Let $X$ be the set of positive odd numbers.
For any formal power series $f(t) = \sum_{j=0}^\infty f_k t^k$, let $[t^k]f(t)$ be the coefficient $f_k$.
When one expand following product into a series of $t$.
$$\left(\sum_{x_1\in X} t^{x_1}\right)\cdots\left(\sum_{x_k\in X} t^{x_k}\right)
= \sum_{x_1,\ldots,x_k \in X} t^{x_1+\ldots + x_k}
= \sum_{n=1}^\infty \alpha_n t^n$$
One will notice there is a one-one correspondence between terms contributing
to $\alpha_n$ and solution $x_1 + \ldots + x_k = n$ for $(x_1,\ldots,x_k) \in X^k$. For example, in the expansion of
$$(t + t^3 + t^5 + \cdots)(t + t^3 + t^5 + \cdots)(t + t^3 + t^5 + \cdots) =
x^3+3x^5+6x^7+10x^9 + \cdots
$$
There are $6$ terms that contribute to $\alpha_7 = 6$. Namely,
$$
\begin{array}{lcr}
(t + t^3 + \color{blue}{t^5} + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots) &\leftrightarrow & 5+1+1 = 7\\
(\color{blue}{t} + t^3 + t^5 + \cdots)
(t + t^3 + \color{blue}{t^5} + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots) &\leftrightarrow & 1+5+1 = 7\\
(\color{blue}{t} + t^3 + t^5 + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots)
(t + t^3 + \color{blue}{t^5} + \cdots) &\leftrightarrow & 1+1+5 = 7\\
(\color{blue}{t} + t^3 + t^5 + \cdots)
(t + \color{blue}{t^3} + t^5 + \cdots)
(t + \color{blue}{t^3} + t^5 + \cdots)&\leftrightarrow & 1+3+3 = 7\\
(t + \color{blue}{t^3} + t^5 + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots)
(t + \color{blue}{t^3} + t^5 + \cdots)&\leftrightarrow & 3+1+3 = 7\\
(t + \color{blue}{t^3} + t^5 + \cdots)
(t + \color{blue}{t^3} + t^5 + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots)&\leftrightarrow & 3+3+1 = 7\\
\end{array}
$$
In general, this leads to
$$\alpha_n = \sum_{n = \sum_{j=1}^k x_j} 1 = a_n
\quad\leftarrow \text{ the number of solutions we seek. }
$$
Since $\displaystyle\;\sum_{x\in X}t^x = \sum_{j=0}^\infty t^{2j+1} =
t \sum_{j=0}^\infty (t^2)^j = \frac{t}{1-t^2},$ the OGF at hand is simply
$$\sum_{n=0}^\infty a_n t^n = \frac{t^k}{(1-t^2)^k}
\quad\implies\quad a_n = [t^n] \frac{t^k}{(1-t^2)^k} = [t^{n-k}]\frac{1}{(1-t^2)^k}
$$
In order for $a_n \ne 0$, we need $n \ge k$ and $n$ and $k$ has same parity.
When this happens, we have
$$a_n = [t^{(n-k)/2}] \frac{1}{(1-t)^k} = \binom{\frac{n-k}{2} + k - 1}{\frac{n-k}{2}} = \binom{\frac{n+k}{2}-1}{\frac{n-k}{2}}$$
Note
Above answer is under the assumption we want the number of solutions for a fixed $k$. If one want the number of solutions for a given $n$ but the number of $k$ can vary, the corresponding OGF will be
$$1 + \frac{t}{1-t^2} + \frac{t^2}{(1-t^2)^2} + \cdots = \sum_{k=0}^\infty \left(\frac{t}{1-t^2}\right)^k
= \frac{1}{1 - \frac{t}{1-t^2}} = 1 + \frac{t}{1 - t - t^2}
$$
The rightmost expression is the famous OGF for
Fibonacci numbers.
$$\frac{t}{1 - t - t^2} = \sum_{n=1}^\infty F_n t^n$$
This means for any $n > 0$, the number of ways of breaking
$n$ into an ordered list of odd numbers $(x_1, \ldots, x_k)$ which sum to $n$ is simply $F_n$.
