I am overall looking to find an explicit expression for $Z(\pi_a,j,0)$ in terms of the $\lambda$ terms in $\pi_a$ (a composition of non-negative integers) and $j$.
$\textbf{These are the conditions for the variables that are known}$ $$ (\lambda_v \ge 0) \text{ and } (a > 2) \\ \pi_a = (\lambda_1,\lambda_2,\ldots,\lambda_a) \\ k = \sum_{m=1}^a \lambda_m \\ 0 \le j \le k- \max\{\pi_a\} \\ \phi(a,j) = \sum_{m=1}^a \lambda_m -j $$ $\textbf{The recurrence relation & other important info}$
Let $Z(\pi_a,j,m) = Z((\lambda_1,\lambda_2,\ldots,\lambda_{a-1},\lambda_a-m),j)$ the recurrence relation is defined as $$ Z((\lambda_1,\lambda_2,\ldots,\lambda_{a-1},\lambda_a) = ((k-1)-(j-1)-(\lambda_a-1))Z((\lambda_1,\lambda_2,\ldots,\lambda_{a-1},\lambda_a),j-1) + Z((\lambda_1,\lambda_2,\ldots,\lambda_{a-1},\lambda_a-1),j) $$ Also known as $$ Z(\pi_a,j,0) = \phi(a-1,j-1)Z(\pi_a,j-1,1) + Z(\pi_a,j,1) $$ These are values of functions at specific points $$ Z(\pi_a,0,m) = 1 \\ Z(\pi_2,j-m,0) = \frac{\lambda_1! \lambda_2!}{(j-m)!(\lambda_1+m-j)!(\lambda_2+m-j)!} $$ when the 3rd dependent variable in the recurrence relation is equal to the $a^{th}$ $\pi_a$ term (or we have $\lambda_a=0$), we evaluate the composition of integers as if it has $a-1$ terms as $\pi_{a-1}$ $$ Z(\pi_a,j,\lambda_a) = Z(\pi_{a-1},j,0) \tag{A} $$ Since the variable depends on a composition of non-negative integers.
$\textbf{What I have done}$
I have begun by expanding out the recurrence relation in two different ways
First way: $$ Z(\pi_a,j,0) = \phi(a-1,j-1) \sum_{w=1}^{\lambda_a} Z(\pi_a,j-1,w) + Z(\pi_{a-1},j,0) \\ = \sum_{s=1}^{a-2} \phi(a-s,j-1) \sum_{w=1}^{\lambda_{a-s}} Z(\pi_{a-s},j-1,w) + Z(\pi_2,j,0) $$ Second way $$ Z(\pi_a,j,0) = \phi(a-1,j-1) Z(\pi_a,j-1,1) + Z(\pi_a,j,1) \\ = \phi(a-1,j-2)\phi(a-1,j-1) Z(\pi_a,j-2,2)+\phi(a-1,j-1) Z(\pi_a,j-1,2) + Z(\pi_a,j,1) \\ = \sum_{q=0}^{\lambda_a} \frac{Z(\pi_a,j-q,1+q)}{\phi(a-1,j)} \prod_{v=0}^q \left[\phi(a-1,j-v)\right] $$ I have tried to manipulate these two expanded versions many times, but it just seems too rigorous.
$\textbf{EDIT}$
I expanded in a third way which may help \begin{align*} z(\pi_a,j,0) = \sum_{m=0}^{\lambda_a} {\lambda_a \choose m} \frac{\phi(a-1,j-m)!}{\phi(a-1,j)!} Z(\pi_{a-1},j-m,0) \tag{3} \end{align*} \begin{align*} Z(\pi_a,j,r) = \phi(a-1,j-1)\sum_{w=1}^{\lambda_3-r}Z(\pi_a,j-1,r+w) + Z(\pi_{a-1},j,0) \therefore \\ Z(\pi_a,j,0) = \phi(a-1,j-1)\sum_{w=1}^{\lambda_3}Z(\pi_a,j-1,w) + Z(\pi_{a-1},j,0) \\ = \phi(a-1,j-1)\phi(a-1,j-2)\sum_{w=1}^{\lambda_3}\sum_{w_1=1}^{\lambda_3-w}Z(\pi_a,j-1,w+w_1)+Z(\pi_{a-1},j-1,0)\phi(a-1,j-1)\sum_{w=1}^{\lambda_3}(1) + Z(\pi_{a-1},j,0) \\ = \sum_{m=0}^{\lambda_a} {\lambda_a \choose m} \frac{\left(\phi(a-1,j)+m\right)!}{\left(\phi(a-1,j)\right)!} Z(\pi_{a-1},j-m,0)\qquad \qquad \end{align*} Since: \begin{align*} \sum_{v=m}^{n} \sum_{\pi_m \in C_v} (1) = {\lambda_v \choose m} \text{ Due to definition of binomial coefficient} \\ \prod_{v=1}^m \phi(a-1,j-v) = \prod_{v=1}^m (\phi(a-1,j)-v) = \frac{\phi(a-1,j)+v)!}{\phi(a-1,j)!} \end{align*} $\textbf{EDIT #2}$
By expanding (3) $a-2$ times, I found that \begin{align*} Z(\pi_a,j,0) = \sum_{m_1=0}^{\lambda_a} \sum_{m_2=0}^{\lambda_{a-1}} \cdots \sum_{m_{a-2}=0}^{\lambda_3} \left(\prod_{v=1}^{a-2} {\lambda_{a-v+1} \choose m_v} \frac{\phi(a-v,j-m_v)!}{\phi(a-v,j)!}\right) Z(\pi_2,j-\sum_{q=1}^{a-2} m_q,0) \\ = \sum_{m=0}^{k-\lambda_1-\lambda_2} \sum_{\gamma_{a-2} \in C_m} \left(\prod_{v=1}^{a-2} {\lambda_{a-v+1} \choose u_v} \frac{\phi(a-v,j-u_v)!}{\phi(a-v,j)!}\right) Z(\pi_2,j-m,0) \\ = \sum_{m=0}^{k-\lambda_1-\lambda_2} Z(\pi_2,j-m,0)\sum_{\gamma_{a-2} \in C_m} \left(\prod_{v=1}^{a-2} {\lambda_{a-v+1} \choose u_v} \frac{\phi(a-v,j-u_v)!}{\phi(a-v,j)!}\right) \end{align*} Where a new composition of integers $\gamma$ with non-negative integer terms is defined: \begin{align*} \gamma_{a-2} = (u_1,u_2,\ldots,u_{a-2}) \\ \sum_{q=1}^{a-2} u_q = m \end{align*} $\textbf{Possible way to derive solution}$ \begin{align*} Z(\pi_a,j,0) = \sum_{m=0}^{k-\lambda_1-\lambda_2} Z(\pi_2,j-m,0)\sum_{\gamma_{a-2} \in C_m} \left(\prod_{v=1}^{a-2} {\lambda_{a-v+_1} \choose u_v} \frac{\phi(a-v,j-u_v)!}{\phi(a-v,j)!}\right) \end{align*} But I am looking for a better looking formula,I also have questions about if the expression converges to a value that enables me to calculate values for $(j < k-\lambda_1-\lambda_2)$. I also want to have this summation simplified if possible $$ \sum_{\gamma_{a-2} \in C_m} \left(\prod_{v=1}^{a-2} {\lambda_{a-v+1} \choose u_v} \frac{\phi(a-v,j-u_v)!}{\phi(a-v,j)!}\right) \\ = \sum_{\gamma_{a-2} \in C_m} \left(\prod_{v=1}^{a-2} {\phi(a-v,j-u_v) \choose u_v} \frac{\lambda_{a-v+1}!}{(\lambda_{a-v+1}-u_v)!}\right) $$ $\textbf{Generating functions Attempt}$
Letting $$ \sigma(j,m)= \sum_{\gamma_{a-2} \in C_m} \left(\prod_{v=1}^{a-2} {\phi(a-v,j-u_v) \choose u_v} \frac{\lambda_{a-v+1}!}{(\lambda_{a-v+1}-u_v)!}\right) $$ \begin{align*} \sigma(j,m)x^{k-\lambda_1-\lambda_2-m} = x^{k-\lambda_1-\lambda_2-m}\sum_{\gamma_{a-2} \in C_m}\prod_{v=1}^{a-2} \left( {\phi(a-v,j-u_v) \choose u_v} \frac{\lambda_{a-v+1}!}{(\lambda_{a-v+1}-u_v)!}\right) \\ \sum_{\gamma_{a-2} \in C_m}\prod_{v=1}^{a-2} \left( {\phi(a-v,j-u_v) \choose u_v} \frac{d^{u_v}}{dx^{u_v}}\left[x^{\lambda_{a-v+1}}\right]\right) \\ = \sum_{\gamma_{a-2} \in C_m}\prod_{v=1}^{a-2} \left( \frac{\phi(a-v,j-u_v)!}{\phi(a-v,j)!} \frac{1}{u_v!}\frac{d^{u_v}}{dx^{u_v}}\left[x^{\lambda_{a-v+1}}\right]\right) \end{align*}