Given an odd natural integer $2a-1$ with $a\geq 1$, associate to it recursively the composition $\psi(1)=\emptyset$ and $\psi(2^{-n}a)+(n+\delta\_{>1}(m))$ if $a=2^n m$ with $m$ odd where $\delta\_{>1}(1)=0$ and $\delta\_{>1}(m)=1$ otherwise.
For $2a-1=37$ we get for example
\begin{align\*}
\psi(2\cdot 19-1)&=\psi(2\cdot 2\cdot 5-1)+(0+\delta\_{>1}(19))\\
&=\psi(2\cdot 3-1)+(1+\delta\_{>1}(5))+1\\
&=\psi(2\cdot 2\cdot 1-1)+2+1\\
&=\psi(2\cdot 1-1)+(1+\delta\_{>1}(1))+2+1\\
&=1+1+2+1\\
\end{align\*}


First values for $\psi$ are given by
$$\begin{array}{r|ccccccccc}
n&1&3&5&7&9&11&13&15&17\\
\hline
\psi(n)&\emptyset&1&1+1&2&1+1+1&1+2&2+1&3&1+1+1+1\\
\end{array}$$


$\psi$ is one-to-one: the composition $n\_1+n\_2+\cdots+n\_l$
is the image of $(\cdots(((2^{1+n\_1}-1)2^{n\_2}-1)2^{n\_3}-1)\cdots)2^{n\_l}-1$.


The partition $1+1+2+1$ corresponds to
$$(((2^{1+1}-1)2^1-1)2^2-1)2^1-1=37.$$


For $n>0$, it sends the $2^{n-1}$ odd integers $2^n+1,2^n+3,\dots,2^{n+1}-1$ to the $2^{n-1}$ compositions of sum $n$ in an order-preserving way for the lexicographic order on compositions.


Restricting $\psi$ to integers $\geq 5$ which are congruent to $1$ modulo $4$ (or to $3$ modulo $4$) and considering summands of compositions as coefficients of continued fraction expansions, we get of course a bijection onto all rationals in $(0,1)$.


*Is there a good reference for the bijection $\psi$ (or some close relative)?*



A bijection between odd natural integers and compositions
More or less a comment. I'd say the closest known bijection is just given by the binary representation, namely, if I understand it correctly, for e.g. $37$ we do $3$ times the last power of $2$ less than $37$, here $3\cdot32=96$, and the binary representation of the difference to $37$, here $96-37=59=2^5+2^4+2^3+2^1+2^0$, and finally the increments in the (decreasing) sequence of exponents, here $(5-4={\bf1},4-3={\bf 1},3-1={\bf 2},1-0={\bf 1})$.


