I am trying to compute the $k$-modules of $\mathcal{As}^¡$, the Koszul dual cooperad of the associative operad $\mathcal{As}$. I am using sections 7.1.4 and 7.2.1 of Algebraic Operads to try to do this. I am also comparing what I am doing with Derived A-infinity algebras in an operadic context that descrbes this cooperad in the proof of proposition 2.7.
So I am tryng to express $\mathcal{As}^¡(n)$ as a $k$-module given by generators. The explicit definition from Loday and Vallette is given in terms of weight. More precisely, since $\mathcal{As}$ is generated by $E=k \cdot m_2$ concentrated in arity 2 and has the relator $R=k\cdot as=k(m_2\circ_1 m_2-m_2\circ_2 m_2)$, concentrated in arity 3, we have a description of some of the direct summands of the Koszul dual in terms of their wieght
$\mathcal{As^¡}^{(0)}=(0, k Id, 0,0,\dots)$
$\mathcal{As^¡}^{(1)}=(0, 0, sE,0,0,\dots)$
$\mathcal{As^¡}^{(2)}=(0, 0, 0,s^2R,0,\dots)$
The $i$-th position represents arity $i$ (starting at $0$) and $s$ is the shift (say downwards) of grading. So it is clear that the generators of $\mathcal{As}^¡(n)$ for $1\leq n\leq 3$ are respectively $\mu_1=Id$ of degree $0$, $\mu_2=sm_2$ of degree $-1$ and $\mu_3=s^2as=sm_2\circ_1 sm_2-sm_2\circ_2 sm_2$ of degree $-2$.
Now I am having problems in finding the generatorss $\mathcal{As}^¡(n)$ for $n>3$ because I am not sure of the explicit description of the whole cooperad. I now that the generators of $\mathcal{As}^¡(n)$ must lie in the weight $n-1$ component since we need to compose the binary operation $n-1$ times to obtain an arity $n$ operation. In particular, they are of degree $1-n$. I suspect that there is only one due to the associativity relation, but I am not sure if we are allowed to use this relation as a rewriting rule since it is being used formaly as a generator.
In the same reference of Loday and Vallette (Section 3.1.3) there is a description for the Koszul dual coalgebra in which the analogue component to the weight $n$ has the general form
$(\bigcap_{i+2+j=n}sE^{\otimes i}\otimes s^2R\otimes sE^{\otimes j}\big)$
but I am not sure how this general term should look like for (co)operads.
One has to be just a bit more careful about what that intersection means. A way to build the generator of $\mathcal {As}^\mathrm{dual}(n)=H^0(B(\mathcal As))$ is by taking a suitably signed sum of all binary trees in $\mathcal As(n)$.
Indeed, in arity two we simply take $m$, the product, and in arity three we take $r_2=m(m,1) -m(1,m)$. In general, the element you want will look like $$ r_n = \sum_{T\in\mathsf{PBT_n}} (-1)^{\varepsilon(T)} m_T$$ where we take the sum over planar binary rooted trees and one has to choose the signs properly, and where by $m_T$ I mean the element in $B(\mathcal As)$ where we put $sm$ in the internal vertices of $T$. In arity four, the following choice works:
$$m(m(m,1),1) - m(m(1,m),1) + m(m,1)(1,m) + m(1,m(m,1)) - m(1,m(1,m))$$
where I write the fork $m(m,1)(1,m)$ since we need levels (since $m$ is actually of degree $1$ here, $m(m,1)(1,m)= -m(1,m)(m,1)$. Maybe you can figure out the correct signs in general! The picture is