Creative telescoping and generating functions of (variants of) multiple zeta values
Kam Cheong Au Steven Charlton

	
	Creative telescoping and generating functions of MZV's]Creative telescoping and generating functions 
 of (variants of) multiple zeta values
	
	
	Kam Cheong Au]Kam Cheong Au
	
	Rheinische Friedrich-Wilhelms-Universität Bonn 
 Mathematical Institute 
 53115 Bonn, Germany 
 	s6kmauuu@uni-bonn.de

       Steven Charlton]Steven Charlton
		Max Planck Institute for Mathematics, Vivatsgasse 7, Bonn 53111, Germany
	charlton@mpim-bonn.mpg.de
	

	[2020]Primary: 11M32. Secondary: 33C20
	
	 
	
	
	 
		We show how to convert the generating series of interpolated multiple zeta values, or multiple t values, with repeating blocks of length 1 into hypergeometric series. Then we invoke creative telescoping on their generating functions, in some known cases for illustration, and in some apparently new cases, reducing them to polynomials in Riemann zeta values. The new evaluations, including ζ^({2}^n,3), ζ^⋆({1,3}^n,1,2) and t^(2,{1}^n,2), resolve some questions raised elsewhere, and seem to be non-trivial using other methods.
	

	
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    25 April 2024
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§ INTRODUCTION


	Multiple zeta values (MZV's) are a multivariable analogue of the Riemann zeta values, defined by the following nested sum
	
    ζ(s_1,…,s_d) ∑_1 ≤ n_1 < n_2 < ⋯ < n_d1/n_1^s_1⋯ n_d^s_d ,

	where s_1,…,s_d are positive integers, and s_d ≥ 2 for convergence.  These values appear in many areas of mathematics, and mathematical physics (associators and knot invariants <cit.>, Feynman amplitudes <cit.>, periods of mixed Tate motives <cit.>).  
	
	Multiple zeta values have a rich algebraic structure, with many known types of relations (including extended double shuffle relations <cit.>, associator relations <cit.>, period polynomial relations <cit.>, or confluence relations <cit.>) and special evaluations (such as of ζ(1,3,…,1,3) <cit.>,  <cit.> which foreshadowed the block decomposition <cit.>, block filtration <cit.>, and block shuffle product <cit.>; of ζ(2, …, 2, 3, 2, …, 2) <cit.> used by Brown to establish the Hoffman basis <cit.> of motivic MZV's and to describe the periods of the category of mixed Tate motives over ℤ; or of ζ(2, …, 2, 4, 2, …, 2) <cit.> which was used to show the period polynomial relations <cit.> follow from the block relations describing the reduced block Lie algebra <cit.>.  One is readily lead to investigate other such special evaluations of MZV's (and many of natural generalisations and variants) as an avenue to uncover deeper structures on the algebra of MZV's and related objects. Beyond the `classical' type of MZV introduced above, the one of the most prominent variants is the so-called alternating MZV's <cit.> (also called Euler-Zagier sums, which allow an alternating sign (-1)^n_i to appear in the numerator of <ref>) or more generally the cyclotomic (or coloured) MZV's of level N (which allow a character σ_i^n_i, σ_i^N = 1, with alternating MZV's being the case N = 2).  In another direction, by relaxing the condition n_i < n_i1 to n_i ≤ n_i1 and introducing a weighting when n_i = n_i+1 we can define so-called interpolated MZV's <cit.>, which bridge between the MZV's and the multiple zeta star values (MZSV's) ζ^⋆, which are obtained by replacing 1 ≤ n_1 < n_2 < ⋯ < n_d by 1 ≤ n_1 ≤ n_2 ≤⋯≤ n_d in <ref>.
	
	Naturally one can combine these two variants, to define the interpolated cyclotomic MZV's as follows.  For s_1,…,s_d integers, and σ_1,…,σ_d ∈{ z ∈ℂz = 1 }, define
	
    ζ^r(s_1,…,s_d ; σ_1,…,σ_d) ∑_1 ≤ n_1 ≤ n_2 ≤⋯≤ n_d r^δ_n_1,n_2 + δ_n_2,n_3 + ⋯ + δ_n_d­1,n_d σ_1^n_1⋯σ_d^n_d/n_1^s_1⋯ n_d^s_d ,

	where δ_a,b is the Kronecker delta, namely δ_a,b = 1 if a = b, and 0 otherwise.  By considering whether each non-strict inequality n_i ≤ n_i1 is n_i < n_i1 or n_i = n_i1, we can expand ζ^r as a sum of 2^d-1 cyclotomic MZV's.  For example
	
    ζ^r(s_1,s_2,s_3; σ_1,σ_2,σ_3)   =     ζ(s_1,s_2,s_3; σ_1,σ_2,σ_3) + 
    		r ·ζ(s_1+s_2,s_3; σ_1·σ_2,σ_3)  
        + r ·ζ(s_1,s_2+s_3; σ_1,σ_2·σ_3) + 
    		r^2 ·ζ(s_1+s_2+s_3; σ_1·σ_2·σ_3)  ;

	note that indices are summed, while characters are multiplied when considering n_i = n_i1 and contracting those terms.  From this, we also see ζ^r(s_1,…,s_d; σ_1,…,σ_d) converges for (s_d,σ_d) ≠ (1,1) (c.f. <cit.>, when r = 0, and expand ζ^r in terms of ζ, for the general case).  In the case of (interpolated) alternating MZV's, it is common to simplify the notation by decorating s_i with a bar (i.e. s_i) if and only if σ_i = -1, for example
	
    ζ^r(a,b,c,d,e; 1,-1,-1,1,-1) = ζ^r(a,b,c,d,e)  .
 
	
	Certain special cases of interpolated MZV's stand out for being of particular interest, including the case r = 0 which recovers the `classical' MZV's ζ^0 = ζ, and the case r = 1, which gives the so-called multiple zeta star values (MZSV's) ζ^1 = ζ^⋆.  At the midpoint r =, the multiple zeta half values (MZV's) have a particularly simple stuffle-product structure, starting with ζ^(a)ζ^(b)= ζ^(a,b) + ζ^(b,a). which involves no depth 1 terms (c.f. <cit.>).  More significantly, Zhao's generalised two-one theorem <cit.> gives a beautiful relation expressing any non-alternating MZSV's in terms of certain special alternating MZV's (those with σ_i = -1 if and only if s_i is even, i.e. exactly the even indices are barred), and vice versa.  (A rephrasing of this bijection via the block decomposition was given by the second named author in <cit.>.)  For example, Zhao's generalised two-one theorem gives
	
    ζ^⋆(1,1,2,2,3,3,2) = 2^4ζ^(1,6,3,4)  .

	In one of the examples below (<ref>), we shall make use of this theorem to rewrite, for a,b ≥ 1:
	
    ζ^⋆({1}^a, {2}^b) = 2^a ζ^({1}^a-1,2b+1)  ,

	in order to be able to write the generating series of this MZSV via hypergeometric functions, and apply the ideas of creative telescoping.
	Here we have employed the notation 
	
    {x_1,…,x_i}^n = x_1,…,x_i,  … , x_1,…,x_i^n repetitions ,

	which means exactly n repetitions of the string x_1,…,x_i.  We will likewise (<ref>) use the generalised two-one theorem to write
	
    ζ^⋆({1,3}^n, 1,2) = 2^2n+1ζ^({2}^2n, 3)  ,

	in order to apply creative telescoping to the resulting hypergeometric generating series, and obtain evaluations for this multiple zeta star value.
	
	In this note, we henceforth focus on the cases r = 0, , 1 of interpolated MZV's. A final object of interest is the (interpolated) multiple t values (MtV's) <cit.>, which changes <ref> to involve a sum only over odd denominators.  Namely
	
    t^r(s_1,…,s_d) ∑_1 ≤ n_1 ≤ n_2 ≤⋯≤ n_dr^δ_n_1,n_2 + δ_n_2,n_3 + ⋯ + δ_n_d­1, n_d/(2n_1 - 1)^s_1⋯ (2n_d - 1)^s_d .

	(Naturally, one can define alternating MtV's; for the current note, we do not need to consider them.)

	The main goal of this note is to explain and apply the ideas of creative telescoping in the context of multiple zeta values.  It seems that these ideas, and in particular the computer algebra implementation available in Christoph Koutschan's  package <cit.> for , are not as well-known as they should be in the MZV context; many conjectural and/or non-trivial multiple zeta value identities can be readily investigated and proven using these techniques, which otherwise are non-trivial to handle.
	
	This note is structured as follows.  We start with some preliminaries: In <ref> we will explain how to write generating series of interpolated multiple zeta values with length 1 repeating blocks via hypergeometric series, which are then amenable to creative telescoping.  In <ref> we will review the ideas of creative telescoping, and how to use the  package.  Then in <ref> we will cover some analytic results necessary to fix the coefficients functions which appear after solving recurrence relations obtained via creative telescoping.
	
	We then start by applying creative telescoping to some known examples, for illustration and background.  We treat Chen and Eie's <cit.> evaluation of ζ^⋆(1,…,1,2,…,2) in <ref>,  Zagier's <cit.> evaluation of ζ(2, …, 2, 3, 2, …, 2) in <ref>, and Murakami's <cit.> evaluation of t(2,…,2,3,2,…,2) in <ref>.  Next, in <ref> we turn to evaluations of ζ^({2}^n, 3) and ζ^(3,{2}^n) which follow from a seemingly non-trivial _6F_5 hypergeometric identity (<ref>).  Via Zhao's generalised two-one theorem we then obtain some apparently new evaluations for multiple zeta star values involving {1,3}^m.  Finally in <ref> we establish an evaluation conjectured by Hoffman and the second named author <cit.>, on t^(2,{1}^n,2). 

  
Acknowledgements  SC is grateful to the Max Planck Institute for Mathematics, Bonn, for support, hospitality and excellent working conditions.



§ PRELIMINARIES


We start with some necessary preliminaries: how to write the generating series of MZV's in a way which we can apply the creative telescoping framework, how the creative telescoping framework works algorithmically, and some analytic properties we need to fix unknown coefficients in the evaluations we derive.



 §.§ Generating functions of MZV variants


We gather some examples of generating series of (interpolated) MZV's, which we shall employ later in the creative telescoping framework. 

We start with generating series which involve a single repeating block of length 1; these follow easily from the definitions of the MZV's and MZSV's.  We have

    ∑_j≥ 0ζ({a}^j) x^j    = ∏_n≥ 1(1+x/n^a)  , 
    ∑_j≥ 0ζ^⋆({a}^j) x^j    = ∏_n≥ 1(1+x/n^a + x^2/n^2a + ⋯) 
    		= ∏_n≥1( 1 - x/n^a)^-1 .

The cases ζ, and ζ^⋆ are somehow better known (c.f. Equation (11), (44) <cit.>), but one can readily generalise this to the interpolated MZV's, to obtain

    ∑_j≥ 0ζ^r({a}^j) x^j = ∏_n≥ 1(1+x/n^a + r x^2/n^2a + r^2 x^3/n^3a + ⋯) 
    	= ∏_n≥1(  
    		1 + (1-r) ·x/n^a/1 - r · x/n^a)  .

Restricting to n = 2m-1 odd immediately gives a generating series for the interpolated MtV's,

    ∑_j≥ 0 t^r({a}^j) x^j     = ∏_m≥ 1(1+x/(2m-1)^a + r x^2/(2m-1)^2a + r^2 x^3/(2m-1)^3a + ⋯)  
        = ∏_m≥ 1(  
    	1 + (1-r) ·x/(2m-1)^a/1 - r · x/(2m-1)^a)  .


Similarly, by introducing the character (-1)^n, we have the generating series for the alternating MZV's

    ∑_j≥ 0ζ^r({a}^j) x^j     = ∏_n≥ 1( 1+(-1)^n x/n^a +   r (-1)^2n x^2/n^2a  +  r^2 (-1)^3n x^3/n^3a + ⋯) 
        = ∏_n≥1(
    		1 + (1-r) ·(-1)^n x/n^a/1 - r ·(-1)^n x/n^a)  .

Likewise, one could obtain a generating series for arbitrary interpolated cyclotomic MZV's.


More generally, we can write the generating function of any family of (interpolated, alternating) MZV's or (interpolated) MtV's, which have repeating blocks of length 1, via some type of hypergeometric series.  For example

    ∑_j,k≥0ζ^r({2}^j, 3, {2}^k)   x^j y^k  
    	
    	    = ∑_ℓ≥ 1∏_n<ℓ[t]
    	   ( 1 + x/n^2 + r x^2/n^4 + ⋯) ·( 1 + r x/ℓ^2 + r^2 x^2/ℓ^4 + ⋯) 
         ·1/ℓ^3·( 1 + r y/ℓ^2 + r^2 y^2/ℓ^4 + ⋯)  ·∏_n>ℓ( 1 + y/n^2 + r y^2/n^4 + ⋯)  
        = ∑_ℓ≥1∏_n < ℓ( 1 + x/n^2 - r x) ·ℓ/(ℓ^2 - r x)(ℓ^2 - r y)·∏_n>ℓ( 1 + y/n^2 - r y)  ;

the additional factors involving ℓ arise from contracting the index 3 with the indices 2 appearing before it, and/or after it. 

The general principle is that, as long as the repeating blocks have length 1, the generating function can be expressed by (single or so-called multiple) hypergeometric series.



 §.§ Creative telescoping
 We now give a brief overview of the creative telescoping approach, and the algorithmic implementation in Christoph Koutschan's  package <cit.>, available for . 


Denote by S_z the shift in z operator: S_z f(z)  f(z+1). Let F(a_1,a_2,…,a_m,n) be a term which is hypergeometric (more generally, holonomic with respect to shift operator) in each variable a_i and n. The algorithm of creative telescoping produces Ore polynomials (polynomials in S_a_i, S_n over the field of rational functions in a_i, n) 𝐐_1,𝐐_2  such that 

    𝐐_1 + (𝐒_n-1) 𝐐_2

 annihilates F and such that 𝐐_1 does not involve n. It follows, formally, that

    𝐐_1 ∑_n=0^∞ F(a_1,a_2,…,a_m,n) + lim_n→∞𝐐_2 F(a_1,a_2,…,a_m,n) - 𝐐_2 F(a_1,a_2,…,a_m,0) = 0

so ∑_n=0^∞ F(a_1,a_2,…,a_m,n) satisfies a (possibly inhomogenous) recurrence relation whose coefficient are rational functions of the a_i. 

The  package  by Christoph Koutschan <cit.> can be used to perform these operations: for a given F, the 𝐐_1, 𝐐_2 can be found using the  routine, and the inhomogenous recurrence of ∑_n=0^∞ F(a_1,a_2,…,a_m,n) in each 𝐒_a_ican be directly found using .  As  applies in a quite general setting, we do have to specify that we are interested in the forward difference operation 𝐒_n - 1 as part of a call to the routine, as well as with respect to which variable(s) we want to find the recurrence/annihilator.


Consider

    F(x,y,n) = y^3 1+xn/1-xn (n-x+1) ((n+1)^2-y^2) ,

where xn x(x+1) ⋯ (x+n-1) denotes the ascending Pochhammer symbol.  If we want to find a recurrence in x satisfied by the sum ∑_n=0^∞ F(x,y,n), we call the  routine on F(x,y,n), specify the forward difference operator 𝐒_n - 1, and ask for the recurrence with respect to x as follows.
[]
 <@@> CreativeTelescoping[ y^3 * Pochhammer[1+x, n] / Pochhammer[1-x, n] / 
            (n+1-x) / ((n+1)^2-y^2), S[n]-1, S[x]] // Factor

Inputting and evaluating this returns the following list 
 
*{ { x (1+x)S_x-(x-y) (x+y) }, { -(-1-n+x) (1+n-y) (1+n+y)/2 x } }       

The first element of this list gives 𝐐_1, and the second element gives 𝐐_2, i.e. 

    𝐐_1    = x (1+x)𝐒_x-(x-y) (x+y)  ,
    𝐐_2    = -(-1-n+x) (1+n-y) (1+n+y)/2 x ,

and indeed 𝐐_1 + (𝐒_n-1) 𝐐_2 annihilates F. 

If we want to find a recurrence with respect to y, we change   to , and make the following call instead.
[]
<@@> CreativeTelescoping[ y^3 * Pochhammer[1+x, n] / Pochhammer[1-x, n] / 
            (n+1-x) / ((n+1)^2-y^2), S[n]-1, S[y]] // Factor

This returns

	*{ { -y^2 (1-x+y)S_y+(1+y)^2 (x+y) } , { -(1+n-x) (1+y)^2 (-1-n+y) (1+2 y)/2 y (-n+y) } }       

This time we read off that

    𝐐_1    = - y^2 (1-x+y)𝐒_y + (1+y)^2 (x+y)  ,
    𝐐_2    = -(1+n-x) (1+y)^2 (-1-n+y) (1+2 y)/2 y (-n+y) ,

and indeed 𝐐_1 + (𝐒_n-1) 𝐐_2 annihilates F, now using the shift in y operator 𝐒_y. 

If one is not interested in seeing the certificate of creative telescoping, 𝐐_2, one can directly input
[language=]
<@@> Annihilator[ Sum[ y^3 * Pochhammer[1+x, n] / Pochhammer[1-x, n] / 
          (n+1-x) / ((n+1)^2-y^2), n, 0, Infinity], 
        S[y], Inhomogeneous -> True] // Factor

This returns a list whose first element gives 𝐐_1, and whose second element gives an explicit, unevaluated form (wrapped in ) of the 
boundary term lim_n→∞𝐐_2 F(a_1,a_2,…,a_m,n) - 𝐐_2 F(a_1,a_2,…,a_m,0) appearing <ref>.  Evaluating this command gives the following, where we have manually suppressed the full expression inside the limit[We have used similar notation «⋯» to that employed by  with the  command.].
 
	*{ { -y^2 (1-x+y)S_y+(1+y)^2 (x+y) } , { 
	1/2 (-y-3 y^2-2 y^3+2   𝙷𝚘𝚕𝚍[ lim_n→∞ «⋯» ]) } }       

Again we find the recurrence relation in y is given via

    𝐐_1 = - y^2 (1-x+y)𝐒_y + (1+y)^2 (x+y)
 and one can determine the inhomogeneous part of the recurrence by evaluating the limit[when finding the limit, one may need to impose certain conditions on the variables from context, such condition usually reflects the convergence of ∑_n≥ 0 F(a_1,⋯,a_m,n)].


Creative telescoping gives us a method to determine the recurrence relation satisfied by certain generating series of interest.  We can also directly find general solutions to these recurrences using different techniques.  After doing so we can then explicitly evaluate the genereating series by using analytic properties, and some initial conditions, to determine the unknown coefficients (which will be a 1-periodic functions in variable of the recurrence) which appear in the general solution.  



 §.§ Moderate growth along the imaginary direction
  We now discuss some of the analytic properties we will need, in order to constrain and fix the 1-periodic coefficient functions in the  general solutions to the creative telescoping recurrence relations. 

We start with an elementary observation.

Let f(z) be a meromorphic function of period 1, that is f(z+1) = f(z), and assume it has poles only at ℤ. Let 0≤ C<π, and suppose f(z) also satisfies the following growth condition:

    |f(x+ y)| = O(e^C |y|)      y→∞

uniformly for x in compact subset of ℝ. Then f(z) is a polynomial in π z.


Let k be the order of pole of f(z) at z=0. Since π z has a simple pole, with residue 1 at z=0, we can choose a degree k polynomial P such that h(z) = f(z) - P(π z) is analytic at z=0. Because cotangent grows slowly along imaginary direction, we see h(x+ y) satisfies the same growth condition in <ref> as y→∞ as above, and h is entire by periodicity. Carlson's Theorem <cit.> implies h=0, so f(z) = P(π z), as claimed.


If we generalise the condition that f has poles only at ℤ, by allowing poles at a_1 + ℤ,…, a_r + ℤ, and easy modification of the above proof shows f(z) is a sum of r polynomials, with variables π(z-a_1),…, π(z-a_r), respectively. 

In this article, a meromorphic function is said to have moderate growth if <ref> is satisfied. A typical example would be the polygamma function ψ^(n)(z) ^n+1/^n+1 zΓ(z). Another example is Γ(a+z)/Γ(b+z) for a,b∈ℂ; more generally, we have the following.

Fix a_0,b_0∈ℝ, then we have  
    Γ(a_0+z)/Γ(b_0+z) = O(|z|^a_0-b_0)  ,     uniformly for (t)∈ℝ, (t)>1 .



For 0≤ m≤ 1, this follows from the well-known (e.g. <cit.>) asymptotic of gamma function: 

    Γ(z+a_0)/Γ(z+b_0)∼ z^a_0-b_0 ,     as z →∞ ,

but we still need to show this holds uniformly for all m∈ℝ. We can assume m>0, as the case m<0 will then follow from the reflection formula of the Γ function.  Write m = m_0 + N, with 0≤ m_0≤ 1, and N a large positive integer, then we have

    Γ(a_0 + m_0 +  t + N)/Γ(b_0 + m_0 + t+N) = ( ∏_k=0^N-1(a_0+k+z)/(b_0+k+z))  · Γ(a_0+z)/Γ(b_0+z) ,      z = m_0 +  t  .

The term Γ(a_0+z)/Γ(b_0+z) is uniformly bounded, along the imaginary direction, when 0≤ m_0 ≤ 1, so it suffices to prove that the product has the required asymptotics. By taking logarithm and exponential, the absolute value of ∏_k=0^N-1(a_0+k+z)/(b_0+k+z) is 

    exp[  ∑_k=0^N-1log(1+a_0/k+z)  - ∑_k=0^N-1log(1+a_0/k+z) ] = exp(  ∑_k=0^N-1log| 1-a_0-b_0/k+z+b_0|  )  .

Since |k+z+b_0|≥ 1 by our assumption, we can use the Taylor expansion log(1-x) = x + O(x^2) to conclude the above is 

    = O(∑_k=0^N-1a_0-b_0/k+z+b_0)  ,

then the elementary estimation ∑_k=1^N 1/k = log N + O(1) gives the claim. 


A less trivial moderate growth example, which is relevant to us, is the hypergeometric function with respect to its parameters.


Fix a,b,c,d,e∈ℂ with (d+e-a-b-c)>0, the hypergeometric function

    32a+z,b,cd+z,e1
 is a meromorphic function of z ∈ℂ. The above lemma implies it is of moderate growth in vertical direction in the variable z. To see this, we write above _3F_2 as 

    Γ(e)Γ(d+z)/Γ(a+z)Γ(b)Γ(c)∑_n≥ 0Γ(a+z+n)Γ(b+n)Γ(c+n)/Γ(1+n)Γ(d+z+n)Γ(e+n)

Since Γ(d+z)Γ(a+z) is of moderate growth, and Γ(a+n+z)Γ(d+n+z) = O(|z+n|^a-d) uniformly for n≥ 0, this is indeed of moderate growth. A similar assertion also holds, say, for

    32a+z,b-z,cd+z,e-z1 .





§ EXAMPLES OF CREATIVE TELESCOPING APPLIED TO KNOWN MZV EVALUATIONS


Now we come to the first main part of this note: we apply the ideas of creative telescoping to a number of known evaluations.  By revisiting known examples, we provide a different viewpoint on how to re-derive the results, which might be useful elsewhere.  



 §.§ Examples zetaˆ⁎({1}ˆm, {2}ˆn+1}), and zetaˆ½({1}ˆa, odd)



In Theorem 1.4 of <cit.>, the authors applied ideas from Yamamoto's 2-coloured poset integrals <cit.> to establish the following evaluation,

    ζ^⋆({1}^m, {2}^n+1) = ∑_p+r = n (-1)^r ζ^⋆({2}^p) ·∑_d⃗ = mζ(d_1 + 2, …, d_r+1 + 2) ∏_j=1^r+1 (d_j+1)  .

It follows that ζ^⋆({1}^m,{2}^n+1) is a polynomial in single zeta values for any n,m ≥ 0 ∈ℤ:  
by the symmetric sum theorem <cit.>, the inner sum over all compositions of m into d parts can be expressed as a polynomial in Riemann zeta values.  (In 6.1 <cit.>, the special cases n = 0, 1 are treated, in 6.2 <cit.>, so are m = 1, 2, 3; handling the general case via the symmetric sum theorem does not seem to be explicitly mentioned.)

Extracting a closed formula for particular cases of fixed m or fixed n, by directly applying the symmetric sum theorem, is easy enough for small values n = 0, 1 or m = 1, 2, 3 (c.f. 6.1, 6.2 <cit.>); in general it seems not too enlightening and can be error-prone. Here we outline a method that directly yields the generating function of ζ^⋆({1}^m, {2}^n+1).  Readers interested in computational details for this example should consult the  notebook attached.

As a starting point, we apply the generalised two-one <cit.> formula, which gives us that for m ≥ 1, 

    ζ^⋆({1}^m, {2}^n+1) = 2^mζ^({1}^m-1, 2n+3)  .

Note, however, that we shall have to excluded the terms with m = 0,


    ζ^⋆({2}^n+1) = (-1)^n 2(2^2n+1 - 1) B_2n+2π^2n+2/(2n+2)!  = -2 ζ^(2n+2)  ,

but this evaluation is well-known <cit.> or <cit.>, and can be incorporated easily into the final result, if desired.

So consider the following generating series, where v = 2x/ℓ, u = 2x/n inside the sum,

    ∑_m,n ≥ 0ζ^r({1}^m, 2n+3) · (2x)^m y^2n+3
       = ∑_ℓ≥1∑_n≥0y^2n+3/ℓ^2n+3 (1 + r + v^2 r^2 + ⋯) ·∏_k < ℓ (1 + u + u^2 t + u^3 t^2 + ⋯ ) 
       = ∑_ℓ≥1∑_n≥0y^2n+3/ℓ^2n+3·1/1- v r·∏_n<ℓ(1 + u/1-u r)

Specialise to r =, and after some straightforward calculations, we find

    f(x,y) ∑_m,n≥0ζ^({1}^m, 2n+3) (2x)^m y^2n+3 = ∑_k≥0y^3 1+xk/((k+1)-x)((k+1)^2 - y^2) 1-xk .

The  (or ) routine in  gives us the relation (valid for x<1):

    -y^2  (-x+y+1) f(x,y+1)+(y+1)^2  (x+y) f(x,y)+-2 y^3-3 y^2-y/2=0  .

The general solution to above recurrence in y, is

    f(x,y) = y/2x + π y^2(1 - 2 x · c) π x Γ(x+y)/2x^2 Γ(x - 1) Γ(x) Γ(1 - x + y) .
 Here c = c(x,y) is a “constant” in y, in the sense that is unchanged under y→ y+1, i.e. c(x,y) is 1-periodic function in y[the recurrence can be solved using the built-in  command of  and simplification afterwards]. 
For generic fixed x, f(x,y) has simple poles at y∈ℤ, and it is of moderate growth in y along imaginary direction, so by  <ref>, 

    c(x,y) = A(x) + B(x) π y ,
 for some meromorphic A(x), B(x) depending only on x.

The functions A(x), B(x) can be fixed by considering the coefficients of y^1, y^2, which must vanish identically since f(x,y) starts with order y^3.  Solving the resulting system of equations tells us

    A(x) = 1 - x - π  x π  x/2(1 - x)x ,      B(x) = -π/2 (1-x) .

Substituting these back into c(x,y), and simplifying, finally gives us our desired generating series
	
    f(x,y) = y/2x + π^2 x y^2 (1 - π x - π y) π xtanπ(x+y)/2(x+y)·Γ(1 + x + y)/Γ(1+x)^2 Γ(1 - x + y) .


In general, we can then extract and calculate (for m≥1,n≥0) that

    ζ^⋆({1}^m, {2}^n+1) = 2 · [x^m-1 y^2(n+1)+1]   f(x,y)  .

Using the Taylor expansion logΓ(1+z) = -γ z + ∑_k=2^∞(-z)^k/kζ(k), we hence conclude that both ζ^⋆({1}^m,{2}^n+1) and ζ^({1}^m-1, 2n+3) are polynomials in Riemann zeta values.  For example (after some work) one obtains,

    ζ^⋆({1}^5,{2}^4) = 2 · [x^5-1 y^2·4 + 1]   f(x,y) 
       -2π ^4/45ζ (3)^3+20 ζ (7) ζ (3)^2+16 ζ (5)^2 ζ (3)-17 π ^10/31185ζ (3)-π ^8 /180ζ (5)-4 π ^6/63ζ (7)-28 π ^4/45ζ (9)+198 ζ (13) .


 


 §.§ Example zeta({2}ˆn, 3, {2}ˆm)



We show how Zagier's celebrated result (<cit.>) evaluating the MZV  ζ({2}^m,3,{2}^n) can be (somewhat directly) re-derived using the creative telescoping methods. Readers interested in computational details for this example should consult the  notebook attached.

First we write down the generating function (very similar to Proposition 1 in <cit.>):

    h(x,y) ∑_n,m≥ 0ζ({2}^m,3,{2}^n) (-x^2)^n (-y^2)^m = ∑_ℓ≥ 11/ℓ^3∏_n<ℓ(1-x^2/n^2) ∏_n>ℓ(1-y^2/n^2)  .

In terms of hypergeometric functions, this is[It can also be written as a derivative of _3F_2 with respect to a parameter, as per <cit.>.]

    = 1/Γ(2-y)Γ(2+y)431,1,1-x,1+x2,2-y,2+y1 .

Our goal is to explicitly find above _4F_3 in terms of polygamma functions, this will give an explicit reduction of ζ({2}^a,3,{2}^b) into single zeta values.  

It turns out that it is more convenient to work with the following modified version[After the relabelling of x,y, the resulting function is of moderate growth w.r.t. each variable by <ref>, whereas the original version is not.]  of h(x,y) obtained by replacing (x,y) by (x-y,x+y), and plus a rational factor.  So consider

    f(x,y)    ∑_m≥ 0(y^2-1) (x-y)^2/(x+y)^2-1·1m1-x+ym1+x-ym/2m2-x-ym2+x+ym_ F(x,y,m)
        = (x-y)^2 (y^2 - 1)/(x+y)^2 - 1431,1,1-x+y,1+x-y2,2-x-y,2+x+y1 .


Using  (or ) in , plus some further human-guided simplification afterwards, we find that f(x,y) satisfies the recurrence

    f(x,y)-f(x+1,y)  = (y^2-1) ( ( 1/4x - 1/x+y - 1/1+x+y + 1/1+2x) + π(x+y) sinπ(x-y)/4x(1+x)(1+2x))  .


The above can be viewed as a first order recurrence in x, and one can verify that

    g(x,y) =    y^2 - 1/4 x y(  (x-y)^2/x+y - (x-y) π(x+y)sinπ(x-y)) 
       [t]  + (y^2 -  1)    { 2 ψ(1 + x + y) - π(x+y)sinπ(x+y)· ( ψ(1+2x) + ψ(1 + 2y) )  
        - π(x+y)sinπ(x-y)·( ψ(1+x) - ψ(1 + 2x) - ψ(1 + y) + ψ(1 + 2y) ) }

also satisfies this recurrence.   (This is easily checked, but beware that  seems to be unreliable and produces incorrect – but closely related results – in this case.)

Therefore f(x,y) = g(x,y) + c(x,y) where c(x,y) is 1-periodic in x. By looking at the definitions of f and g, we see that for generic y, f(x,y) and g(x,y) could have poles only at x∈ -y +  ℤ. Note f is analytic at x=-y; it can be checked easily that g(x,y) is also analytic at x=-y.  Therefore c(x,y) is analytic at this point, thus by periodicity, it is an entire function of x.  Being of moderate growth in x, we see c(x,y) depends only on y, so we have

    c(x,y) = c(0,y) = f(0,y) - g(0,y)  .

However, f(0,y) can be easily calculated, because F(x,0,m) reduces a rational function in x,m.  We find

    f(0,y) = 1/2 (y^2 - 1) ( 2γ - y^-1 + ππ y + 2 ψ(1+y) ) ,

and this agrees with g(x,y) as x → 0.  Therefore c(x,y) is identically 0, so we have proved f(x,y) = g(x,y)

 If one has not so presciently guessed the form of g(x,y), one would take any other solution to the recurrence, and find by <ref>, that for generic y, c(x,y) = A(y) + B(y) π(x+y), as f(x,y) - g(x,y) is 1-periodic, of moderate growth in x, with at most a simple pole at x ∈ -y + ℤ.  Analyticity of f(x,y) at x = -y, and considering x = 0 would lead to equations for A(y), B(y), which, when solved, again prove the evaluation of f(x,y).


We then recover

    h(x+y,x-y) = ((x+y)^2 - 1) · f(x,y)/Γ(2-x-y)Γ(2+x+y)(x-y)^2 (y^2 - 1)  = -sinπ(x+y)· f(x,y)/π(x-y)^2 (x+y)(y^2 - 1) ,

or by replacing (x,y) by (x+y2, -x+y2), we directly have

    h(x,y) = 4 sinπ y/π x^2 (4 - (x-y)^2) f( x+y/2, -x+y/2)  .

Using the Taylor series expansion ψ(1+z) = -γ + ∑_k≥ 2ζ(k) (-z)^k-1, we therefore see the MZV's ζ({2}^n,3,{2}^m) can indeed be expressed via Riemann zeta values.



 §.§ Example t({2}ˆn, 3, {2}ˆm})



Murakami <cit.> showed that t({2}^n,3,{2}^m) can be expressed in terms of Riemann zeta values, and from this deduced { t^𝔪(k_1,…,k_d) | k_i = 2, 3 } form a basis of the space of motivic multiple zeta values (as well as showing generally any MtV with all indices >1 enjoys a Galois descent to MZV's).  We give an alternative proof of Murakami's evaluation, using similar manipulations as above.  Readers interested in computational details for this example should consult the  notebook attached[we remark that the formulas involved in this example on t({2}^m,3,{2}^n) are much shorter than those involved in ζ({2}^m,3,{2}^n) above.].  

Consider the following generating series of the MtV's

    ∑_n,m≥ 0   t({2}^n,3,{2}^m)    · 2^3 (-4x^2)^n (-4y^2)^m 
       = ∑_ℓ≥ 01/(ℓ+12)^3∏_0≤ n<ℓ(1-x^2/(n+12)^2) ∏_n>ℓ( 1-y^2/(n+12)^2)
    
       = ∑_ℓ≥ 02 cosπ  x Γ(ℓ-x+12) Γ(ℓ+x+12)/(2 ℓ+1) Γ(ℓ-y+32) Γ(ℓ+y+32) .

As before, it is more convenient to work with the following version, where  (x,y) is replaced by (x-y,x+y).  So our goal is to express

    f(x,y) ∑_k≥ 02 cosπ  (x-y)  Γ(k+x-y+1/2) Γ(k-x+y+1/2)/(2 k+1)  Γ(k-x-y+3/2) Γ(k+x+y+3/2)

 in terms of polygamma functions. 

Using , we find that f(x,y) satisfies the following recurrence in x

    f(x+1,y) (-x^2-2 x     +y^2-1 ) 
    +f(x,y) ( y^2-x^2 ) 
         +(8 x^2+8 x+1) cosπ  (x+y) - cosπ  (x-y)/4 x (x+1) (2 x+1)=0  .

The general solution to this is 

    f(x,y) = 
    	[t] 
    	 cosπ(x-y)/4(x^2 - y^2)   ( 4(1-y^2) · c + x^-1 + 3 + 4 ψ(x) - 4 ψ(2x) ) 
        + cosπ(x+y)/4(x^2 -y^2)( -x^-1 + 5 - 4 γ - 4 ψ(2x) )  .

where c(x+1,y) = c(x,y).  (This is easily checked, however beware that  seems to be unreliable and produces incorrect – but closely related – results also in this case.)  

For generic y, f(x,y)/cosπ(x-y) is of moderate growth in x, it has poles at x∈ y+(ℤ+1/2), and the various digamma functions above also produce a simple pole at x∈ℤ+1/2. Therefore <ref> tells us

    c(x,y) = A(y) + B(y) tanπ(x-y) + C(y)tan (π x)  ,

with A(y), B(y), C(y) to be determined. 

To determine A(y), B(y), C(y), we expand the tentative solution f(x,y)  at x = ± y, 0.  At x = ± y, it must analytic, so the coefficient of (x± y)^-1 must vanish identically.  At x = 0, one can directly evaluate <ref>, as the summand becomes a simple rational function (and cosπ y), so the tentative solution must also give

    f(0,y) = -cosπ y/y^2( γ - ψ(y) + 2 ψ(2y)  ) + πsinπ y/2 y^2 .

From these conditions, we find

    A(y) = 1 - cos2π y/4y(y^2 - 1) + 1/y^2 - 1( 3/4  + 5/4cos2 π y + (ψ(y) + γ) - 2 (ψ(2y) + γ) cos^2π y) 
    
    		B(y) = sin2π y/y^2 - 1( 1-5y/4y + γ + ψ(2y) ) 
    
    		C(y) = 0  .


Substituting these into the tentative solution, and simplifying, gives the following final closed form solution

    f(x,y) = [t]
    	sinπ xsinπ y/2 x y (x + y)    + cosπ(x-y)/x^2 - y^2( ψ(1 + x) - ψ(1 + y)) 
        - 2cosπ xcosπ y/x^2 - y^2( ψ(1 + 2x) - ψ(1 + 2y) ) .

The original generating series we wanted, h(x,y), is then recovered simply as h(x,y) = f(x+y/2, -x+y/2).  This shows that t({2}^n, 3, {2}^m) can indeed be expressed by Riemann zeta values, by applying the Taylor series expansion ψ(1+z) = -γ + ∑_k≥ 2ζ(k) (-z)^k-1.



	——————



We now apply creative telescoping methods to some (apparently) new MZV-type evaluations.  With the new results, we settle some conjectures or questions raised elsewhere.


§ RESULTS ON ZETAˆ½({-2}ˆM,3), ZETAˆ½(3,{-2}ˆM), AND ZETA STARS INVOLVING {1,3}ˆM



	In this section, we show an evaluation for a combination of _6F_5 hypergeometric series, and use this to give explicit evaluations of ζ^({2}^n, 3), and ζ^(3,{2}^n) in terms of Riemann zeta values.  By applying the generalised two-one theorem we convert these into four families of multiple zeta star value evaluations. 

    As a starting point, using the principal of <ref>, we can write down the following generating series, where u=(-1)^n x/n^2, v = (-1)^ℓ x/l^2 inside the summation.
	
    ∑_k≥ 0ζ^r({2}^k,3) x^k    = ∑_ℓ≥ 11/ℓ^3 (1+v r+v^2 r^2 + ⋯)∏_n<ℓ  (1+u+u^2 r + u^3 r^2 +⋯ ) 
       = ∑_ℓ≥ 11/ℓ^31/1-vr ∏_n<ℓ(1+u/1-ur)  ,
       
    ∑_k≥ 0ζ^r(3,{2}^k) x^k    = ∑_ℓ≥ 11/ℓ^3 (1+vr+v^2r^2 + ⋯)  ∏_ℓ>n  (1+u+u^2 r + u^3 r^2 +⋯ ) 
       = ∑_ℓ≥ 11/ℓ^31/1-vr ∏_n>ℓ(1+u/1-ur)  .

    It is most convenient to work with the generating series weighted by (8x^2)^k instead of just x^k; we must also specialise to r =.  If we split the resulting sum into ℓ = 2m+2 even, and ℓ = 2m+1 odd terms (m≥0), we can write
    
    ∑_k≥0ζ^({2}^n, 3) (8x^2)^k = 
        	-1/x^2∑_m≥0( F_even(x, - x, m) + F_odd(x, - x, m) )  , 
    ∑_k≥0ζ^(3,{2}^n) (8x^2)^k = 
        	π xtanhπ x/x^2∑_m≥0( F_even( x, x, m) + F_odd( x, x, m) )  ,

    where
    
    F_even(x,y,m)     = -(4x^2-1)y^2/8(x^2-1)(4y^2-1)·1m^2 32-xm32+xm1-ym1+ym/2m2-x2+xm3/2-ym3/2+ym·1/m! ,
       
    
       F_odd(x,y,m)     = 
       	-y^2/4y^2-1·12m1m12-x12+xm1-ym1+ym/32m1-xm1+xm32-ym32+ym·1/m! ,

   and the factor (π x) tanh(π x) comes from  
    (π x) tanh(π x) = ∏_n=1^∞( 1 + u/1 - 1/2 u)  ,      u = (-1)^n x/n^2 .

	Definitionally then, one has
	
    ∑_m≥0 F_even(x,y,m)    = -(4x^2-1)y^2/8(x^2-1)(4y^2-1)·651,1,32-x,32+x,1-y,1+y2,2-x,2+x,32-y,32+y1 , 
    ∑_m≥0 F_odd(x,y,m)    = -y^2/4y^2-1·6512,1,12-x, 12+x, 1-y, 1+y32, 1-x, 1+x, 32-y,32+y1 .


	On account of the following theorem, we can evaluate this combination of _6F_5 functions through digamma and trigonometric functions.


For all x,y∈ℂ, the combination of _6F_5 functions

    ∑_m≥ 0( F_even(x,y,m) + F_odd(x,y,m) )
 
is equal to the following 

    1/8ψ (1+x+y)
    +1/8ψ(1+x-y)
    -1/4ψ(1+x)
    +1/4ψ(1+y)
    -1/2ψ(1+2 y)
    +1/8 x
    -γ/4
    +π/8tanπ y
        
    +π  xtanπ  y-1/16 (x-y)
    -π  xtanπ y+1/16 (x+y)
    +1/8π  xtanπ  y·(ψ(1+x+y)-ψ(1+x-y))
     .



Using the Taylor series expansion ψ(1+z) = -γ + ∑_k≥ 2ζ(k) (-z)^k-1, we directly obtain the following explicit evaluations of ζ^({2}^n,3) and ζ^(3,{2}^n).

 
	We have the following explicit reduction of ζ^({2}^n,3) and ζ^(3,{2}^n) into Riemann's zeta, with the convention that ζ(0) = -1/2:

    ζ^({2}^n,3) =
    [t]
      (-δ_neven/2 · 8^n +    δ_nodd/2 · (-4)^n + 2/(-2)^n) ζ(2n+3) 
        + 8/3ζ(2)∑_a + b + 2c = n+1 
     a,b,c ≥ 0(-1)^a+c(1 - 2^2a)/2^3a + 3b + 4cζ(2a)ζ(2b) ζ(4c+3)  ,
     
       ζ^(3,{2}^n) = [t]
        -^n  δ_neven/2^2n+1ζ(2n+3) 
         + 1/6ζ(2)∑_a + b + c = n + 1 
     a,b,c≥0[t] 
     (-1)^a (1 -2^2a)/8^n   ζ(2a)ζ(2b) ζ(2c+3) 
       ·( 2  δ_ceven+ (2)^c+1 δ_codd  - 8 · 4^c  )  .


Here we extend the Kronecker delta so that δ_∙ = 1 if ∙ is true, and δ_∙ = 0 otherwise.[Note: Since the factor 1 - 2^2a = 0 for a = 0, each term of the sum necessarily contains a factor of π^2, which can be cancelled against the ζ(2) in the denominator.  Although seeing division by ζ(2) in evaluations is somewhat unusual, this form seems to be the best way to write the results. Alternatively one could simplify ζ(2a)/ζ(2) using the known evaluation, at the expense of introducing Bernoulli numbers and factorials.]


By the generalised two-one theorem <cit.> (c.f. also <cit.>, for a description via the block decomposition, which gives a more direct way to pass from ζ^⋆ to ζ^ arguments), we have

    ζ^⋆({1,3}^n,1,2)    = 2^1+2nζ^({2}^2n,3) 
    ζ^⋆(2,{1,3}^n,1,2)    = -2^2+2nζ^({2}^2n+1,3) 
    ζ^⋆(1,2,{1,3}^n)    = 2^1+2nζ^(3,{2}^2n) 
    ζ^⋆(2,3,{1,3}^n)    = -2^2+2nζ^(3,{2}^2n+1)  .

Therefore these multiple zeta star values can also be expressed as polynomials in Riemann zeta values.  In particular we have the following explicit evaluations.


	The following evaluations hold for all n ≥ 0, with the convention that ζ(0) = -1/2:
	
    ζ^⋆({1,3}^n, 1,2)    = (4 - 1/16^n) ζ(4n+3) + 8/3ζ(2)∑_a + b + 2c = 2n+1 
     a,b,c ≥0(-1)^a+c (1 - 2^2a) /4^a+b+cζ(2a)ζ(2b)ζ(4c+3)  , 
    ζ^⋆(2,{1,3}^n,1,2)    = ( 4 + 1/2 ·(-4)^n) ζ(4n+5) - 8/3ζ(2)∑_a + b + 2c = 2n+2 
     a, b, c ≥ 0(-1)^a+c (1-2^2a)/4^a+b+cζ(2a)ζ(2b)ζ(4c+3)  , 
     andζ^⋆(1,2,{1,3}^n)    = -ζ(4n+3)/(-4)^n + 1/3 ζ(2)∑_a + b + c = 2n + 1 
     a,b,c ≥ 0[t]  (-1)^a (1 - 2^2a)/16^n   ζ(2a)ζ(2b)ζ(2c+3)  
       · (2  δ_ceven + (2)^c+1 δ_codd - 8 · 4^c)  , 
    ζ^⋆(2,3,{1,3}^n)    = -1/12 ζ(2)∑_a + b + c = 2n + 2 
     a,b,c ≥ 0[t]
    		  (-1)^a (1 - 2^2a) /16^n   ζ(2a)ζ(2b)ζ(2c+3) 
       · (2  δ_ceven + (2)^c+1 δ_codd - 8 · 4^c)   .





	
		Some parts of these evaluations can be handled with more standard MZV techniques; here is a brief outline.  Using the stuffle-antipode (c.f. <cit.>, <cit.>, <cit.>), entailing stuffle-regularisation, we have that
		
    ζ^⋆(1,{1,3}^n) = 
       ∑_k=0^n ζ^⋆({1,3}^k)  _∗,Tζ({3,1}^n-k, 1) - ∑_k=0^n-1ζ^⋆(3,{1,3}^k)  _∗,Tζ(1,{3,1}^n-1-k, 1)  .

		
		From <cit.>, we can deduce that ζ^⋆({1,3}^k) and ζ^⋆(3,{1,3}^k) are polynomials in Riemann zeta values (the former is given explicitly in Theorem B therein, for the latter we need to know that ζ(3,{1,3}^k) is a polynomial in Riemann zeta values, which is given in Theorem 1 of <cit.>).
		
		By considering shuffle regularisation, and the shuffle products of ζ({2}^i) with ζ_1({2}^j), the dual of ζ({2}^j,1), and of  ζ({2}^k,1) with itself (c.f. <cit.>),  we have
		
    ∑_i=0^2n+1 (-1)^i _,T ζ({2}^i) ·ζ_1({2}^2n+1-i)    = (-1)^n 2^2n_,T ζ({3,1}^n, 1)  , 
    ∑_i=0^2n (-1)^i _,T ζ({2}^i,1) ·ζ({2}^2n-i,1)    = (-1)^n 2^2n+1_,T ζ({1,3}^n,1, 1)  .
 
		As ζ({2}^j,1) is a polynomial in single-zetas (c.f. <cit.>), thence ζ_1({2}^j) by duality, we see that the shuffle regularised ζ({3,1}^n, 1) and ζ({3,1}^n, 1, 1) are as well.  This also holds true after switch from shuffle regularisation to stuffle regularisation (via the recipe established in  <cit.>).   (Alternatively, one can differentiate the generating series expression <cit.> for ∑_n _{1,3}^n(x) t^4n, and apply know asymptotics of hypergeometric functions, c.f. <cit.>, and <cit.>.)   Ultimately, one finds
		 
			
    ζ^⋆(1,{1,3}^n+1) = -1/2 ·(-4)^nζ(4n+5) + 8/3ζ(2)∑_a + b + 2c = 2n+2 
     a, b, c ≥ 0(-1)^a+c (1-2^2a)/4^a+b+cζ(2a)ζ(2b)ζ(4c+3)  .

		Finally by applying the cyclic sum theorem <cit.> to the word (x^2y^2)^n, one has (as was pointed out to SC by Michael Hoffman)
		
    ζ^⋆(2,{1,3}^n,1,2) = -ζ^⋆(1,{1,3}^n+1) + 4 ζ(4n+5)  ,

		so ζ^⋆(2,{1,3}^n,1,2) is also a polynomial in single-zeta values.  
		
		Using the stuffle-antipode of zeta-half values and the expressions from 
		
			 #1Equations (#1)<ref>
		
		–(<ref>) (alternatively using a generalisation of techniques from <cit.>), one can derive formulae for ζ^⋆(1,2,{1,3}^n) and ζ^⋆({2,3,{1,3}^n) in terms of ζ^⋆(2,{1,3}^n,1,2) and ζ^⋆({1,3}^n,1,2), so the only essential evaluation remaining is that of ζ^⋆({1,3}^n,1,2).
	
    
	
		From the explicit formula above for ζ^⋆({1,3}^n, 1,2), and the following formula for ζ^⋆(3,{1,3}^n) derivable from <cit.> (c.f. also Theorem 9 <cit.>),
		
    ζ^⋆(3,{1,3}^n) = - 8/3ζ(2)∑_a + b + 2c = 2n+1 
     a,b,c ≥0(-1)^a+c (1 - 2^2a) /4^a+b+cζ(2a)ζ(2b)ζ(4c+3)  ,

		we see that the following combination simplifies dramatically
		
    ζ^⋆({1,3}^n, 1,2) + ζ^⋆(3,{1,3}^n) =  (4 - 1/16^n) ζ(4n+3) = -4 ζ(4n+3) .

		This is similar in form (but nevertheless different) to the cyclic sum identity arising in <ref>.  We venture to suggest there is some extension to the cyclic sum theorem in the setting of alternating interpolated MZV's, for which this is a special case.
	

	
		Broadhurst <cit.> showed that
		
    ∑_n=0^∞_{1,3}^n(x) t^4n = 211+2 t, -1+2 t1x211-2 t, -1-2 t1x ,

		as both sides are annihilated by the differential operator
		
    ( (1-x)/ x)^2 ( x / x)^2 - t^4  ,

		and have the same series expansion to order O(x^5).  By taking / x, and applying the asymptotic formula (c.f. <cit.>) for x → 1^-,
		
    211+z,1-z2x = 2 γ - ψ(1+z) - ψ(1-z) - log(1-x)/Γ(1+z)Γ(1-z) + O((1-x)log(1-x))   ,

		and taking x → 1^-, one obtains a generating series expression for
		
    ∑_n=0^∞ζ({1,3}^n, 1,2) t^4n+4

		in terms of trigonometric and digamma functions, showing ζ({1,3}^n, 1,2) is a polynomial in Riemann zeta values.  Can one give a similar differential equation proof for the evaluation of ζ^⋆({1,3}^n) and more importantly ζ^⋆({1,3}^n,1,2), by establishing a hypergeometric^(?) evaluation for
		
    f(x,y) ∑_n=0^∞^⋆_{1,3}^n(x) t^4n ,

		which satisfies
		
    ( (1-x)x / x)^2  ( x / x)^2 f(x,y) - t^4 f(x,y) = -t^4 (1-x)  .

		(Here ^⋆ is the natural “star-ification” of the classical multiple polylogarithms, c.f. Le_s_!,…,s_d(t) in 10.6, the proof of Lemma 10.6.1, & Equation (10.14) <cit.>.)
		This could allow a more conceptual proof for the evaluation of ζ^⋆({1,3}^n,1,2) bypassing the non-conceptual creative telescoping approach above.
	

	Using the above evaluations, and properties of MZV's, we can derive another apparently new MZSV evaluation involving the repeating {1,3}^n string.


	The following evaluation holds
	
    ζ^⋆(1,2,{1,3}^n,1,2) =  
       16/3ζ(2)∑_a + b + 2d + 2j = 2n+2 
     a, b, d, ≥ 0 , j ≥ 1(-1)^b+d(2 - (-4)^-j)(1 - 4^a)/2^2a + 2b + 2dζ(2a)ζ(2b)ζ(4d+3) ·ζ(4j+1) 
        - 8/3ζ(2)∑_a + b + 2d + 2j = 2n+1 
     a, b, d, j ≥ 0(-1)^b+d(4 - 4^-2j)(1 - 4^a)/2^2a + 2b + 2dζ(2a)ζ(2b)ζ(4d+3) ·ζ(4j + 3) 
       -2∑_j+k = n+1 
     j,k ≥ 1 (2 - (-4)^-j)(2 - (-4)^-k) ζ(4j+1)ζ(4k+1) 
       + 1/2∑_j+k = n+1 
     j,k ≥ 0{ (4 - 4^-2j)(4 - 4^-2k) + (-4)^-j-k}ζ(4j+3)ζ(4k+3)   .

	
	
		From the stuffle-product of zeta-half values (c.f. <cit.>), we have
		
    2 δ_nevenζ^(3,{2}^n, 3) = ∑_i=0^n (-1)^i ζ^({2}^i, 3) ζ^({2}^n-i, 3)  .

		By <ref>, ζ^({2}^i, 3) is a polynomial in Riemann zeta values, and therefore so is the whole right hand side.  Finally, by the generalised two-one theorem (c.f. <cit.>, <cit.>), we have
		
    ζ^(3,{2}^2n,3) = 2^-2n-2ζ^⋆(1,2,{1,3}^n,1,2)

		so the result follows.
	



	A similar result for ζ^(3,{2}^2n+1, 3) = -2^-2n-3ζ^⋆(2,{3,1}^n+1,2), fails, as already 
    ζ^(3,2,3) = -2^-3ζ^⋆(2,3,1,2) = -27/40ζ (3,5)+1/16π ^2 ζ (3)^2-9/4ζ (5) ζ (3)+1769 π ^8/10368000
 involves the (apparently) irreducible weight 8 depth 2 element ζ(3,5).



	 Recall the well-known results <cit.> that ζ^⋆({1,3}^n), ζ^⋆(3,{1,3}^n) are polynomials in single-zeta values (c.f. <ref>), and a corresponding result  <cit.> for ζ^⋆(2,{1,3}^n) = -2^2n+1ζ^({2}^2n+1) (which then also follows by <cit.>; the conversion to MSV's is via the generalised two-one theorem <cit.>).  In <ref> we also presented a similar result for ζ^⋆(1,{1,3}^n).
	Combining these, with the results of <ref> and <ref> we obtain the following list seems, which seems to exhaust the convergent MZSV's involving repeating 1,3 arguments which evaluate as single-zeta polynomials.
	
    [t]
    			    ζ^⋆({1,3}^n)  ,   ζ^⋆(3,{1,3}^n)  ,   ζ^⋆(2,{1,3}^n)  ,   ζ^⋆(1,{1,3}^n)  , 
    
    			   ζ^⋆({1,3}^n,1,2)  ,  ζ^⋆(2,{1,3}^n,1,2)  ,  ζ^⋆(1,2,{1,3}^n)  ,  ζ^⋆(2,3,{1,3}^n)  ,   
       ζ^⋆(1,2,{1,3}^n,1,2)

	
	There are a number of further evaluations, which can be obtained for stuffle regularised MZSV's which have trailing 1's.  In particular, we have that
	
    [t]
    		    _∗,Tζ^⋆({1,3}^n,1)  ,   _∗,Tζ^⋆(3,{1,3}^n,1)  ,   
       _∗,Tζ^⋆(2,{1,3}^n,1)  ,   _∗,Tζ^⋆(1,{1,3}^n,1)

	are polynomials in Riemann zeta values and T, the regularisation parameter.  Again this list seems to be exhaustive, for stuffle regularised MZSV's with at least one trailing 1.  We briefly sketch how to show these results.

  
Case _∗,Tζ^⋆({1,3}^n,1):  Apply the stuffle-antipode to ζ^⋆({1,3}^n), to write
	
    ∑_i=0^n _∗,Tζ({3,1}^i) ·ζ^⋆({1,3}^n-i) - ∑_i=0^n-1ζ(3,{1,3}^i) ·_∗,Tζ^⋆({1,3}^n-1-i, 1) = 0  .

	From the known evaluations of _∗,Tζ({3,1}^j)  , ζ^⋆({1,3}^j), and _∗,Tζ({1,3}^ℓ,1), the claim follows. 

  
Case _∗,Tζ^⋆(3,{1,3}^n-1,1) Apply the stuffle-antipode to _∗,Tζ^⋆({3,1}^n) to write
	
    ∑_i=0^nζ({1,3}^i) ·_∗,Tζ^⋆({3,1}^n-i) - ∑_i=0^n-1_∗,Tζ({1,3}^i,1) ·ζ^⋆(3,{1,3}^n-1-i) = 0  .

	From the known evaluations of ζ({1,3}^j) , _∗, Tζ^⋆({3,1}^k), and _∗,Tζ({1,3}^ℓ,1), the claim follows. 

	

  
Case _∗, Tζ^⋆(1, {1,3}^n, 1: Apply the stuffle-antipode to _∗,Tζ^⋆(1,{1,3}^n,1) to write
	
    ∑_i=0^n ζ({1,3}^i) ·_∗,Tζ^⋆(1, {1,3}^n-i,1)  - ∑_i=1^n _∗,Tζ({1,3}^i, 1) ·ζ^⋆(1,{1,3}^n-i)    
    
    		+ ζ({1,3}^n, 1, 1)     = 0  .

	From the known evaluations of ζ({1,3}^j) , _∗,Tζ({1,3}^k,1) , ζ^⋆(1,{1,3}^ℓ) and _∗,Tζ({1,3}^m, 1,1) (see <ref> above), the claim follows.

  
Case _∗,Tζ^⋆(2,{1,3}^n, 1): Apply the stuffle-antipode to ζ^⋆(2,{1,3}^n, 1) to write
		
    ∑_i=0^n ζ({1,3}^i) ·_∗,Tζ^⋆(2, {1,3}^n-i,1)  - ∑_i=1^n _∗,Tζ({1,3}^i, 1) ·ζ^⋆(2,{1,3}^n-i)    
    
    		+ ζ({1,3}^n, 1, 2)     = 0  .

		From the known evaluations of ζ({1,3}^j) , _∗,Tζ({1,3}^k,1) , ζ^⋆(2,{1,3}^ℓ) and ζ({1,3}^m, 1,2), the dual of ζ(3,{1,3}^m), the claim follows.


We now turn to the proof of <ref>, which implies all of the above MZSV and MZV evaluations.


	
We give a proof of this Theorem via Wilf-Zeilberger pairs, which is a special case of creative telescoping. 

Introduce

    F(x,y,m)     F_even(x,y,m) + F_odd(x,y,m) 
       = [t]
    	   y^2(4m+3)(4x^2 - 1)(y^2-1)/8(m+1)(2m+1)(x^2 - 1)(4y^2-1)· ((2m+1)^2-4x^2) · ((m+1)^2-y^2)
       · (4m^2 + 6m - 4x^2 + 3) ·32-xm32+xm2-ym2+ym/2-xm2+xm32-ym32+ym .

Let g(x,y) denote the digamma and trigonometric expression in statement of <ref>. We need to prove that f(x,y) := ∑_m≥ 0 F(x,y,m) = g(x,y). To this end, let C(x,y) = f(x,y) - g(x,y). 


The functions f(x,y) and g(x,y) are analytic with possible isolated poles only at (x,y)∈ℤ×(ℤ+12).


For g(x,y) this is easy to check. For f, its poles come from the terms 2± xm, 32± ym in denominator of F(x,y,m) (or from x^2 - 1, or 4y^2 - 1) resulting in a pole whenever 2 ± x, 32± y are negative integers.  


We remark that the series ∑_m≥ 0 F(x,y,m) is absolutely convergent for all x,y∈ℂ, (ignoring issue near those poles), as F(x,y,m) is O(m^-3) for large m. 


The function C(x,y) = f(x,y) - g(x,y) is analytic at (x,y)=(0,12). 


We calculate the Laurent expansion of f and g at (x,y)=(0,12).  For g this is easy,

    g(x,y) = 1/2 π ^2-1/8/y-1/2 + O(1)  .

For f, note that we already have 12y-1 (from the denominator factor 4y^2-1) in the expression of f, so around (x,y) = (0, 12) we have

    f(x,y) = 1/2y-1∑_m≥ 0- 3 (4 m+3) (4 m^2+6 m+3) /256 (m+12) (m+1) (m+32) (2 m+1)^3 ·32m^3 52m/1m2m^3_ a_m + O(1)  .

one checks

    a_m = Δ_m ( - 3 (m+1)^3 (4 m^2+2 m+1) /4 (2 m+1)^4 (2 m+3)·32m^3 52m/1m2m^3 ) := Δ_m b_m  ,

where Δ_m 𝐒_m - 1 denotes the forward difference operator, i.e. Δ_m h(m) = h(m+1) - h(m).  So the sum telescopes and one finds

    ∑_m≥ 0 a_m = lim_m→∞ b_m - b_0 = -1/4 - 1/π^2 .

Thus around (x,y) = (0, 12), we also have 
    f(x,y) = 1/2 π ^2-1/8/y-1/2 + O(1)

whence C(x,y) = f(x,y) - g(x,y) is indeed analytic there as claimed.


By <ref>, both g(x,y),f(x,y) have moderate growth along the imaginary direction if we fix one variable, and let the other vary.  The following observation is the core of our argument.

	The function C(x,y) is 1-periodic in each variable, i.e.

    C(x+1,y) = C(x,y)  ,      C(x,y+1) = C(x,y)  .



Let Δ_x = 𝐒_x - 1 be the forward difference operator, i.e. Δ_x h(x) = h(x+1)-h(x). We need to prove Δ_x f(x,y) = Δ_x g(x,y) and Δ_y f(x,y) = Δ_y g(x,y). Write 
    r_1(x,y,n) =     -(m+1) (2 m+1) (m-x+1) ((2m+1)^2 - 4y^2)/(4 m+3) x (2 m-2 x-1) (4 m^2+6 m-4 x^2+3) 
       ·( 4m^2 + 2m - 4x^2 - 8x - 9/(x+1)^2 - y^2 + 2m^2 + 4m^2 x + 2m x + m - 5x - 8y^2 - 4xy^2 - 1/(x^2 - y^2)((x+1)^2 - y^2)) 
    
    r_2(x,y,n) =    (m+1) (2 m+1) (m-x+1) (m+x+1) (2 m-2 y+1)/(4 m+3) y (4 m^2+6 m-4 x^2+3 ) 
       ·( 4m^2 + 2m - 4x^2 + 4y + 3/(x^2 - (y+1)^2)(m-y) - 4y^2 + 4m y + 4y + 2m + 1/(x^2 - y^2)(x^2 - (y+1)^2))  .

One checks 
    Δ_x F(x,y,m) = Δ_m (r_1(x,y,m) F(x,y,m))  ,     Δ_y F(x,y,m) = Δ_m (r_2(x,y,m) F(x,y,m))

so 

    Δ_x f(x,y) = ∑_m≥ 0Δ_x F(x,y,m)    = ∑_m≥ 0Δ_m (r_1(x,y,m) F(x,y,m)) 
       = lim_m→∞ r_1(x,y,m)F(x,y,m) - r_1(x,y,0)F(x,y,0)  .

One calculates the limit (it gives factor π x tanπ y that appears in g(x,y)).  Then by comparing the above to Δ_x g(x,y), which is readily calculable,  one sees – after some simplification of digamma functions – that is agrees with Δ_x f(x,y).  So Δ_x g(x,y) = Δ_x f(x,y) or equivalently C(x+1,y) = C(x,y). For the variable y the calculation is analogous.  This completes the proof of the lemma.


Finally we argue C(x,y) = 0 identically.  The 1-periodicity and holomorphicity at (0,12) implies C(x,y) is holomorphic at each (x,y) ∈ℤ× (ℤ+12), so C is an entire function. Fix a generic y, consider C(x,y) as an entire function of x; C(x,y) being of moderate growth implies it is a constant in x. Similarly, it is constant in y.  Thus C(x,y) is a constant independent of x or y. 

Since f(x,0) = 0 (note the factor y^2 in front of F(x,y,m)) and g(x,0) = 0, this constant must be 0. This completes the proof of <ref>.




§ RESULTS ON HOFFMAN'S MULTIPLE T-VALUE Tˆ½(2, {1}ˆN, 2)



In this last section, we will prove the following evaluation on interpolated multiple t-values.

	The following evaluation holds for any n ≥ 0 ∈ℤ,

    t^(2,{1}^n,2) = n+3/2^n+2 t(n+4)    =  n+3/2^n+2·(1 - 1/2^n+4) ζ(n+4)   .




	This resolves Conjecture 3.10 from <cit.>, which posed the evaluation in the case of n odd.  The case n even was already handled therein, using a symmetry theorem for MtV's to establish a general generating series expression for t^(2ℓ+2, {1}^2n, 2ℓ+2).  In the case of n odd, t(n+4) is expected to be irreducible, so the symmetry theorem seemed to be insufficient to resolve this case.  It was also noted in <cit.> that evaluations of t^(2ℓ+2, {1}^2n+1, 2ℓ+2), ℓ > 0, apparently involved higher depth irreducibles, so no simple formulae are expected beyond this case.



Readers interested in computational details for this example should consult the  notebook attached. To begin with, we first write down the generating function of our MtV's of interest (or rather a small modification thereof), using the strategy of <ref>.  We have

    ∑_k≥ 0 2^k+4( t^r(2,{1}^k,2) - r^k+1  t(4+k) ) (2x)^k 
        = ∑_n = 0 
     m = 0^∞[t]  1/(m + 12)^2    (1+ur+u^2r^2+⋯) 
       ·∏_m<ℓ<n(1+s+s^2r+s^3r^2+⋯) 
    ·1/(n + 12)^2(1+vr+v^2r^2+⋯)  
        = ∑_n = 0 
     m = 0^∞1/(m+12)^2 (1 - r u)·∏_m < ℓ < n( 1 - s/1 - s r) ·1/(n+12)^2 (1 - r v) ,

where inside the summation

    u=2x/m + 12,    v=2x/n  + 12,    s=2x/ℓ + 12 .

When the interpolation parameter is r=, the right hand side further simplifies into the following double hypergeometric series

    ∑_n>0∑_m=0^n-11/(n+12)(m+12)Γ(12+m-x)Γ(12+n+x)/Γ(32+n-x)Γ(32+m+x) F(x)


First we note that the above series (in n) converges absolutely[if we view F(x) as a double series, then absolute convergence is only for |(x)|<1/2.] for (x) < 1/2. Indeed, we have 
    Γ(12+m-x)Γ(12+n+x)/Γ(32+n-x)Γ(32+m+x)
     = O(m^-1-2x) O(n^-1+2x)  ,

so the inner sum is O(n^-1+2x) ∑_m≥ 0^n-1O(m^-1-2x)/m+ 1/2 = O(n^-1+2x) O(n^max(0,-1-2x)).  Finally, if it holds that 2x-1 + max(0,-1-2x) < 0, (which is the case for x<12), we have

    ∑_n>01/n+12 O(n^-1+2x) O(n^max(0,-1-2x)) < ∞ .

 
 We also need asymptotic behaviour of the inner sum.

Let 

    g(n,x) 1/n+12Γ(12+n+x)/Γ(32+n-x)∑_m = 0^n-1Γ(m+12-x)/(m+12)Γ(32+m+x)

then for (x) sufficiently negative, 

    g(n,x) = -1/(1+2x)n^3 + 2+x/(1+2x)n^4 + O(n^-5)  .



This follow easily by Euler-Maclaurin summation. For (x)>1, we have 
    ∑_m=0^n-1 m^x = n^x+1/x+1 + n^x/2 + O(n^x-1)

From

    Γ(m+1/2-x)/(m+1/2)Γ(3/2+m+x) = m^-2x-2 + (-1-x)m^-2x-3 + O(m^-2x-4)

it follows that, if (x) is sufficiently negative (i.e. (-2x-3) > 0), we have

    ∑_m= 0^n-1Γ(m+1/2-x)/(m+1/2)Γ(3/2+m+x)
       = (-x-1) (n/-2 x-2+1/2) n^-2 x-3+(n/-2 x-1+1/2) n^-2 x-2 + O(n^-2x-3)

multiplying with the asymptotic expansion of Γ(1/2+n+x)/(n+1/2)Γ(3/2+n-x) gives the result. 


Finally we come to the algebraic part of the proof.  Creative telescoping (calling  to find the annihilator ideal of the inner sum, and then  to compute the creative telescoping relations for this annihilator ideal) says the operator

    (x+2)^3 𝐒_x^2-2 (x+1)^3 𝐒_x+x^3 + (𝐒_n-1)P
 kills g(n,x), where 

    P =    [t] 
    	1+2n/n( 
    	2n^3 - 6n^2 + 3n -2/8    - (7 + 14n+4n^2)x/16 - nx^2/4 - x^3/4
        + 2n^3 + 2n^2 - n - 1/8(3+2x) - 4n^3 - 8n^2 + 5n - 1/4(3-2n+2x)) 𝐒_x 
        - [t] 
    	(1+2n)(1+2n-2x)/n( 
    	   8n^2 + 16n-1/32 + (n-1)(2n-1)/2(3 - 2n+2x) + 3x/16 - x^2/8 - 4n^2-1/64(x+1)^2
        - 12n^2+1/64(x+1) - n^2/8(2x+1) + n^2-1/4(3+2x))   .


Therefore applying this to g(n,x), and summing over n leads to 

    x^3 F(x)-2(x+1)^3 F(x+1)+(x+2)^3 F(x+2) + lim_n→∞ (Pg)(n,x) - (Pg)(1,x) = 0  ,

via the telescoping in n.  Using the asymptotic behaviour of g(n,x) given in <ref>, one can evaluate the limit (under the assumption that (x) is sufficiently negative), obtaining

    lim_n→∞ (Pg)(n,x) = 4(1+4x+2x^2)/(1+2x)^2(3+2x)^2 .

Overall, we therefore have that F satisfies the recurrence

    x^3 F(x)-2(x+1)^3 F(x+1)+(x+2)^3 F(x+2) + 4 (2 x^2+4 x+1)/(2 x+1)^2 (2 x+3)^2 = 0  ,

this holds when (x) is sufficiently negative. As F(x) is analytic for (x)<1/2, the above implies F can be meromorphically continued to all of ℂ. This second order recurrence has the general solution

    2 ψ(x+1/2)-x  ψ'(x+1/2)/4 x^3+c_1(x)/x^3+c_2(x)/x^2 ,

where c_i(x) are 1-periodic functions.  The recursion also implies that F(x) has a double pole at x=12.  Since F has moderate growth by <ref>, we see both c_i are of moderate growth and both have at most double poles at 12 + ℤ. Therefore by <ref>, there exists constants u_i,v_i,w_i ∈ℂ such that

    c_i(x) = u_i + v_i tan(π x) + w_i tan^2(π x)  .

We know, from first few values of t^(2,{1}^k,2) - 1/2^k+1 t(k+4), obtained using the alternating MZV Data Mine <cit.>, say, that

    F(x) = π^4/24 + 31/2ζ(5) x+ π^6/20x^2 + ⋯ .

These three terms (and the analyticity of F(x) at x=0) are already enough to allow us to fix these six constants.  We find, namely

    6      u_1 = 0 ,        v_1 = -π/2 ,          w_1 = 0  , 
          u_2 = 2 + π^2/4 ,         v_2 = 0 ,        w_2 = π^2/4 .

So overall we have proven that

    F(x) = γ + 2 log(2) + ψ(12+x)/2x^3 - ψ'(12+x)/4x^2 + 3π^2/8x^2 - πtanπ x/2x^3 + π^2 tan^2π x/4x^2 .

Since t(n) = (1 - 12^n) ζ(n), this is readily seen to be equivalent to the statement of <ref>.



	The authors (the second in particular) would be interested to see a more “conceptual” proof of this evaluation.  The approach taken by SC in a recent note <cit.> on evaluations of multiple S and T values[Warning: the note  <cit.> employs the opposite MZV convention, wherein ζ(2,1) converges.] of the form S([.1pt]2, 1, …, 1, [.1pt]1) and T([.1pt]2, 1, …, 1, [.1pt]1), resolving Conjectures of Xu, Yan, and Zhao presented in <cit.>, suggests looking at the following iterated integral representation for the interpolated MtV,
	
    2^n+2  t^(2,{1}^n,2) 
       = ∫_0^1 (-1/t-1  - 1/t+1)dt ∘dt/t∘(( 1/t-1 + 1/t+1 - 1/t) dt )^n+1∘dt/t
       = ∫_0 < t_1 < t_2 < t_3 < 1-2/(t_1-1)(t_1+1)1/t_2·( log(t_2^-1 - t_2) - log(t_3^-1 - t_3) )^n+1/(n+1)!·1/t_3  t_1  t_2  t_3  ,

	and trying to explicitly write down a primitive via fixed depth multiple polylogarithms.  Unfortunately the current form of the integral does not seem to permit such a direct fixed depth evaluation.


	habbrv2 
	 

