 𝑘 is even) to replace 𝑗. In the two cases, we will get the expression (17) for 𝑎^(𝑁)_𝑘−1. By Theorem 1, we have 𝐸()=2𝑛,𝑘_{[(𝑘−1)/2]𝑗=0}(−1)^𝑗𝜋^2𝑗2𝑘−1−2𝑗!2^{2𝑘−2−2𝑗}!!𝜁,𝑘!𝑘−1−2𝑗2𝑗+12𝑛−2𝑗 (18) which reproduces Hoffman’s formula (10).

3. Proofs of the Theorems

Proof of Theorem 1. The left side of (12) is _{𝑗1+⋯+𝑗𝑘𝑗=𝑛1,…,𝑗𝑘≥1}𝜁2𝑗₁,2𝑗₂,…,2𝑗_𝑘=_{𝑗1+⋯+𝑗𝑘𝑗=𝑛1,…,𝑗𝑘≥1𝑛1>𝑛2>⋯>𝑛𝑘>0}1𝑛^2𝑗1₁𝑛^2𝑗2₂⋯𝑛^2𝑗𝑘_𝑘=_{𝑛1>𝑛2>⋯>𝑛𝑘>0𝑗1+⋯+𝑗𝑘𝑗=𝑛1,…,𝑗𝑘≥1}1𝑛^2𝑗1₁𝑛^2𝑗2₂⋯𝑛^2𝑗𝑘_𝑘. (19) The second sum in (19) is the coefficient of 𝑥^2𝑛 in the formal power series ^∞_𝑗=1𝑥²𝑛²₁^𝑗∞_𝑗=1𝑥²𝑛²₂^𝑗⋯^∞_𝑗=1𝑥²𝑛²_𝑘^𝑗=𝑥^2𝑘𝑛²₁−𝑥²𝑛²₂−𝑥²⋯𝑛²_𝑘−𝑥²=^𝑘_𝑗=1𝑥^2𝑘𝑛²_𝑗−𝑥²_{1≤𝑚≤𝑘𝑚≠𝑗}1𝑛²_𝑚−𝑛²_𝑗. (20) It follows that the coefficient of 𝑥^2𝑛 earlier is ^𝑘_𝑗=11𝑛_𝑗^{2𝑛−2𝑘+2}_{1≤𝑚≤𝑘𝑚≠𝑗}1𝑛²_𝑚−𝑛²_𝑗. (21)
Hence, the sum (19) is _{𝑛1>𝑛2>⋯>𝑛𝑘}^𝑘_>0𝑗=11𝑛_𝑗^{2𝑛−2𝑘+2}_{1≤𝑚≤𝑘𝑚≠𝑗}1𝑛²_𝑚−𝑛²_𝑗. (22)
Now, consider the function 𝑓_𝑁(𝑥)=_{𝑟=1𝑟≠𝑁}𝑥1+𝑟²−𝑁². (23) We partition 𝑓_𝑁(𝑥) into two parts. Let _𝑟>𝑁𝑥1+𝑟²−𝑁²=𝑃₀^(𝑁)+𝑃₁^(𝑁)𝑥+𝑃₂^(𝑁)𝑥²+⋯,_{1≤𝑟<𝑁}𝑥1+𝑟²−𝑁²=𝑄₀^(𝑁)+𝑄₁^(𝑁)𝑥+𝑄₂^(𝑁)𝑥²+⋯. (24) Then, we have 𝑃₀^(𝑁)=𝑄₀^(𝑁)≜1, 𝑄_𝑚^(𝑁)=0, for all𝑚≥𝑁, and 𝑃^(𝑁)_𝑗−1=_{𝑛1>𝑛2>⋯>𝑛𝑗−1>𝑁}1𝑛²₁−𝑁²⋯𝑛²_𝑗−1−𝑁²,𝑄∀1<𝑗≤𝑘,_{(𝑁)𝑘−𝑗}=_{𝑁>𝑛𝑗+1>𝑛𝑗+2>⋯>𝑛𝑘≥1}1𝑛²_𝑗+1−𝑁²⋯𝑛²_𝑘−𝑁²,∀1≤𝑗<𝑘. (25)
Consider the sum (22). For 𝑗∈{1,2,…,𝑘}, we treat each sum in (22) with respect to 𝑛_𝑗 as follows: _{𝑛1>𝑛2>⋯>𝑛𝑘>0}1𝑛_𝑗^{2𝑛−2𝑘+2}_{1≤𝑚≤𝑘𝑚≠𝑗}1𝑛²_𝑚−𝑛²_𝑗=^∞_{𝑛𝑗=𝑘−𝑗+1}1𝑛_𝑗^{2𝑛−2𝑘+2}×_{𝑛1>𝑛2>⋯>𝑛𝑗−1>𝑛𝑗1≤𝑚<𝑗}1𝑛²_𝑚−𝑛²_𝑗×_{𝑛𝑗>𝑛𝑗+1>⋯>𝑛𝑘≥1𝑗<𝑚≤𝑘}1𝑛²_𝑚−𝑛²_𝑗=^∞_{𝑛𝑗=𝑘−𝑗+1}1𝑛_𝑗^{2𝑛−2𝑘+2}𝑃^(𝑛𝑗)_𝑗−1𝑄^(𝑛𝑗)_𝑘−𝑗=^∞_{𝑁=𝑘−𝑗+1}1𝑁^{2𝑛−2𝑘+2}𝑃^(𝑁)_𝑗−1𝑄^(𝑁)_𝑘−𝑗=^∞_𝑁=11𝑁^{2𝑛−2𝑘+2}𝑃^(𝑁)_𝑗−1𝑄^(𝑁)_𝑘−𝑗. (26) In the last step, 𝑁 begins with 1 since 𝑄^(𝑁)_𝑘−𝑗=0 for 1≤𝑁<𝑘−𝑗.
It follows that the sum (22) becomes that ^∞_𝑁=1𝑃₀^(𝑁)𝑄^(𝑁)_𝑘−1+𝑃₁^(𝑁)𝑄^(𝑁)_𝑘−2+⋯+𝑃^(𝑁)_𝑘−1𝑄₀^(𝑁)𝑁^{2𝑛−2𝑘+2}. (27) Clearly, the sum 𝑃₀^(𝑁)𝑄^(𝑁)_𝑘−1+𝑃₁^(𝑁)𝑄^(𝑁)_𝑘−2+⋯+𝑃^(𝑁)_𝑘−1𝑄₀^(𝑁) in (27) is the coefficient of 𝑥^𝑘−1 in the Cauchy product of 𝑃₀^(𝑁)+𝑃₁^(𝑁)𝑥+𝑃₂^(𝑁)𝑥²+⋯⋅𝑄₀^(𝑁)+𝑄₁^(𝑁)𝑥+𝑄₂^(𝑁)𝑥²+⋯; (28) that is, it is the coefficient of 𝑥^𝑘−1 in the power series 𝑓_𝑁(𝑥)=𝑎₀^(𝑁)+𝑎₁^(𝑁)𝑥+⋯+𝑎^(𝑁)_𝑘−1𝑥^𝑘−1+⋯. (29) Therefore, the sum (27) is ^∞_𝑁=1𝑎^(𝑁)_𝑘−1𝑁^{2𝑛−2𝑘+2}. (30) The proof is completed.

Remark 5. If we take 𝑥 to be a complex variable, then the series ^∞_{𝑟=1𝑟≠𝑁}𝑥𝑟²−𝑁² (31) is absolutely and uniformly convergent for 𝑥 in any compact set in the complex plane; thus, the function 𝑓_𝑁(𝑥)=^∞_{𝑟=1𝑟≠𝑁}𝑥1+𝑟²−𝑁² (32) is analytic in the complex plane. Hence, it may be expanded as a Taylor series.

Proof of Theorem 2. First we recall Euler’s classical formula sin(𝜋𝑧)=𝜋𝑧^∞_𝑟=1𝑧1−²𝑟²,𝑧∈ℂ. (33) Similar to Euler’s formula, Eisenstein studied a product of two variables and proved that for (𝜔,𝑧)∈(ℂ⧵ℤ)×ℂ the following formula holds (see [10, page 17]): sin[𝜋(𝜔−𝑧)]()=sin𝜋𝜔^∞_𝑟=−∞𝑧1−𝑟+𝜔≜lim^𝑛_{𝑛→∞𝑟=−𝑛}𝑧1−.𝑟+𝜔 (34) Let 𝑁≥1 be temporarily fixed. By (34), for 𝜔∉ℤ we have ^∞_{𝑟=1𝑟≠𝑁}𝑧1−²−2𝜔𝑧𝑟²−𝜔²=sin[𝜋(𝜔−𝑧)]()𝑧sin𝜋𝜔1−𝜔⁻¹𝑧1−²−2𝜔𝑧𝑁²−𝜔²⁻¹. (35) Now, let 𝜔→𝑁. We get ^∞_{𝑟=1𝑟≠𝑁}𝑧1−²−2𝑁𝑧𝑟²−𝑁²=2𝑁²(−1)^𝑁sin[𝜋(𝑁−𝑧)]𝜋(𝑁−𝑧)𝑧².−2𝑁𝑧 (36) We write 𝑧²−2𝑁𝑧=−𝑥. Or equivalently, let 𝑧=𝑁±𝑁²−𝑥. Then, we get ^∞_{𝑟=1𝑟≠𝑁}𝑥1+𝑟²−𝑁²=2𝑁²(−1)^𝑁+1sin𝜋𝑁²−𝑥𝜋𝑥𝑁².−𝑥 (37)

Proof of Theorem 3. Taking logarithms of both sides of (32), we get that log𝑓_𝑁(𝑥)=^∞_{𝑟=1𝑟≠𝑁}𝑥log1−𝑁²−𝑟². (38) By Remark 5, the series may be differentiated term-by-term; hence, we have 𝑓^′_𝑁(𝑥)𝑓_𝑁(𝑥)=^∞_{𝑟=1𝑟≠𝑁}−1𝑁²−𝑟²−𝑥=−^∞∞_{𝑟=1𝑟≠𝑁𝑗=0}𝑥^𝑗𝑁²−𝑟^2𝑗+1=−^∞_𝑗=0𝑥^𝑗∞_{𝑟=1𝑟≠𝑁}1𝑁²−𝑟^2𝑗+1=−^∞_𝑗=0𝑏_𝑗+1𝑥^𝑗, (39) where we denote 𝑏_𝑗+1=^∞_{𝑟=1𝑟≠𝑁}1𝑁²−𝑟^2𝑗+1,for𝑗=0,1,2,…. (40) The order of the summation can be changed since the series ∑^∞_{𝑟=1𝑟≠𝑁}(−1/(𝑁²−𝑟²−𝑥)) is dominated by ∑^∞_𝑛=1(𝐿/𝑛²) for some positive constant 𝐿. From (39), we get that 𝑓^′_𝑁(𝑥)=−𝑓_𝑁(𝑥)^∞_𝑗=0𝑏_𝑗+1𝑥^𝑗, (41) or ^∞_𝑘=1𝑘𝑎_𝑘^(𝑁)𝑥^𝑘−1=−^∞_𝑖=0𝑎_𝑖^(𝑁)𝑥^𝑖∞_𝑗=0𝑏_𝑗+1𝑥^𝑗. (42) Write out the Cauchy product in the right side of (42), then compare the coefficient of 𝑥^𝑘−1 on both sides. We get that 𝑘𝑎_𝑘^(𝑁)=−_{𝑖+𝑗=𝑘−1𝑖,𝑗≥0}𝑏_𝑗+1𝑎_𝑖^(𝑁)=−^𝑘−1_𝑗=0𝑏_𝑘−𝑗𝑎_𝑗^(𝑁). (43)

Proof of Theorem 4. We now study the the generating function 𝑓_𝑁(𝑥)=2𝑁²(−1)^𝑁+1sin𝜋𝑁²−𝑥𝜋𝑥𝑁².−𝑥 (44) We may use L’Hospital’s rule to verify that lim_𝑥→02𝑁²(−1)^𝑁+1sin𝜋𝑁²−𝑥𝜋𝑥𝑁²−𝑥=1. (45) Now we expand out 𝑓_𝑁(𝑥). We have 𝑓_𝑁(𝑥)=2𝑁²(−1)^𝑁+1𝑥^∞_𝑚=0(−1)^𝑚𝜋^2𝑚𝑁²−𝑥^𝑚()!=2𝑚+12𝑁²(−1)^𝑁+1𝑥^∞_𝑚=0(−1)^𝑚𝜋^2𝑚()!×2𝑚+1_𝑚𝑘=0𝑚𝑘𝑁^{2𝑚−2𝑘}(−1)^𝑘𝑥^𝑘=2𝑁²(−1)^∞𝑁+1_𝑘=0(−1)^𝑘𝑁^2𝑘×^∞_𝑚=𝑘(−1)^𝑚(𝑁𝜋)^2𝑚()!𝑚𝑘𝑥2𝑚+1^𝑘−1. (46) By (11) and (13), we have 𝑎^(𝑁)_𝑘−1=2𝑁^2−2𝑘(−1)^𝑁+𝑘+1×^∞_𝑚=𝑘(−1)^𝑚(𝑁𝜋)^2𝑚()!𝑚𝑘2𝑚+1,∀𝑘=1,2,3,…. (47)
Consider the function sin𝑁𝜋√𝑥𝑁𝜋√𝑥=^∞_𝑚=0(−1)^𝑚(𝑁𝜋)^2𝑚()!𝑥2𝑚+1^𝑚. (48) Clearly, the sum in (47) can be rewritten as ^∞_𝑚=𝑘(−1)^𝑚(𝑁𝜋)^2𝑚()!𝑚𝑘=12𝑚+1√𝑘!sin𝑁𝜋𝑥𝑁𝜋√𝑥_𝑥^(𝑘)_𝑥=1, (49) where ()_𝑥^(𝑘) means the 𝑘th derivative of a function with respect to 𝑥.
We denote 𝜙(𝑥)=sin√𝑥/√𝑥. Then, we have 𝜙𝑁²𝜋²𝑡=sin𝑁𝜋√𝑡𝑁𝜋√𝑡, (50) and, hence, 𝑑^𝑘𝜙𝑁²𝜋²𝑡𝑑𝑡^𝑘=(𝑁𝜋)^2𝑘𝑑^𝑘𝜙(𝑥)𝑑𝑥^𝑘_{𝑥=𝑁2𝜋2𝑡},∀𝑘=0,1,2,…, (51) which implies that sin𝑁𝜋√𝑡𝑁𝜋√𝑡_𝑡^(𝑘)_𝑡=1=(𝑁𝜋)^2𝑘sin√𝑥√𝑥_𝑥^(𝑘)_{𝑥=(𝑁𝜋)2}. (52) Finally, from (47) (49) we get that 𝑎^(𝑁)_𝑘−1=2𝑁²𝜋^2𝑘(−1)^𝑁+𝑘+1√𝑘!sin𝑥√𝑥_𝑥^(𝑘)_{𝑥=(𝑁𝜋)2},𝑘≥1. (53)
We may apply Hoffman’s result [6, Lemma 1.3] to get the direct formula for sin√𝑥√𝑥_𝑥^(𝑘)_{𝑥=(𝑁𝜋)2}. (54) Here, we use some simple properties of the Bessel functions of the first kind to give its direct expression.
Lemma 6. Let 𝑘≥0 be an integer and let 𝑥>0. Then one has𝑑^𝑘𝑑𝑥^𝑘sin√𝑥√𝑥=𝜋2(−1)^𝑘2^−𝑘𝑥^{−(2𝑘+1)/4}𝐽_𝑘+1/2√𝑥, (55)where 𝐽_𝑘+1/2(𝑥) denotes the Bessel function of the first kind of index 𝑘+1/2.
The Bessel functions with a half-integer index can be represented by elementary functions. The following lemma is well known.
Lemma 7. Let 𝑘≥0 be an integer, and let 𝑥>0. Then, one has𝐽_𝑘+1/2(𝑥)=2𝜋1√𝑥sin𝑥−𝑘𝜋2^[𝑘/2]_𝑗=0(−1)^𝑗𝑘+2𝑗!!!⋅12𝑗𝑘−2𝑗(2𝑥)^2𝑗+cos𝑥−𝑘𝜋2×^{[(𝑘−1)/2]}_𝑗=0(−1)^𝑗𝑘+2𝑗+1!!!⋅12𝑗+1𝑘−2𝑗−1(2𝑥)^2𝑗+1. (56)
From Lemmas 6 and 7, and (53), we get that 𝑎^(𝑁)_𝑘−1=𝜋^𝑘−1𝑘!(2𝑁)^𝑘−1sin𝑘𝜋2^[𝑘/2]_𝑗=0(−1)^𝑗𝑘+2𝑗!!!⋅12𝑗𝑘−2𝑗(2𝑁𝜋)^2𝑗−cos𝑘𝜋2×^{[(𝑘−1)/2]}_𝑗=0(−1)^𝑗𝑘+2𝑗+1!!!⋅12𝑗+1𝑘−2𝑗−1(2𝑁𝜋)^2𝑗+1. (57) This completes the proof of Theorem 4.

4. Examples

The direct formula for 𝑎_𝑘^(𝑁) can be found from Theorem 4. However, we would like to use Theorem 3 to present some concrete examples to show how to calculate 𝑎_𝑘^(𝑁) for small 𝑘. The difficult part of the recursion formula (14) is for 𝑗≥1 to calculate the sum 𝑏_𝑗=^∞_{𝑟=1𝑟≠𝑁}1𝑁²−𝑟^2𝑗=12𝑁^𝑗∞_{𝑟=1𝑟≠𝑁}1+1𝑁−𝑟𝑁+𝑟^𝑗=𝜆^𝑗∞_{𝑟=1𝑟≠𝑁}𝐴_𝑟+𝐵_𝑟^𝑗, (58) where we denote 𝐴_𝑟=1/(𝑁−𝑟),𝐵_𝑟=1/(𝑁+𝑟), and 𝜆=1/2𝑁.

It follows from 𝐴_𝑟𝐵_𝑟=𝜆[𝐴_𝑟+𝐵_𝑟] that 𝐴_𝑟+𝐵_𝑟²=𝐴²_𝑟+𝐵²_𝑟+2𝜆𝐴_𝑟+𝐵_𝑟,𝐴_𝑟+𝐵_𝑟³=𝐴³_𝑟+𝐵³_𝑟+3𝜆𝐴²_𝑟+𝐵²_𝑟+6𝜆²𝐴_𝑟+𝐵_𝑟. (59) Generally, we can use induction on 𝑗 to prove that if for 𝑗≥2 we have gotten some positive integers 𝑐₁,𝑐₂,…,𝑐_𝑗−1 such that 𝐴_𝑟+𝐵_𝑟^𝑗=𝐴^𝑗_𝑟+𝐵^𝑗_𝑟+𝑐₁𝜆𝐴_𝑟^𝑗−1+𝐵_𝑟^𝑗−1+⋯+𝑐_𝑗−1𝜆^𝑗−1𝐴_𝑟+𝐵_𝑟, (60) then the expression for [𝐴_𝑟+𝐵_𝑟]^𝑗+1 is 𝐴_𝑟+𝐵_𝑟^𝑗+1=𝐴_𝑟^𝑗+1+𝐵_𝑟^𝑗+1+1+𝑐₁𝜆𝐴^𝑗_𝑟+𝐵^𝑗_𝑟+1+𝑐₁+𝑐₂𝜆²𝐴_𝑟^𝑗−1+𝐵_𝑟^𝑗−1+⋯+1+𝑐₁+𝑐₂+⋯+𝑐_𝑗−1𝜆^𝑗−1𝐴²_𝑟+𝐵²_𝑟+21+𝑐₁+𝑐₂+⋯+𝑐_𝑗−1𝜆^𝑗𝐴_𝑟+𝐵_𝑟. (61)

Note that if 𝑖≥2 is an even integer, then we have ^∞_{𝑟=1𝑟≠𝑁}𝐴^𝑖_𝑟+𝐵^𝑖_𝑟=_{1≤𝑟<𝑁}1(𝑁−𝑟)^𝑖+1(𝑁+𝑟)^𝑖+_𝑟>𝑁1(𝑁−𝑟)^𝑖+1(𝑁+𝑟)^𝑖=2𝜁(𝑖)−1+2^𝑖(2𝑁)^𝑖=2𝜁(𝑖)−1+2^𝑖𝜆^𝑖. (62) Similarly, if 𝑖≥1 is an odd integer, then we have ^∞_{𝑟=1𝑟≠𝑁}𝐴^𝑖_𝑟+𝐵^𝑖_𝑟=−1+2^𝑖(2𝑁)^𝑖=−1+2^𝑖𝜆^𝑖. (63)

From (58)–(63), we get that𝑏₁=−𝑎₁^(𝑁)=𝜆^∞_{𝑟=1𝑟≠𝑁}𝐴_𝑟+𝐵_𝑟=−3𝜆²,𝑏₂=𝜆^2∞_{𝑟=1𝑟≠𝑁}𝐴²_𝑟+𝐵²_𝑟+2𝜆𝐴_𝑟+𝐵_𝑟=𝜆²⋅2𝜁(2)−5𝜆²−6𝜆²=−11𝜆⁴+2𝜁(2)𝜆²,𝑏₃=𝜆^3∞_{𝑟=1𝑟≠𝑁}𝐴³_𝑟+𝐵³_𝑟+3𝜆𝐴²_𝑟+𝐵²_𝑟+6𝜆²𝐴_𝑟+𝐵_𝑟=𝜆³⋅−9𝜆³+3𝜆2𝜁(2)−5𝜆²−18𝜆³=−42𝜆⁶+6𝜁(2)𝜆⁴,𝑏₄=𝜆^4∞_{𝑟=1𝑟≠𝑁}𝐴⁴_𝑟+𝐵⁴_𝑟+4𝜆𝐴³_𝑟+𝐵³_𝑟+10𝜆²𝐴²_𝑟+𝐵²_𝑟+20𝜆³𝐴_𝑟+𝐵_𝑟=−163𝜆⁸+20𝜁(2)𝜆⁶+2𝜁(4)𝜆⁴. (64)

From formula (14), we get that 𝑎₂^(𝑁)1=−2𝑎₀^(𝑁)𝑏₂+𝑎₁^(𝑁)𝑏₁1=−2−11𝜆⁴+2𝜁(2)𝜆²−3𝜆²²=10𝜆⁴−𝜁(2)𝜆²=58𝑁⁴−𝜁(2)4𝑁²,𝑎₃^(𝑁)1=−3𝑎₀^(𝑁)𝑏₃+𝑎₁^(𝑁)𝑏₂+𝑎₂^(𝑁)𝑏₁1=−3−42𝜆⁶+6𝜁(2)𝜆⁴+3𝜆²−11𝜆⁴+2𝜁(2)𝜆²−3𝜆²10𝜆⁴−𝜁(2)𝜆²=35𝜆⁶−5𝜁(2)𝜆⁴=3564𝑁⁶−5𝜁(2)16𝑁⁴,𝑎₄^(𝑁)1=−4𝑎₀^(𝑁)𝑏₄+𝑎₁^(𝑁)𝑏₃+𝑎₂^(𝑁)𝑏₂+𝑎₃^(𝑁)𝑏₁=126𝜆⁸−21𝜁(2)𝜆⁶−12𝜁(4)𝜆⁴+12𝜁²(2)𝜆⁴=63128𝑁⁸−21𝜁(2)64𝑁⁶+3𝜁(4)64𝑁⁴. (65)

For 𝑘=2,3,4, using 𝑎₁^(𝑁), 𝑎₂^(𝑁), and 𝑎₃^(𝑁) in formula (12), respectively, we will get identities (6) and (8). Moreover, for 𝑛≥5, we have 𝐸(2𝑛,5)=63𝜁()−1282𝑛21𝜁(2)𝜁()+3642𝑛−2𝜁(4)𝜁().642𝑛−4 (66)

Acknowledgment

This work is supported by the National Natural Science Foundation of China (1127208).