The following lemma should help you out (or at least partially) for small cases of $n$:
The number of compositions of n into odd parts equals the number of compositions
of n + 1 into parts greater than one.
proof: We'll construct a bijection. We'll start with a composition of $n$ of length $l$, $n=a_1+a_2+\dots+a_l$. We will represent this composition with a binary sequences of $n-l-1$ 0's and $l-1$ 1's, as follows:
$a_1-1$ zeros, $1$, $a_2-1$ zeros, $1$, $\dots$ $a_l-1$ zeros
(Thus, for example, the associated binary sequence to $13=3+1+5+3+1$ is $001100001001$)
Notice that a composition of $n$ where each $a_i$ is odd, then the associated binary sequence must have strings of zeros of even length. And thus if we take the conjugate composition (Which is, turn every zero in the corresponding binary sequence of the composition into a 1 and vice versa), we obtain a composition of $n$ of lenth $n-l+1$ with strings of $1$'s of even length. Denote this conjugate composition by $n-l+1=c_1+c_2+\dots+c_{n+l-1}$
(in our example, we have the conjugate composition $13=1+1+3+1+1+1+2+1+2$)
Next, we define $b_i=c_{2i-1}+c_{2i}$ for all $1\leq i \leq (n-l)/2$ and $b_{\frac{n-l}{2}+1}=c_{n-l+1}+1$ (notice how $n$ and $l$ must have the same parity)
Then $n+1=b_1+b_2+\dots +b_{\frac{n-l}{2}}+b_{\frac{n-l}{2}+1}$ is a composition of $n+1$ and length $\frac{n-l}{2}+1$, and moreover, all $b_i>1$!
(In our example: $14=2+4+2+3+3$)
To conclude the bijective relation, we have to check if all steps are reversible, but this is easy: Begin with a composition of n + 1 into parts greater
than one, reduce the last part by 1, split each part $j$ (other than the last part) into the pair of parts $j − 1, 1$, and calculute the conjugate composition of the precedent composition. $\Box$
in the following article, Cayley showed that the number of compositions of n + 1 into parts greater than one equals the $n$th Fibonacci number $F_n$.
A. Cayley, Theorems in trigonometry and on partitions, Collected Mathematical Papers, vol. 10, 16.
Quite surprisingly, it follows that the number of compositions into odd parts equals the $n$'th fibonnaci number.
Which is suggested in this OEIS entry: http://oeis.org/A000045:
"$F(n) =$ number of compositions of n into odd parts; e.g., $F_6$ counts 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 1+5, 3+1+1+1, 3+3, 5+1. - Clark Kimberling, Jun 22 2004"
Best Answer
Rather than write up a sanitized answer, I’m going to show how I actually attacked the problem.
It never hurts to start by gathering some data. The compositions of $1,3,5,7$, and $9$ into odd numbers of odd parts are:
$$\begin{align*} 1&=1\\ 3&=3=1+1+1\\ 5&=5=3+1+1=1+3+1=1+1+3=1+1+1+1+1\\ 7&=7=5+1+1=1+5+1=1+1+5\\ &=3+3+1=3+1+3=1+3+3=3+1+1+1+1\\ &=1+3+1+1+1=1+1+3+1+1=1+1+1+3+1\\ &=1+1+1+1+3=1+1+1+1+1+1+1\\ 9&=9=7+1+1=1+7+1=1+1+7=5+3+1\\ &=5+1+3=3+5+1=3+1+5=1+5+3=1+3+5\\ &=3+3+3=5+1+1+1+1=1+5+1+1+1=\\ &=1+1+5+1+1=1+1+1+5+1=1+1+1+1+5\\ &=3+3+1+1+1=3+1+3+1+1=3+1+1+3+1\\ &=3+1+1+1+3=1+3+3+1+1=1+3+1+3+1\\ &=1+3+1+1+3=1+1+3+3+1=1+1+3+1+3\\ &=1+1+1+3+3=3+1+1+1+1+1+1\\ &=1+3+1+1+1+1+1=1+1+3+1+1+1+1\\ &=1+1+1+3+1+1+1=1+1+1+1+3+1+1\\ &=1+1+1+1+1+3+1=1+1+1+1+1+1+3\\ &=1+1+1+1+1+1+1+1+1 \end{align*}$$
If $a_n$ is the number of compositions of $n$ into an odd number of odd parts, we have the following little table:
$$\begin{array}{rcc} n:&1&3&5&7&9\\ a_n:&1&2&5&13&34 \end{array}$$
We haven’t a lot of data, but it is still noticeable that the numbers $a_n$ so far are all Fibonacci numbers. In fact, they are alternate Fibonacci numbers:
$$0,\color{crimson}1,1,\color{crimson}2,3,\color{crimson}5,8,\color{crimson}{13},21,\color{crimson}{34}$$
That makes me wonder what happens if we count compositions of even numbers into odd parts:
$$\begin{align*} 2&=1+1\\ 4&=3+1=1+3=1+1+1+1\\ 6&=5+1=1+5=3+3\\ &=3+1+1+1=1+3+1+1=1+1+3+1=1+1+1+3\\ &=1+1+1+1+1+1 \end{align*}$$
If we redefine $a_n$ to be simply the number of compositions of $n$ into odd parts, we now have $a_2=1=F_2$, $a_4=3=F_4$, and $a_6=8=F_6$. Moreover, there is no composition of $0$ into odd parts, so $a_0=0=F_0$. The conjecture that $a_n=F_n$ for all $n\in\Bbb N$ is starting to look very good. To prove it, we need only find some argument to show that $a_n=a_{n-1}+a_{n-2}$ for $n\ge 2$, since we already know that $a_0=F_0$ and $a_1=F_1$.
Suppose that $n\ge 2$.
Added: Sarah has pointed out to me in the comments that I failed to read the question carefully enough: the foregoing answer gives the total number of compositions of an odd number into odd parts, but it doesn’t break that down by the number of parts. Let me remedy that. Since we have the data, let’s take a look at a table showing the number of compositions of $n$ into $k$ odd parts:
$$\begin{array}{c|cc} n\backslash k&1&2&3&4&5&6&7&8&9\\ \hline 1&1\\ 2&0&1\\ 3&1&0&1\\ 4&0&2&0&1\\ 5&1&0&3&0&1\\ 6&0&3&0&4&0&1\\ 7&1&0&6&0&5&0&1\\ 8&0&(4)&0&(10)&0&(6)&0&(1)\\ 9&1&0&10&0&15&0&7&0&1 \end{array}$$
Blank entries are $0$. The four numbers in parentheses weren’t previously calculated; I added them to complete the table through $n=k=9$.
If you get rid of the zero diagonals, the remaining diagonals seem to be just the rows of Pascal’s triangle. If this is actually the case, and we let $a_n(k)$ be the number of compositions of $n$ into $k$ odd parts, then
$$a_n(k)=\binom{\frac12(n+k)-1}{k-1}\;.\tag{1}$$
(To see this, it’s helpful to realize that $n+k$ is constant along each of the diagonals in question.) Since it’s clear that
$$a_n(1)=\begin{cases} 1,&\text{if }n\text{ is odd}\\ 0,&\text{if }n\text{ is even}\\ \end{cases}$$
and $a_n(n)=1$ for all $n\in\Bbb Z^+$, $(1)$ will follow if we can show that
$$a_n(k)=a_{n-1}(k-1)+a_{n-2}(k)$$
holds in general; this is the counterpart of Pascal’s identity. And this recurrence has a fairly simple combinatorial justification. The last part of a composition of $n$ into $k$ odd parts is either $1$ or greater than $1$.
In the first case we remove the last part to get a composition of $n-1$ into $k-1$ odd parts and note that every composition of $n-1$ into $k-1$ odd parts can be extended to a composition of $n$ into $k$ odd parts by adding a $1$ part at the end.
In the second case we subtract $2$ from the last part to get a composition of $n-2$ into $k$ parts and note that every composition of $n$ into $k$ odd parts with a last part greater than $1$ can be obtained from a composition of $n-2$ into $k$ odd parts by adding $2$ to the last part.
Note that it’s still helpful to consider partitions of even integers into odd parts as well as partitions of odd integers.