Algorithms for Some Euler-Type Identities for Multiple Zeta Values

. . . , s k are positive integers with s 1 > 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)/4was 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.

Following [6], for  ≤ , let (2, ) be the sum of all multiple zeta values with even arguments whose weight is 2 and whose depth is ; that is, In [7], Gangl et al. proved the following identities: −1 Recently, using harmonic shuffle relations, Shen and Cai proved the following results in [8]: In [6], applying the theory of symmetric functions, Hoffman established the generating function for the numbers (2, ).He proved that Based on this generating function, some formulas for (2, ) for arbitrary  ≥  are given.For example, Hoffman obtained that where  2 is the 2th Bernoulli number.
In this paper we use a technique introduced by Granville [9] to present an elementary recursion algorithm to calculate (2, ), we also give some direct formula for (, ) for arbitrary  ≥ .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.

Statements of the Theorems
Theorem 1.Let  denote a positive integer.Let  ()  0 ,  () 1 ,  () 2 , . . ., be a series of numbers defined by Then, for any two positive integers  and  with  ≥ , one has Theorem 2. Given a positive integer , we have When  is not large, we may use the following recursion algorithm to calculate  ()  then use Theorem 1 to get the formula for (2, ).Theorem 3. The coefficients  ()  0 ,  () 1 , . . .,  ()  , . . ., can be calculated recursively by the following formulas: where  1 ,  2 , . . ., are the numbers defined by In [6], Hoffman established an interesting result [6, Lemma 1.3] to obtain his formula (10) for (2, ).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  ()   .
Theorem 4. For  ≥ 1, one has To deduce (17) from ( 16), we only need to write the expression of  ()  −1 , respectively, according to whether  is odd or even, and use In the two cases, we will get the expression (17) for  ()  −1 .By Theorem 1, we have which reproduces Hoffman's formula (10).

Proofs of the Theorems
Proof of Theorem 1.The left side of (12) is The second sum in ( 19) is the coefficient of  2 in the formal power series It follows that the coefficient of  2 earlier is Hence, the sum (19) is Now, consider the function We partition   () into two parts.Let Then, we have  () 0 =  () 0 ≜ 1,  ()  = 0, for all  ≥ , and Consider the sum (22).For  ∈ {1, 2, . . ., }, we treat each sum in (22) with respect to   as follows: In the last step,  begins with 1 since  () − = 0 for 1 ≤  <  − .
It follows that the sum (22) becomes that Clearly, the sum that is, it is the coefficient of  −1 in the power series Therefore, the sum ( 27) is The proof is completed.
Remark 5.If we take  to be a complex variable, then the series is absolutely and uniformly convergent for  in any compact set in the complex plane; thus, the function is analytic in the complex plane.Hence, it may be expanded as a Taylor series.