# When is a multiple sum given in closed form?

Let $d$ be a positive integer and $a>0$. Consider a following multiple sum: $${\mathcal S}^{(d)}_a(j) := \sum\limits_{0 \le j_1 \le j_2 \le \dots \le j_d \le j} \prod\limits_{l=1}^d f_a\left(j_l + \frac{2}{3} (l-1)\right)$$ where $$f_a(j) := \binom{j-\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j-\frac{1}{3}+a}{-\frac{1}{3}}$$ If $a=1/3$ then the product of binomial coefficients in the definition of $f_a(j)$ cancels out and produces one binomial coefficient only. Using this fact along with the Pascal triangle identity we have shown that our multiple sum has a following closed form: $${\mathcal S}^{(d)}_{\frac{1}{3}}(j) := \sum\limits_{p=0}^d \binom{j+ \frac{2 d+p-1}{3}}{\frac{p}{3}} \cdot {\mathcal A}_p^{(d)}$$ Here the coefficients ${\mathcal A}$ do not depend on $j$ and they satisfy the following recursion relations: \begin{eqnarray} {\mathcal A}^{(d+1)}_p &=& {\mathcal P} \cdot {\mathcal A}^{(d)}_{p-1} \cdot \binom{\frac{p-3}{3}}{-\frac{2}{3}} \quad \mbox{where $p=1,\dots,d+1$} \\ {\mathcal A}^{(d+1)}_0 &=& - {\mathcal P} \sum\limits_{p=0}^d {\mathcal A}^{(d)}_p \cdot \binom{\frac{p-2}{3}}{-\frac{2}{3}} \cdot \binom{\frac{2 d}{3} + \frac{p-1}{3}}{\frac{p+1}{3}} \end{eqnarray} and ${\mathcal P} := (-\frac{2}{3})!/((-\frac{1}{3})!)^2$.

Now, the question is the following. From When is a sum given in closed form? we know that ${\mathcal S}^{(1)}_a(j)$ admits a closed form iff $a = 1/3 + h$ where $h$ is a non-negative integer. Is it true that in this case also the multiple sum ${\mathcal S}^{(d)}_a(j)$ (for $d=1,2,..$) admits a closed form? If not what are the possible values of $a$ so that the multiple sum has a closed form and what is the closed form in those cases ?

We have: $$f_a(j) = \bar{P} \sum\limits_{l=0}^h \alpha^{(h)}_l \cdot \binom{j-\frac{1}{3}+l}{-\frac{2}{3}-h+l}$$ where \begin{eqnarray} \bar{P} &:=& \frac{h!}{(-\frac{1}{3})!^2 (-\frac{2}{3})!} \\ \alpha_l^{(h)} &:=& \frac{(-\frac{2}{3}+h-l)!(-\frac{2}{3}-h+l)!}{l! (h-l)!} \\ \end{eqnarray} Using the above we easily find ${\mathcal S}_a^{(1)}(j)$. It reads: $${\mathcal S}_a^{(1)}(j) = \bar{P} \left(\sum\limits_{l=0}^h \alpha^{(h)}_l \binom{j+\frac{2}{3}+l}{\frac{1}{3} -h +l} +\beta_1^{(h)} \right)$$ where \begin{eqnarray} \beta_1^{(h)} &:= &\frac{2 \pi \left(\binom{-\frac{1}{3}}{-\frac{2}{3}}-2 \binom{\frac{2}{3}}{\frac{1}{3}}\right) \Gamma \left(\frac{4}{3}\right)}{\sqrt{3} \Gamma \left(h+\frac{2}{3}\right)} \end{eqnarray} Therefore for $d=1$ the sum in question has a closed form in the sense that it is given as a sum of a fixed number of hypergeometric terms plus a constant.
Now take $d=2$. We have: \begin{eqnarray} {\mathcal S}^{(2)}_a(j) &=& \sum\limits_{j_1=0}^j {\mathcal S}_a^{(1)}(j_1) \cdot f_a(j_1 + \frac{2}{3}) \\ &=& \bar{P}^2 \sum\limits_{l=0}^h \sum\limits_{l_1=0}^h \alpha_l^{(h)} \alpha_{l_1}^{(h)} \cdot \left( \underbrace{\sum\limits_{j_1=0}^j \binom{j_1+\frac{2}{3}+l}{\frac{1}{3}-h+l} \binom{j_1+\frac{1}{3}+l_1}{-\frac{2}{3}-h+l_1}}_{{\mathfrak S}_j} \right)+\\ && \beta_1^{(h)} \bar{P}^2 \sum\limits_{l_1=0}^h \alpha_{l_1}^{(h)} \left( \binom{j+1+\frac{1}{3}+l_1}{\frac{1}{3}-h+l_1} - \binom{0+\frac{1}{3}+l_1}{\frac{1}{3}-h+l_1}\right) \end{eqnarray} The first term on the right hand side contains a nasty sum. Let us use Gosper's algorithm to find out whether that sum can be expressed in closed form. This happens iff one of the top free term differs from the the difference of the adjacent top free term and the adjacent bottom free term by a nonnegative integer $m \in {\mathbb N}_+$. This means that we have two cases: $$(A): \quad \frac{2}{3} + l = \left( \frac{1}{3} + l_1 \right) - \left(-\frac{2}{3}-h+l_1\right) + m = 1+ h + m \\ \Longrightarrow \mbox{False}$$ or $$(B): \quad \frac{1}{3} + l_1 = \left( \frac{2}{3} + l \right) - \left(\frac{1}{3}-h+l\right) + m = \frac{1}{3}+ h + m \\ \Longrightarrow \mbox{False, except l_1=h}$$ Thus we conclude that for $d \ge 2$ the multiple sum in question is not Gosper summable.