Im a little out of my depth here and am not well versed in combinatorics. Im not sure if this problem is too hard to solve or if there exists well known results to prove it. Here is part 1 which might be able to help with the second part of the problem.
1) Let $n\in\mathbb{N}$ be even and $m_i\in\mathbb{N}$ such that $1\leq m_i\leq n$. Is there a closed form solution for how many ways can $n$ be decomposed into $k$ elements such that $\sum^k_{i=1}m_i=n$ for $1\leq k \leq n$?
For example. Let $n=2$. Then for $k=1$, we get that $m_1=2$ is the only solution, and for $k=2$, then $m_1=1$, and $m_2=1$. So the answer would be that for $n=2$ and $k=1$ there is one solution and for $k=2$ there is one solution.
Im thinking maybe I can use this in some way to help solve the following conjecture.
2) For $n=2,6,10, \dots, 2(2l+1)$, $k$ and $m_i$ defined above then
\begin{eqnarray} \sum_{k_{odd}}\frac{1}{k+1}\big(\sum\frac{1}{\Pi^{k}_{i=1}m_i}\big) < \sum_{k_{even}}\frac{1}{k+1}\big(\sum\frac{1}{\Pi^{k}_{i=1}m_i}\big)\;, \end{eqnarray} where the second summations are over all possible solutions of the first problem stated above. In my example, this more complicated inequality holds ($\tfrac{1}{4}<\tfrac{1}{3}$) and is related to the BCH multiplication in Lie Group\algebra theory. Is this problem hopeless at the moment?
Appreciate any and all the input!
If I have understood your question correctly, you want to find the number of integral solutions to the equation
$$x_1+x_2+x_3+\cdots+x_k=n$$
with the constraint that $x_i\ge1$
Let us substitute $t_i=x_i+1$ with $t_i\ge0$
$$t_1+t_2+t_3+\cdots+t_k=n-k$$
Imagine $n-k$ apples placed in a row, and you want to divide it into $k$ groups. This can easily be done by placing $k-1$ partitions between them.
Thus we have a permutation at hand,
$$\frac{(n-1)!}{(n-k)!(k-1)!} = \binom{n-1}{k-1}$$
Note: This considers $(3,2)$ and $(2,3)$ as two ways of decomposing $5$