2 Multiple Hurwitz polylogs

Several authors have introduced multiple Hurwitz zeta values (MHZs) in specific contexts (e.g., [Reference Kamano8, Reference Matsumoto15]). In this section, we define a MHP, which generalizes both MPs (in one variable) and MHZs. Furthermore, we introduce an interpolated variant. Analytical properties, such as convergence and differential equations satisfied by interpolated MHPs, are then discussed.

Let ${\boldsymbol k}=(k_1,\dots ,k_d)\in {\mathbb {N}}^d$ be an index for $d\ge 1$ and $a\in {{\mathbb {C}}}$ with $a\notin {{\mathbb {Z}}}_{\le 0}$ . The multiple Hurwitz zeta (MHZ) value $\zeta ({\boldsymbol k};a)$ for the admissible ${\boldsymbol k}$ discussed in this paper is defined by the series

(2.1) $$ \begin{align} \zeta({\boldsymbol k};a)=\sum_{m_{1}>\cdots>m_d\ge 0} \dfrac{1}{(m_{1}+a)^{k_{1}}\cdots (m_d+a)^{k_d}}, \end{align} $$

and the power series $\widetilde {\mathrm {Li}}_{{\boldsymbol k}}(z ;a)$ is defined as

(2.2) $$ \begin{align} \widetilde{\mathrm{Li}}_{{\boldsymbol k}}(z ;a)=\sum_{m_{1}>\cdots>m_d\ge0} \dfrac{z^{m_{1}}}{(m_{1}+a)^{k_{1}}\cdots (m_d+a)^{k_d}}. \end{align} $$

Because $a\notin {\mathbb {Z}}_{\le 0}$ , the denominators in the RHSs of equations (2.1) and (2.2) never vanish.

Proof (i) Let $|z|\le 1$ . Applying $M=\min \{m\in {\mathbb {Z}}_{\ge 0}\mid {\Re }(m+a)\geq 1\},$ we have

(2.3) $$ \begin{align} &\left|\widetilde{\mathrm{Li}}_{{\boldsymbol k}}(z ;a)\right| \leq \sum_{m_{1}>\cdots>m_d\ge 0}\dfrac{|z|^{m_1}}{|m_{1}+a|^{k_{1}}\cdots|m_d+a|^{k_d}}\notag\\ &=\sum_{i=0}^d \sum_{m_{1}>\cdots>m_i\ge M>m_{i+1}>\cdots>m_d\ge 0}\dfrac{|z|^{m_1}}{|m_{1}+a|^{k_{1}}\cdots|m_d+a|^{k_d}}\notag\\ &=\sum_{i=0}^d \Big(\sum_{m_{1}>\cdots>m_i\ge M}\dfrac{|z|^{m_1}}{|m_{1}+a|^{k_{1}}\cdots|m_i+a|^{k_i}}\notag \\ &\qquad\quad \sum_{M>m_{i+1}>\cdots>m_d\ge 0}\dfrac{1}{|m_{i+1}+a|^{k_{i+1}}\cdots|m_d+a|^{k_d}}\Big)\notag\\ &\le \sum_{i=0}^d \Big(\sum_{m_{1}>\cdots>m_i\ge M}\dfrac{|z|^{m_1}}{|m_{1}+a|^{k_{1}}\cdots|m_i+a|^{k_i}}\cdot C_i \Big) \end{align} $$

for certain positive constants $C_i \ (0\le i\le d)$ (where $C_d$ can be 1). Because

$$ \begin{align*} |m+a|\ge {\Re }(m+a)\ge m-M+1, \end{align*} $$

for $m\ge M$ based on the definition of M, the following holds:

(2.4) $$ \begin{align} (3) &\le \sum_{i=0}^d \Big(\sum_{m_{1}>\cdots>m_i\ge M}\dfrac{|z|^{m_1}}{(m_1-M+1)^{k_{1}}\cdots (m_i-M+1)^{k_i}}\cdot C_i \Big)\notag\\ &\le \sum_{i=0}^d \Big(\sum_{m_{1}>\cdots>m_i>0}\dfrac{|z|^{m_1+M-1}}{m_{1}^{k_{1}}\cdots m_i^{k_i}}\cdot C_i\Big)\notag\\ &=|z|^{M-1}\sum_{i=0}^d C_i {\mathrm{Li}}_{k_1, \ldots, k_i}(|z|). \end{align} $$

It is well known that the usual MP ${\mathrm {Li}}_{{\boldsymbol k}}(z)$ , defined as

$$ \begin{align*} \mathrm{Li}_{{\boldsymbol k}}(z)=\sum_{m_{1}>\cdots>m_d>0} \dfrac{z^{m_{1}}}{{m_{1}}^{k_{1}}\cdots {m_d}^{k_d}}, \end{align*} $$

converges absolutely at $|z|<1$ for any ${\boldsymbol k}$ , and even at $|z|=1$ for an admissible ${\boldsymbol k}$ . Indeed, ${\mathrm {Li}}_{{\boldsymbol k}}(1)$ agrees with MZVs $\zeta ({{\boldsymbol k}})$ . Hence, equation (2.4) is bounded from above. This concludes (i).

1. (ii) Because $\widetilde {\mathrm {Li}}_{{\boldsymbol k}} (1 ;a)=\zeta ({\boldsymbol k};a)$ for an admissible ${\boldsymbol k}$ , the assertion directly follows from (i).

The multiple Hurwitz polylog (MHP) is defined as

$$ \begin{align*} \mathrm{Li}_{{\boldsymbol k}}(z ;a):=z^{a}\, \widetilde{\mathrm{Li}}_{{\boldsymbol k}}(z ;a) =\sum_{m_{1}>\cdots>m_d\ge0}\dfrac{z^{m_{1}+a}}{(m_{1}+a)^{k_{1}}\cdots (m_d+a)^{k_d}}. \end{align*} $$

This yields a multivalued function in z on $|z|\le 1$ caused by the complex power $z^a$ . To make this a single-valued function, let R (respectively, $\overline {R}$ ) be a unit disc (respectively, a closed unit disc) with a slit

$$ \begin{align*} R=\{z\in {{\mathbb{C}}}\mid |z|<1\}- {\mathbb{R}}_{\le 0},\qquad \overline{R}=\{z\in {{\mathbb{C}}}\mid |z|\le1\}- {\mathbb{R}}_{\le 0}. \end{align*} $$

In addition, we choose the principal branch $\pi <\arg z\le \pi $ for $z\in R$ (respectively, $z\in \overline {R}$ ). Lemma 2.1(i) implies that $\mathrm {Li}_{{\boldsymbol k}}(z ;a)$ is single-valued and holomorphic in R for any index ${\boldsymbol k}$ and converges in $\overline {R}$ for an admissible ${\boldsymbol k}$ .

When $a = 1$ , $\mathrm {Li}_{{\boldsymbol k}}(z ;a)$ reduces to the usual MP

$$\begin{align*}\mathrm{Li}_{{\boldsymbol k}}(z ;1)=\sum_{m_{1}>\cdots>m_d\ge0}\dfrac{z^{m_{1}+1}}{(m_{1}+1)^{k_{1}}\cdots (m_d+1)^{k_d}} =\sum_{m_{1}>\cdots>m_d>0}\dfrac{z^{m_{1}}}{{m_{1}}^{k_{1}}\cdots {m_d}^{k_d}}=\mathrm{Li}_{{\boldsymbol k}}(z), \end{align*}$$

and to the MHZs when $z = 1$ for an admissible ${\boldsymbol k}$ , that is, $\mathrm {Li}_{{\boldsymbol k}}(1 ;a)=\zeta ({\boldsymbol k};a)$ . The differential formula behaves better for $\mathrm {Li}_{{\boldsymbol k}}(z ;a)$ than $\widetilde {\mathrm {Li}}_{{\boldsymbol k}}(z ;a)$ . The lemma below is generalized in Proposition 2.3.

Lemma 2.2 Let ${\boldsymbol k}=(k_1, \ldots , k_d)\in {\mathbb {N}}^d$ . The MHP satisfies the following:

$$\begin{align*}\dfrac{d}{dz}\mathrm{Li}_{{\boldsymbol k}}(z ;a) =\left\{\!\!\begin{array}{ll} \dfrac{1}{z}\ \mathrm{Li}_{(k_{1}-1,k_{2},\ldots,k_d)}(z; a), & (k_{1}\geq2, \ d\ge 1),\\[6pt] \dfrac{1}{1-z}\,\mathrm{Li}_{(k_{2},k_{3},\ldots,k_d)}(z; a), & (k_{1}=1, \ d\geq2), \end{array} \right. \end{align*}$$

and $\frac {d}{dz}\mathrm {Li}_{(1)}(z; a)=z^a/(1-z)$ for $z\in R$ .

Except for formula of $\mathrm {Li}_{1}(z; a)$ , the differential formula of the usual MP $\mathrm {Li}_{{\boldsymbol k}}(z)$ is the same in appearance as that of Lemma 2.2. Therefore, we expect the generating function of the Ohno–Zagier type for the MHPs to behave similar to that of the MPs.

For generalization, using Yamamoto’s method in [Reference Yamamoto21], let us define the t-multiple Hurwitz polylog (t-MHP), and t-multiple Hurwitz zeta value (t-MHZ) as

(2.5) $$ \begin{align} \mathrm{Li}^t_{{\boldsymbol k}}(z;a)=\sum_{{\boldsymbol p}}t^{\sigma({{\boldsymbol p}})}\mathrm{Li}_{{{\boldsymbol p}}}(z;a), \qquad {\zeta}^t({\boldsymbol k};a)=\sum_{{\boldsymbol p}}t^{\sigma({{\boldsymbol p}})}{\zeta}({\boldsymbol p};a), \end{align} $$

where, in each summation, ${\boldsymbol p}$ runs over all indices of the form ${\boldsymbol p}=(k_1\Box k_2\Box \cdots \Box k_d)$ obtained by filling in boxes $\Box $ with symbols “ ” or “ $+$ ,” and $\sigma ({{\boldsymbol p}}):=d-\mathrm {dep}\,{\boldsymbol p}$ , or the number of boxes in which “ $+$ ” is filled in.

The star-type MHP (MHSP) and the star-type MHZ (MHSZ) are, respectively, defined for $a\notin {\mathbb {Z}}_{\le 0}$ as

$$ \begin{align*} \mathrm{Li}^{\star}_{{\boldsymbol k}}(z ;a) &=\sum_{m_{1}\ge \cdots\ge m_d\ge0}\dfrac{z^{m_{1}+a}}{(m_{1}+a)^{k_{1}}\cdots (m_d+a)^{k_d}}, \\{\zeta}^{\star}({\boldsymbol k} ;a) &=\sum_{m_{1}\ge \cdots\ge m_d\ge0}\dfrac{1}{(m_{1}+a)^{k_{1}}\cdots (m_d+a)^{k_d}}. \end{align*} $$

The t-MHP (respectively, t-MHZ) is a polynomial in t and interpolates the MHP (respectively, MHZ) at $t=0$ and MHSP (respectively, MHSZ) at $t=1$ , i.e., $\mathrm {Li}^0_{{\boldsymbol k}}(z;a)=\mathrm {Li}_{{\boldsymbol k}}(z ;a)$ and $\mathrm {Li}^1_{{\boldsymbol k}}(z;a)=\mathrm {Li}^{\star }_{{\boldsymbol k}}(z ;a)$ (respectively, ${\zeta }^0({\boldsymbol k} ;a)={\zeta }({\boldsymbol k} ;a)$ and ${\zeta }^1({\boldsymbol k} ;a)={\zeta }^{\star }({\boldsymbol k} ;a)$ ). See the summary in Table 1.

Table 1

When ${\boldsymbol k}$ is an index of depth $1$ , $\mathrm {Li}^t_{{\boldsymbol k}}(z;a)=\mathrm {Li}_{{\boldsymbol k}}(z;a)$ clearly holds by definition.

The t-MHP satisfies the following differential relation.

Proposition 2.3 Let ${\boldsymbol k}=(k_1, \ldots , k_d)\in {\mathbb {N}}^d$ . The t-MHP satisfies

$$\begin{align*}\dfrac{d}{dz}\mathrm{Li}^{t}_{{\boldsymbol k}}(z ;a) =\left\{\!\!\begin{array}{ll} \dfrac{1}{z}\ \mathrm{Li}^{t}_{(k_{1}-1,k_{2},\ldots,k_d)}(z; a), & (k_{1}\geq2, \ d\ge 1),\\[6pt] \left(\dfrac{t}{z}+\dfrac{1}{1-z}\right)\,\mathrm{Li}^{t}_{(k_{2},k_{3},\ldots,k_d)}(z; a), & (k_{1}=1, \ d\geq2), \end{array} \right. \end{align*}$$

and $\frac {d}{dz}\mathrm {Li}^{t}_{(1)}(z; a)=z^a/(1-z)$ for $z\in R$ .

Proof The former case follows easily from the former formula provided in Lemma 2.2 and equation (2.5). For the latter case, we have

$$ \begin{align*} \frac{d}{dz}\mathrm{Li}^t_{(1,k_2,\ldots,k_d)}(z; a) &=\sum_{{\boldsymbol p}=(1\Box k_2\Box\cdots\Box k_d)}\frac{d}{dz}\mathrm{Li}_{{\boldsymbol p}}(z; a)t^{\sigma({\boldsymbol p})}\\ &=\sum_{{\boldsymbol p}=(1+k_2\Box\cdots\Box k_d)}\frac{d}{dz}\mathrm{Li}_{{\boldsymbol p}}(z; a)t^{\sigma({\boldsymbol p})} +\sum_{{\boldsymbol p}=(1,k_2\Box\cdots\Box k_d)}\frac{d}{dz}\mathrm{Li}_{{\boldsymbol p}}(z; a)t^{\sigma({\boldsymbol p})}\\ &=\sum_{{\boldsymbol p}=(k_2\Box\cdots\Box k_d)}\frac{1}{z}\,\mathrm{Li}_{{\boldsymbol p}}(z; a)t^{\sigma({\boldsymbol p})+1} +\sum_{{\boldsymbol p}=(k_2\Box\cdots\Box k_d)}\frac{1}{1-z}\mathrm{Li}_{{\boldsymbol p}}(z; a)t^{\sigma({\boldsymbol p})}\\ &=\left(\frac{t}{z}+\dfrac{1}{1-z}\right)\mathrm{Li}^t_{(k_2,\dots,k_d)}(z; a). \end{align*} $$

Because $\mathrm {Li}^{t}_{(1)}(z; a)=\mathrm {Li}_{(1)}(z; a)$ , the last assertion follows from Lemma 2.2.

3 Generating function of MHPs

Following previous studies [Reference Li11, Reference Li and Wakabayashi14, Reference Ohno and Zagier17], we introduce the generating function of the MHPs as follows:

For $i\in {\mathbb N}$ , we define the i-th height of ${{\boldsymbol k}}=(k_{1},\ldots ,k_d)\in {\mathbb {N}}^d$ as i-ht $({{\boldsymbol k}})= \sharp \{l\, |\, k_{l}> i\}$ . The usual height coincides with the $1$ -height, that is, $1$ -ht $({{\boldsymbol k}})=$ ht $({\boldsymbol k})$ . For a fixed $r\in {\mathbb N}$ , and any integers $k,d,h_{1},\ldots ,h_{r}\geq 0$ , and $0\le j\le r-1$ , we define the set of indices as

$$ \begin{align*} I(k,d,h_{1},h_{2},\ldots,h_{r})&=\{{{\boldsymbol k}}\,|\,\mathrm{wt}({{\boldsymbol k}})=k,\mathrm{dep}({{\boldsymbol k}})=d, i\mbox{-ht}({{\boldsymbol k}})=h_{i}\quad (i=1,\ldots, r)\}, \\ I_{j}(k,d,h_{1},h_{2},\ldots,h_{r})&=\{{{\boldsymbol k}}=(k_1,\ldots, k_d)\,|\,{{\boldsymbol k}}\in I(k,d,h_{1},\ldots,h_{r}),k_{1}\geq j+2\}. \end{align*} $$

Clearly, $\{I_j\}$ gives a decreasing filtration of I:

$$\begin{align*}I(k,d,h_{1},\ldots,h_{r})\supset I_{0}(k,d,h_{1},\ldots,h_{r})\supset I_{1}(k,d,h_{1},\ldots,h_{r})\supset\cdots \supset I_{r-1}(k,d,h_{1},\ldots,h_{r}).\end{align*}$$

Note that

$$ \begin{align*} I(k,d,h_{1},\ldots,h_{r})\neq\emptyset &\Longrightarrow k\geq d+\textstyle \sum_{j=1}^{r}h_{j}, \quad d\geq h_{1}\geq\cdots\geq h_{r}\geq 0, \\ I_{j}(k,d,h_{1},\ldots,h_{r})\neq\emptyset &\Longrightarrow k\geq d+\textstyle \sum_{j=1}^{r}h_{j}, \quad d\geq h_{1}\geq\cdots\geq h_{r}\geq 0, \quad h_{j+1}\geq 1. \end{align*} $$

For $k,d,h_{1},\ldots ,h_{r}\geq 0$ , the sums of the MHPs of a fixed weight, depth, and ith heights for all $1\le i \le r$ , are defined as

$$ \begin{align*} G^t(k,d,h_1,\ldots,h_r; z ;a)&=\sum_{{{\boldsymbol k}}\in I(k,d,h_{1},\ldots,h_{r})}\mathrm{Li}_{{\boldsymbol k}}^t(z ;a), \\ G_j^t(k,d,h_{1},\ldots,h_{r}; z ;a)&=\sum_{{{\boldsymbol k}}\in I_{j}(k,d,h_{1},\ldots,h_{r})}\mathrm{Li}_{{\boldsymbol k}}^t(z ;a)\quad (0\le j\le r-1). \end{align*} $$

Each sum is understood to be zero if $I(k,d,h_{1},\ldots ,h_{r})$ or if $I_j(k,d,h_{1},\ldots ,h_{r})$ is empty except when $(k,d,h_{1},\ldots ,h_{r})=(0,0,0, \ldots , 0)$ for $G^t$ . In this case, we conventionally apply

(3.1) $$ \begin{align} G^t(0,0,\ldots,0; z ;a)=c^t(z) &:= \left. \frac{d}{dz}\mathrm{Li}^t_{(1)}(z ;a)\right/ \left(\frac{t}{z}+\frac{1}{1-z}\right) \notag \\ &=\left. \frac{z^{a}}{1-z}\right/ \left(\frac{t}{z}+\frac{1}{1-z}\right) =\frac{z^{a+1}}{z+t(1-z)}. \end{align} $$

It should be noted that $c^0(z)=z^{a}$ and $c^1(z)=z^{a+1}$ .

For a fixed $r\in {\mathbb {N}}$ , we define the generating functions

$$\begin{align*}\Phi^t=\Phi^t(x_1,\dots,x_{r+2}; z; a), \quad \Phi^t_j=\Phi^t_j(x_1,\dots,x_{r+2}; z; a)\quad (1\leq j\leq r-1)\end{align*}$$

as a formal power series in $x_1, \ldots , x_{r+2}$ by

$$ \begin{align*} \Phi^t&=\sum_{k\geq d+h \atop d\geq h_{1}\geq\cdots\geq h_{r}\geq0}G^t(k,d,h_{1},\ldots,h_{r}; z; a)x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+1}^{h_{r-1}-h_{r}}x_{r+2}^{h_{r}},\\ \Phi^t_{j}&=\sum_{k\geq d+h \atop d\geq h_{1}\geq\cdots\geq h_{r}\geq0}G^t_{j}(k,d,h_{1},\ldots,h_{r}; z; a)x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+1}^{h_{r-1}-h_{r}}x_{r+2}^{h_{r}}, \end{align*} $$

where $\sum _{j=1}^r h_j$ is abbreviated as h to shorten the equation. The notations $\Phi ^{t, (r)}:=\Phi ^t$ and $\Phi ^{t, (r)}_{j}:=\Phi ^t_{j}$ are used to emphasize the dependence on r. Note that for $r\in {\mathbb {N}}$ ,

$$ \begin{align*} \Phi^{t, (r)}=\Phi^{t, (r+1)}\big|_{x_{r+3}=x_1x_{r+2}},\qquad \Phi^{t, (r)}_j=\Phi^{t, (r+1)}_j\big|_{x_{r+3}=x_1x_{r+2}} \quad (1\leq j\leq r-1). \end{align*} $$

These generating functions satisfy the following differential equations. Let $D=z\frac {d}{dz}$ be the Euler operator.

Proof (1) and (2) can be shown in the same way as Lemma 2.2 in [Reference Li11] (or Lemma 6 in [Reference Li and Wakabayashi14]) by replacing their function $\Phi ^t$ (respectively, $\Phi ^t_j$ ) with our Hurwitz-type $\Phi ^t(x_1,\dots ,x_{r+2}; z; a)$ (respectively, $\Phi ^t_j(x_1,\dots ,x_{r+2}; z; a)$ ). We will next demonstrate (3) and (4). In the following, $h=\sum _{j=1}^r h_j$ . For (3), because

$$ \begin{align*} \Phi^t_{0}&-\Phi^t_{1}\\ =&\hspace{-2ex} \sum_{k\geq d+h, h_1\ge 1\atop d\geq h_{1}\geq\cdots\geq h_{r}\geq0} \hspace{-3ex}\{G^t_{0}(k,d,h_{1},\ldots,h_{r}; z; a)-G^t_{1}(k,d,h_{1},\ldots,h_{r}; z; a)\} x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+2}^{h_{r}}, \end{align*} $$

by applying Lemma 3.1(3), we have

$$ \begin{align*} D&(\Phi^t_{0}-\Phi^t_{1})\\ =&\hspace{-2ex}\sum_{k\geq d+h, h_1\ge 1\atop d\geq h_{1}\geq\cdots\geq h_{r}\geq0} \hspace{-3ex} \{G^t(k-1,d,h_{1}-1,h_{2},\ldots,h_{r}; z; a)-G^t_{0}(k-1,d,h_{1}-1,h_{2},\ldots,h_{r}; z; a)\} \\ &\times x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+2}^{h_{r}}\\ =&\hspace{-2ex}\sum_{k, d, h_{1}, \ldots, h_{r}\geq0} \hspace{-3ex} \{G^t(k-1,d,h_{1}-1,h_{2},\ldots,h_{r}; z; a)-G^t_{0}(k-1,d,h_{1}-1,h_{2},\ldots,h_{r}; z; a)\} \\ &\times x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+2}^{h_{r}} -G^t(0,0,0,0,\ldots,0;a;z)\dfrac{x_3}{x_2}. \end{align*} $$

By shifting the indices $k-1$ to k, and $h_1-1$ to $h_1$ in the sum above,

$$ \begin{align*} &=\sum_{k, d, h_{1}, \ldots, h_{r}\geq0} \hspace{-3ex} \{G^t(k,d,h_{1},h_{2},\ldots,h_{r}; z; a)-G^t_{0}(k,d,h_{1},h_{2},\ldots,h_{r}; z; a)\} \\ &\quad \times x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+2}^{h_{r}}\dfrac{x_3}{x_2} -c^t(z)\dfrac{x_3}{x_2}\\ &=\dfrac{x_{3}}{x_{2}}(\Phi^t-\Phi^t_{0}-c^t(z)). \end{align*} $$

Hence, (3) can be proven. For (4), because

$$ \begin{align*} \Phi&^t-\Phi^t_{0}\\ =&\hspace{-2ex}\sum_{k, d, h_1,\ldots, h_{r}\geq0} \hspace{-3ex}\{G^t(k,d,h_{1},\ldots,h_{r}; z; a)-G^t_{0}(k,d,h_{1},\ldots,h_{r}; z; a)\} x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+2}^{h_{r}}, \end{align*} $$

by applying Lemma 3.1(4) and based on the fact mentioned in Remark (ii) above, we have

$$ \begin{align*} &D(\Phi^t-\Phi^t_{0})\\&= \sum_{k, d, h_1,\ldots, h_{r}\geq0} \hspace{-4ex}D \{G^t(k,d,h_{1},\ldots,h_{r}; z; a)-G^t_{0}(k,d,h_{1},\ldots,h_{r}; z; a)\} x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+2}^{h_{r}} \\&=Dc^t(z)+\hspace{-3ex}\sum_{ (k,d,\ldots,h_r)\atop \not=(0,0,\ldots,0)} \hspace{-1ex}\left(t+\frac{z}{1-z}\right)G^t(k-1,d-1,h_{1},\ldots,h_{r}; z; a) x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+2}^{h_{r}}. \end{align*} $$

By shifting the indices $k-1$ to k, and $d-1$ to d in the sum above,

$$ \begin{align*} &=Dc^t(z)+\hspace{-3ex}\sum_{k, d, h_1,\ldots, h_{r}\geq0} \hspace{-1ex}\left(t+\frac{z}{1-z}\right)G^t(k,d,h_{1},\ldots,h_{r}; z; a) x_{1}^{k-d-h}x_{2}^{d-h_{1}}x_{3}^{h_{1}-h_{2}}\cdots x_{r+2}^{h_{r}}x_2\\ &=Dc^t(z)+\left(t+\frac{z}{1-z}\right)x_2 \Phi^t. \end{align*} $$

This shows that (4) holds and thus completes the proof.

Let us define

$$\begin{align*}y_{0}=\Phi^t_{r-1}, \quad y_{j}=\Phi^t_{r-1-j}-\Phi^t_{r-j}\quad(1\le j\le r-1), \quad y_{r}=\Phi^t-\Phi^t_{0}.\end{align*}$$

Proof Because $\Phi ^t=\sum _{j=0}^ry_j$ and $\Phi ^t_0=\sum _{j=0}^{r-1}y_j$ , (1)–(4) are rewriting of Proposition 3.2. In addition, (5) is proven through the induction of j and (2). To prove (6), using (5) and (3), we have

$$ \begin{align*} y_r-c^t(z)&=\frac{x_2}{x_3}Dy_{r-1}=\frac{x_2}{x_3}\left( \frac{x_3}{x_{r+2}}D^ry_0- \frac{x_1x_3}{x_{r+2}}D^{r-1}y_0\right)\\[3pt] &=\frac{x_2}{x_{r+2}}D^ry_0-\frac{x_1x_2}{x_{r+2}}D^{r-1}y_0. \\[-40pt] \end{align*} $$

Proposition 3.4 Here, $y_{0}=\Phi ^t_{r-1}$ satisfies the following condition:

$$ \begin{align*} &\Big[ D^{r+1}-(x_1+tx_2)D^r-t\sum_{j=0}^{r-1}(x_{r+2-j}-x_1x_{r+1-j})D^j\\[-1ex] &-z\big\{ D^{r+1}-(x_1+(t-1)x_{2})D^{r}-(t-1)\sum_{j=0}^{r-1}(x_{r+2-j}-x_{1}x_{r+1-j})D^{j} \big\} \Big] y_{0}=x_{r+2}z^{a}. \end{align*} $$

Proof Proposition 3.3(6) implies that

(3.2) $$ \begin{align} D(y_{r}-c^t(z))=\dfrac{x_{2}}{x_{r+2}}D^{r+1}y_{0}-\dfrac{x_{1}x_{2}}{x_{r+2}}D^{r}y_{0}. \end{align} $$

Furthermore, Proposition 3.3(5) and (6) give

(3.3) $$ \begin{align} \sum_{j=0}^r y_j-c^t(z) &=y_{0}+\sum_{j=1}^{r}\left(\dfrac{x_{r+2-j}}{x_{r+2}}D^{j}y_{0}-\dfrac{x_{1}x_{r+2-j}}{x_{r+2}}D^{j-1}y_{0}\right) \notag\\ &=\sum_{j=0}^{r}\dfrac{x_{r+2-j}}{x_{r+2}}D^{j}y_{0} -\sum_{j=0}^{r-1}\dfrac{x_{1}x_{r+1-j}}{x_{r+2}}D^{j}y_{0}\notag \\ &=\sum_{j=0}^{r-1}\dfrac{x_{r+2-j}-x_{1}x_{r+1-j}}{x_{r+2}}D^{j}y_{0}+\dfrac{x_{2}}{x_{r+2}}D^{r}y_{0}. \end{align} $$

Substituting equations (3.2) and (3.3) into the equation

$$ \begin{align*} D(y_{r}-c^t(z))=\left(t+\dfrac{z}{1-z}\right)x_{2} \sum_{j=0}^r y_j, \end{align*} $$

in Proposition 3.3(4), we have

$$ \begin{align*} \frac{x_{2}}{x_{r+2}}&D^{r+1}y_{0}-\dfrac{x_{1}x_{2}}{x_{r+2}}D^{r}y_{0}\\ &=\left(t+\dfrac{z}{1-z}\right)x_{2} \left(\sum_{j=0}^{r-1}\dfrac{x_{r+2-j}-x_{1}x_{r+1-j}}{x_{r+2}}D^{j}y_{0}+\dfrac{x_{2}}{x_{r+2}}D^{r}y_{0}+c^t(z)\right). \end{align*} $$

By multiplying both sides by $(1-z)x_2^{-1}x_{r+2}$ ,

$$ \begin{align*} (1-z)&(D^{r+1}y_{0}-x_{1}D^{r}y_{0})\\ &=\left(t-(t-1)z\right)\left(\sum_{j=0}^{r-1}(x_{r+2-j}-x_{1}x_{r+1-j})D^{j}y_{0}+x_{2}D^{r}y_{0}+x_{r+2}c^t(z)\right)\\ &=\left(t-(t-1)z\right)\left(\sum_{j=0}^{r-1}(x_{r+2-j}-x_{1}x_{r+1-j})D^{j}y_{0}+x_{2}D^{r}y_{0}\right)+x_{r+2}z^{a}. \end{align*} $$

Hence, we obtain the following desired equation:

$$ \begin{align*} &D^{r+1}y_0-(x_1+tx_2)D^ry_0-t\sum_{j=0}^{r-1}(x_{r+2-j}-x_1x_{r+1-j})D^jy_0\\[-1ex] &-z\big\{D^{r+1}y_0-(x_1+(t-1)x_{2})D^{r}y_0-(t-1)\sum_{j=0}^{r-1}(x_{r+2-j}-x_{1}x_{r+1-j})D^{j}y_0 \big\} =x_{r+2}z^{a}. \\[-22pt] \end{align*} $$

Let $\mathcal {P}_1(X)=\mathcal {P}^{\, t}_1(X;x_1,\ldots , x_{r+2})$ and $\mathcal {P}_2(X)=\mathcal {P}^{\, t}_2(X;x_1,\ldots , x_{r+2})$ be polynomials of degree $r+1$ in X defined by

$$ \begin{align*} \mathcal{P}_1(X)&=X^{r+1}-(x_1+tx_2)X^r-t\sum_{j=0}^{r-1}(x_{r+2-j}-x_1x_{r+1-j})X^j, \\ \mathcal{P}_2(X)&=X^{r+1}-(x_1+(t-1)x_{2})X^{r}-(t-1)\sum_{j=0}^{r-1}(x_{r+2-j}-x_{1}x_{r+1-j})X^{j}. \end{align*} $$

Note that

$$ \begin{align*} \mathcal{P}^{\, t-1}_1(X;x_1,\ldots, x_{r+2})&=\mathcal{P}^{\, t}_2(X;x_1,\ldots, x_{r+2}), \\ \mathcal{P}^{\, st}_1(X;x_1,x_2, \ldots, x_{r+2})&=\mathcal{P}^{\, t}_1(X;x_1,sx_2, \ldots, sx_{r+2}) \end{align*} $$

are valid for any values of parameters s and t. The polynomial

$$ \begin{align*} \mathcal{L}(X):=\mathcal{P}_1(X)-z \mathcal{P}_2(X) \end{align*} $$

satisfies

(3.4) $$ \begin{align} \mathcal{L}(D)\Phi^t_{r-1}=x_{r+2}z^{a} \end{align} $$

by virtue of Proposition 3.4. Let $\{-\alpha _j\}$ and $\{-\beta _j\}$ be the zeros of $\mathcal {P}_2(X)$ and $\mathcal {P}_1(X)$ respectively, that is,

(3.5) $$ \begin{align} \mathcal{P}_1(X)=\prod_{j=1}^{r+1} (X+\beta_j), \qquad \mathcal{P}_2(X)=\prod_{j=1}^{r+1} (X+\alpha_j). \end{align} $$

Because the identity as an operator $ \mathcal {L}(D)z^{a}=z^{a} \mathcal {L}(D+a) $ holds, by applying $\widetilde {\Phi }^t_{r-1}:=z^{-a}\Phi ^t_{r-1}$ and dividing equation (3.4) by $z^{a}$ , we obtain

(3.6) $$ \begin{align} \mathcal{L}(D+a)\widetilde{\Phi}^t_{r-1}=x_{r+2}. \end{align} $$

Applying D to both sides to make this a homogeneous equation, we have

$$ \begin{align*} D\mathcal{L}(D+a)\widetilde{\Phi}^t_{r-1}=0, \end{align*} $$

or equivalently,

(3.7) $$ \begin{align} \{D(D+a+\beta_1) &\cdots (D+a+\beta_{r+1})\notag \\ & -z(D+1)(D+a+\alpha_1)\cdots (D+a+\alpha_{r+1}) \} \widetilde{\Phi}^t_{r-1}=0. \end{align} $$

The generalized hypergeometric function

$$\begin{align*}_{r+2}\,F_{r+1}\left(\!\!\! \begin{array}{c} b_1,\ldots, b_{r+2}\\ c_1,\ldots, c_{r+1} \end{array} \!;\, z\right)\! =\sum_{m=0}^{\infty}\dfrac{[b_1]_m\dots[b_{r+2}]_m}{[c_1]_m\dots [c_{r+1}]_m}\frac{z^m}{m!}, \end{align*}$$

where $[b]_m=\prod _{i=0}^{m-1} (b+i)$ , is a solution to the differential equation

(3.8) $$ \begin{align} \left[ D(D+c_1-1)\cdots (D+c_{r+1}-1)-z(D+b_1)\cdots (D+b_{r+2}) \right] F=0, \end{align} $$

which is holomorphic at $z=0$ . See [Reference Andrews, Askey and Roy1, Reference Gasper and Rahman6] for further details.

Theorem 3.5 Letting $z\in R$ and $a\in {\mathbb {C}}$ with $a\not \in {\mathbb {Z}}_{\leq 0}$ , we have

(3.9) $$ \begin{align} &\Phi^t_{r-1}(x_{1},\ldots,x_{r+2}; z; a)= \frac{x_{r+2}}{\mathcal{P}_1(a)}\,z^{a}\, {}_{r+2}F_{r+1}\left(\!\!\! \begin{array}{lcl} a+\alpha_{1},&\hspace{-2ex}\ldots, &\hspace{-1.5ex} a+\alpha_{r+1},\ 1\\ a+\beta_1+1,&\hspace{-1.5ex} \ldots,&\hspace{-1.5ex} a+\beta_{r+1}+1 \end{array} ;z\right) \end{align} $$

(3.10) $$ \begin{align} &=\frac{x_{r+2}}{\mathcal{P}_2(a-1)}\,z^{a-1}\, \left\{{}_{r+2}F_{r+1}\left(\!\!\! \begin{array}{lcl} a+\alpha_{1}-1,&\hspace{-2ex}\ldots, &\hspace{-1.5ex} a+\alpha_{r+1}-1,\ 1\\ a+\beta_1,&\hspace{-1.5ex} \ldots,&\hspace{-1.5ex} a+\beta_{r+1} \end{array} ;z\right)-1\right\}, \end{align} $$

where $\{\alpha _{j}\}$ and $\{\beta _{j}\}$ are defined as in equation ( 3.5 ).

Proof Applying $\Phi ^{t}_{r-1}=\Phi ^{t}_{r-1}(x_{1},\ldots ,x_{r+2}; z; a)$ and $\widetilde {\Phi }^{t}_{r-1}=\widetilde {\Phi }^t_{r-1}(x_{1},\ldots ,x_{r+2}; z; a)$ , from equations (3.7) and (3.8), we have

$$ \begin{align*} \widetilde{\Phi}^t_{r-1}=C\cdot{_{r+2}F_{r+1}}\left(\!\!\!\begin{array}{lcl} a+\alpha_{1},&\hspace{-2ex}\ldots, &\hspace{-1.5ex} a+\alpha_{r+1},\ 1\\ a+\beta_1+1,&\hspace{-1.5ex} \ldots,&\hspace{-1.5ex} a+\beta_{r+1}+1\end{array} ;z\right),\end{align*} $$

for a constant C, which is independent of z. This shows that C is the constant term of $\widetilde {\Phi }^t_{r-1}$ . Equation (3.6) shows that

(3.11) $$ \begin{align} [\mathcal{P}_1(D+a)-z \mathcal{P}_2(D+a)]\widetilde{\Phi}^t_{r-1}=x_{r+2}. \end{align} $$

The constant terms of both sides of equation (3.11) give the equation $\mathcal {P}_1(a)C=x_{r+2}$ , and thus $C=x_{r+2}/\mathcal {P}_1(a)$ . Because $\Phi ^t_{r-1}=z^{a}\widetilde {\Phi }^t_{r-1}$ , this completes the proof of equation (3.9). Equation (3.10) can be easily obtained by shifting the parameters in the hypergeometric function in equation (3.9).

Theorem 3.6 Letting $z\in R$ and $a\in {\mathbb {C}}$ with $a\not \in {\mathbb {Z}}_{\leq 0}$ , we have

$$ \begin{align*} \Phi^t_{0}(x_{1},\ldots,x_{r+2}; z; a) =\frac{1}{\mathcal{P}_1(a)}& \Bigg\{ \sum_{j=0}^{r-1}A_{j} \sum_{l=0}^{j} \begin{pmatrix} a \\ j-l\\ \end{pmatrix}j!z^{a+l} \prod_{i=1}^{r+1}\dfrac{[a+\alpha_{i}]_{l}}{[ a +\beta_{i}+1]_{l}} \\ &\ {}_{r+2}F_{r+1}\left(\!\!\! \begin{array}{lcl} a+\alpha_{1}+l,&\hspace{-3ex}\ldots, &\hspace{-1.5ex} a+\alpha_{r+1}+l,\ l+1\\ a+\beta_1+l+1,&\hspace{-1.5ex} \ldots,&\hspace{-1.5ex} a+\beta_{r+1}+l+1 \end{array} ;z\right) \kern-1pt\!\Bigg\}, \end{align*} $$

where

$$ \begin{align*} A_{j}&=\sum_{i=j}^{r-1}(x_{r+2-i}-x_{1}x_{r+1-i})\displaystyle\genfrac{\{}{\}}{0pt}{}{i}{j}+x_{1}x_{2}\displaystyle\genfrac{\{}{\}}{0pt}{}{r-1}{j} \qquad (0\leq j\leq r-1), \end{align*} $$

$\genfrac {\{}{\}}{0pt}{}{n}{m}$ is the Stirling number of the second kind, and $\{\alpha _j\}$ and $\{\beta _j\}$ are defined in equation ( 3.5 ).

Proof Applying $\Phi ^t_{0}=\Phi ^t_{0}(x_{1},\ldots ,x_{r+2}; z; a)$ , the equation

$$ \begin{align*} \Phi^t_{0} &=\sum_{j=0}^{r-1}\dfrac{x_{r+2-j}-x_1x_{r+1-j}}{x_{r+2}}D^{j}y_{0} +\dfrac{x_1x_2}{x_{r+2}}D^{r-1}y_{0} =\sum_{j=0}^{r-1}\dfrac{A_{j}}{x_{r+2}}z^{j}\left(\dfrac{d}{dz}\right)^{j}y_{0}, \end{align*} $$

obtained in the same way as in Theorem 2.2 in [Reference Li11], and equation (3.9) demonstrate the theorem.

Note that the product on the right-hand side is computed as follows:

$$\begin{align*}\prod_{i=1}^{r+1}\dfrac{[a+\alpha_{i}]_{l}}{[ a +\beta_{i}+1]_{l}} =\frac{{\mathcal P}_2(a)\cdots {\mathcal P}_2(a+l-1)}{{\mathcal P}_1(a+1)\cdots {\mathcal P}_1(a+l)}. \end{align*}$$

4 Special cases and applications

4.1 The case of $r=1$

4.1.1 The case of height one: $h_1=1$

When $r=1$ , Theorems 3.5 and 3.6 reduce to

(4.1) $$ \begin{align} \hspace{-27pt}\Phi^t_{0}(x_{1}, x_2, x_{3}; z; a)&= \frac{x_{3} z^{a}}{\mathcal{P}_1(a)}\, {}_{3}F_{2}\left(\!\!\! \begin{array}{ll} a+\alpha_{1},&a+\alpha_{2},\ 1\\ a+\beta_1+1,&a+\beta_{2}+1 \end{array} ;z\right) \end{align} $$

(4.2) $$ \begin{align} &\qquad\qquad\qquad\qquad\, = \frac{x_{3} z^{a-1}}{\mathcal{P}_2(a-1)}\, \left\{{}_{3}F_{2}\left(\!\!\! \begin{array}{ll} a+\alpha_{1}-1,&a+\alpha_{2}-1,\ 1\\ a+\beta_1,&a+\beta_{2} \end{array} ;z\right)-1\right\} , \end{align} $$

where

$$ \begin{align*} {\mathcal{P}_1(X)}&=X^2-(x_1+tx_2)X-t(x_3-x_1x_2)=(X+\beta_1)(X+\beta_2), \\ {\mathcal{P}_2(X)}&=X^2-(x_1+(t-1)x_2)X-(t-1)(x_3-x_1x_2)=(X+\alpha_1)(X+\alpha_2). \end{align*} $$

To examine the terms of height one in $\Phi ^t_{0}(x_{1}, x_2, x_{3}; z; a)$ , dividing both sides of equation (4.1) (resp. equation (4.2)) by $x_3$ , and applying $x_3=0$ , ${\mathcal {P}_1(X)}=(X-x_1)(X-tx_2)$ and ${\mathcal {P}_2(X)}=(X-x_1)(X-(t-1)x_2)$ yield the following.

Proposition 4.1 We then have

$$ \begin{align*} &\sum_{k>d>0}\mathrm{Li}^t_{(k-d+1, \{1\}_{d-1})}(z; a)x_1^{k-d-1}x_2^{d-1}= \sum_{i, j>0}\mathrm{Li}^t_{(i+1, \{1\}_{j-1})}(z; a)x_1^{i-1}x_2^{j-1} \\&\quad= \frac{z^{a}}{(a-x_1)(a-tx_2)} {}_{3}F_{2}\left(\!\!\!\begin{array}{ll} a-x_1,&a-(t-1)x_2,\ 1\\ a-x_1+1,&a-tx_2+1 \end{array} ;z\right) \\&\quad= \frac{z^{a-1}}{(a-x_1-1)(a-(t-1)x_2-1)} \left\{{}_{3}F_{2}\left(\!\!\! \begin{array}{ll} a-x_1,&a-(t-1)x_2,\ 1\\ a-x_1+1,&a-tx_2+1 \end{array} ;z\right)-1\right\}. \end{align*} $$

In particular, when $z=1$ , we have

(4.3) $$ \begin{align} &\sum_{i, j>0}\zeta^t(i+1, \{1\}_{j-1}; a)x_1^{i-1}x_2^{j-1}\notag \\ &\quad= \frac{1}{(a-x_1)(a-tx_2)} {}_{3}F_{2}\left(\!\!\! \begin{array}{ll} a-x_1,&a-(t-1)x_2,\ 1\\ a-x_1+1,&a-tx_2+1 \end{array} ;1\right) \end{align} $$

(4.4) $$ \begin{align} &= \frac{1}{(a-x_1-1)(a-(t-1)x_2-1)} \left\{{}_{3}F_{2}\left(\!\!\! \begin{array}{ll} a-x_1-1,&a-(t-1)x_2-1,\ 1\\ a-x_1,&a-tx_2 \end{array} ;1\right)-1\right\}. \end{align} $$

By substituting $a=0$ and $t=0$ into equations (4.3) and (4.4), respectively, we obtain the following:

(4.5) $$ \begin{align} 1+\sum_{i, j>0}\zeta(i+1, \{1\}_{j-1})x_1^{i}x_2^{j} &=1+ \frac{x_1x_2}{1-x_1} {}_{3}F_{2}\left( \begin{array}{ll} -x_1+1,&x_2+1,\ 1\\ -x_1+2,&2 \end{array} ;1\right) \end{align} $$

(4.6) $$ \begin{align} &= {}_{2}F_{1}\left( \begin{array}{l} -x_1, \ x_2\\ -x_1+1 \end{array} ;1\right)=\frac{\Gamma(1-x_1)\Gamma(1-x_2)}{\Gamma(1-x_1-x_2)}. \end{align} $$

The left-hand side of equation (4.5) should be symmetric in $x_1$ and $x_2$ based on the duality for MZVs of height 1; that is, $\zeta (i+1, \{1\}_{j-1})=\zeta (j+1, \{1\}_{i-1})$ for $i, j\in {\mathbb {N}}$ . The corresponding symmetry on the right-hand side of equation (4.5) is given by equation (4.6) or is given directly by the known transformation (see e.g., [Reference Gasper and Rahman6])

$$ \begin{align*} {}_{3}F_{2}\left( \begin{array}{l} b_1, b_2, b_3\\ c_1, c_2 \end{array} ;1\right)= \frac{\Gamma(c_2)\Gamma(c_1+c_2-b_1-b_2-b_3)}{\Gamma(c_2-b_1)\Gamma(c_1+c_2-b_2-b_3)} {}_{3}F_{2}\left( \begin{array}{l} b_1, \ c_1-b_2, \ c_1-b_2\\ c_1, c_1+c_2-b_2-b_3 \end{array} ;1\right) \end{align*} $$

for ${}_{3}F_{2}$ , where $\Re (c_2-b_1)>0$ and $\Re (c_1+c_2-b_1-b_2-b_3)>0$ when substituting

$$ \begin{align*} (b_1, b_2, b_3, c_1, c_2)=(1, 1-x_1, 1+x_2, 2, 2-x_1). \end{align*} $$

4.1.2 Application to t-values

The generating functions associated with the “t-values” defined below are described by hypergeometric functions ${}_{3}F_{2}$ as a simple application of Proposition 4.1.

Let $a=p/q\in {\mathbb {Q}}$ with $-1<a\le 0$ , where p and q are coprime integers, and $q>0$ . We introduce a variant of MZVs for an admissible index ${\boldsymbol k}\in {\mathbb {N}}^d$ as

$$ \begin{align*} t^{(p/q)}({\boldsymbol k})=t^{(p/q)}(k_1,\ldots, k_d)=\sum_{m_1>\cdots >m_d>0 \atop \forall m_i\equiv p\ (\mathrm{mod} q)} \frac{1}{m_1^{k_1}\cdots m_d^{k_d}}. \end{align*} $$

Note that

$$ \begin{align*} t^{(p/q)}({\boldsymbol k})&=\sum_{m_1>\cdots >m_d>0} \frac{1}{(qm_1+p)^{k_1}\cdots (qm_d+p)^{k_d}}\\ &=q^{-\mathrm{wt} ({\boldsymbol k})}\sum_{m_1>\cdots >m_d>0} \frac{1}{(m_1+\frac{p}{q})^{k_1}\cdots (m_d+\frac{p}{q})^{k_d}} =q^{-\mathrm{wt} ({\boldsymbol k})}\zeta({\boldsymbol k}; p/q). \end{align*} $$

By substituting $z=1$ , $a=p/q$ , $t=0$ , $qx_1=X_1$ , and $qx_2=X_2$ into Proposition 4.1, we obtain the following.

Proposition 4.2

$$ \begin{align*} \sum_{i, j>0}t^{(p/q)}(i+1, \{1\}_{j-1})X_1^{i-1}X_2^{j-1} = \frac{1}{(p+q-X_1)(p+q)} {}_{3}F_{2}\kern-1pt\left( \!\!\kern-1pt\begin{array}{ll} \frac{p+q-X_1}{q},&\frac{p+q+X_2}{q},\ 1\\ \frac{p+2q-X_1}{q},&\frac{p+2q}{q} \end{array} ;1\kern-1pt\right). \end{align*} $$

For example, cases $q\le 4$ are given by

$$ \begin{align*} \sum_{i, j>0}t^{(-1/2)}(i+1, \{1\}_{j-1})X_1^{i-1}X_2^{j-1} &= \frac{1}{1-X_1} {}_{3}F_{2}\left( \begin{array}{ll} \frac{1-X_1}{2},&\frac{1+X_2}{2},\ 1\\[4pt] \frac{3-X_1}{2},&\frac{3}{2} \end{array} ;1\right), \\[4pt] \sum_{i, j>0}t^{(-1/3)}(i+1, \{1\}_{j-1})X_1^{i-1}X_2^{j-1} &= \frac{1}{2(2-X_1)} {}_{3}F_{2}\left( \begin{array}{ll} \frac{2-X_1}{3},&\frac{2+X_2}{3},\ 1\\[4pt] \frac{5-X_1}{3},&\frac{5}{3} \end{array} ;1\right), \\[4pt] \sum_{i, j>0}t^{(-2/3)}(i+1, \{1\}_{j-1})X_1^{i-1}X_2^{j-1} &= \frac{1}{1-X_1} {}_{3}F_{2}\left( \begin{array}{ll} \frac{1-X_1}{3},&\frac{1+X_2}{3},\ 1\\[4pt] \frac{4-X_1}{3},&\frac{4}{3} \end{array} ;1\right), \\[4pt] \sum_{i, j>0}t^{(-1/4)}(i+1, \{1\}_{j-1})X_1^{i-1}X_2^{j-1} &= \frac{1}{3(3-X_1)} {}_{3}F_{2}\left( \begin{array}{ll} \frac{3-X_1}{4},&\frac{3+X_2}{4},\ 1\\ \frac{7-X_1}{4},&\frac{7}{4} \end{array} ;1\right), \\[4pt] \sum_{i, j>0}t^{(-3/4)}(i+1, \{1\}_{j-1})X_1^{i-1}X_2^{j-1} &= \frac{1}{1-X_1} {}_{3}F_{2}\left( \begin{array}{ll} \frac{1-X_1}{4},&\frac{1+X_2}{4},\ 1\\[4pt] \frac{5-X_1}{4},&\frac{5}{4} \end{array} ;1\right). \end{align*} $$

4.2 The case of ${\boldsymbol k}=(n, \ldots , n)$

Letting $n\in {\mathbb {N}}$ with $n\ge 2$ , when $r=n-1$ , we substitute $x_1=\cdots =x_n=0$ into equations (3.9) and (3.10).

Theorem 4.3 For $n\ge 2$ , we have

(4.7) $$ \begin{align} \sum_{d=1}^{\infty} \mathrm{Li}^t_{(\{n\}_{d})}(z; a)x_{n+1}^d &= \frac{x_{n+1}}{\mathcal{P}_1(a)}\,z^{a}\, {}_{n+1}F_{n}\left(\!\!\!\begin{array}{lcl} a+\alpha_{1},&\hspace{-2ex}\ldots, &\hspace{-1.5ex} a+\alpha_{n},\ 1\\ a+\beta_1+1,&\hspace{-1.5ex} \ldots,&\hspace{-1.5ex} a+\beta_{n}+1 \end{array}\!\!;z\right) \notag \\[4pt]&=\frac{x_{n+1}}{\mathcal{P}_2(a-1)}\,z^{a-1} \left\{{}_{n+1}F_{n}\left( \!\!\!\begin{array}{lcl} a+\alpha_{1}-1,&\hspace{-2ex}\ldots, &\hspace{-1.5ex} a+\alpha_{n}-1,\ 1\\ a+\beta_1,&\hspace{-1.5ex} \ldots,&\hspace{-1.5ex} a+\beta_{n} \end{array}\!\!;z\right)-1\!\right\}, \end{align} $$

where

$$ \begin{align*} \mathcal{P}_1(X)=X^{n}-tx_{n+1}=\prod_{j=1}^n(X+\beta_j), \quad \mathcal{P}_2(X)=X^{n}-(t-1)x_{n+1}=\prod_{j=1}^n(X+\alpha_j). \end{align*} $$

As an alternative proof, equation (4.7) can be proven by computing the coefficients of the hypergeometric series on the right-hand side. Although the computation is slightly complicated, it is worth noting. For $m\ge 1$ , the following holds:

$$ \begin{align*} \prod_{j=1}^n \frac{[a+\alpha_j-1]_m}{[a+\beta_j]_m} &=\prod_{i=0}^{m-1} \prod_{j=1}^n\frac{a+\alpha_j+i-1}{a+\beta_j+i} =\prod_{i=0}^{m-1} \frac{{\mathcal P}_2(a+i-1)}{{\mathcal P}_1(a+i)}\\[4pt] &=\prod_{i=0}^{m-1} \frac{(a+i-1)^n-(t-1)x_{n+1}}{(a+i)^n-tx_{n+1}}\\[4pt] &=\frac{(a-1)^n-(t-1)x_{n+1}}{(a+m-1)^n-tx_{n+1}}\prod_{i=0}^{m-2} \left(1+\frac{x_{n+1}}{(a+i)^n-tx_{n+1}}\right). \end{align*} $$

Hence, we have

$$ \begin{align*} \mbox{RHS of equation (23) } &=\sum_{m>0}\frac{x_{n+1}}{(a+m-1)^n-tx_{n+1}} \prod_{i=0}^{m-2} \left(1+\frac{x_{n+1}}{(a+i)^n-tx_{n+1}}\right)z^{m+a-1}\\ &=\sum_{m>0}\sum_{\ell=1}^{m} \sum_{m-1=m_1>\cdots >m_{\ell} \ge0} \prod_{s=1}^{\ell} \frac{x_{n+1}}{(a+m_s)^n-tx_{n+1}} z^{m+a-1}\\ &=\sum_{\ell>0}\sum_{m_1>\cdots >m_{\ell} \ge0} \prod_{s=1}^{\ell} \frac{x_{n+1}}{(a+m_s)^n-tx_{n+1}} z^{m_1+a}\\ &=\sum_{\ell>0}\sum_{m_1>\cdots >m_{\ell} \ge0} \prod_{s=1}^{\ell} \sum_{d_s>0} \frac{t^{d_s-1}x_{n+1}^{d_s}}{(a+m_s)^{nd_s}}z^{m_1+a}\\ &=\sum_{\ell>0}\sum_{d\ge \ell}\sum_{d_1+\cdots +d_{\ell}=d}\sum_{m_1>\cdots >m_{\ell} \ge0} \prod_{s=1}^{\ell} \frac{1}{(a+m_s)^{nd_s}}z^{m_1+a}t^{d-\ell}x_{n+1}^{d}\\ &=\sum_{d>0}\sum_{\ell=1}^{d}\sum_{d_1+\cdots +d_{\ell}=d}\mathrm{Li}_{(nd_1,\ldots, nd_{\ell})}(z; a) t^{d-\ell}x_{n+1}^{d}\\ &=\sum_{d>0}\mathrm{Li}^t_{(\{n\}_d)}(z; a)x_{n+1}^{d}=\mbox{LHS of (23)}. \end{align*} $$