An integer partition is a sequence $\lambda=(\lambda_1\geq\lambda_2\geq\dotsb\geq\lambda_k)$ of positive integers, for some $k\geq1$. Consider the following two sets of partitions of $n$. Fix a positive integer $s>1$.
Let $\mathcal{O}_{n,s}=\{\lambda\vdash n: \lambda_i\in\{1,3,5,\dots,2s-1\}\}$. Parts are odd integers.
Also, let
$\mathcal{A}_{n,s}=\{a_1+2a_2+2a_3+\cdots+2a_s=n: a_1\geq a_2\geq a_3\geq\cdots\geq a_s\geq0\}$.
I would like ask:
QUESTION. Is there a combinatorial or bijective proof of the equinumerosity $\#\mathcal{O}_{n,s}=\#\mathcal{A}_{n,s}$?
Postscript. Fix $\ell>1$. Define
$\mathcal{L}_{n,s}=\{\lambda\vdash n: \lambda_i\equiv 1\mod \ell, \,\,i=1,\dots,s\}$ and
$\widetilde{\mathcal{L}}_{n,s}=\{a_1+\ell a_2+\ell a_3+\cdots+\ell a_s=n: a_1\geq a_2\geq a_3\geq\cdots\geq a_s\geq0\}$.
Then, Per Alexandersson's solution below appears to prove that $\#\mathcal{L}_{n,s}=\#\widetilde{\mathcal{L}}_{n,s}$.
Best Answer
Take a partition (Young diagram) in $O_{n,s}$. Number of elements in first column is $a_1$. Removing the first column now gives a partition with only even parts. Divide all parts by $2$, and take conjugate. This gives $a_2 \geq a_3 \geq \dotsc \geq a_s$, and you now have an element in the second set.
It is a nice problem, I will most likely steal it!