$\newcommand{\d}{\mathrm{d}}\newcommand{\H}{\mathrm{H}}$Consider the space, $\Omega^n(M)$, of differential $n$-forms on a smooth, torsion-free manifold, $M$, without boundary, of dimension $D$, and construct the space of mixed differential forms up to degree $k$: $$\Omega^{(k)}(M) := \bigoplus_{n=0}^k \Omega^n(M).$$ The motivation for considering this arises from physics, from the BV formulation of gauge theories, where if the physical field is a $k$-form field, the tower below it represents ghost-for-ghost-for-...-ghost fields. Such mixed differential forms give a neat and tidy way of packing all this tower in a single field.
One can then turn this into a complex $$0\to \Omega^{(0)}(M) \overset{\d}{\to} \Omega^{(1)}(M) \overset{\d}{\to} \cdots \overset{\d}{\to} \Omega^{(D)}(M) \to 0,$$ where $\d:\Omega^{(k)}(M)\to\Omega^{(k+1)}(M)$ can be defined as $$ \d\equiv \d_{(k)} := \d_k \oplus \d_{k-1} \oplus \cdots \oplus \d_0 \oplus \iota,$$ with $\d_k:\Omega^k(M)\to\Omega^{k+1}(M)$ the ordinary exterior derivative and $\iota:0\to \Omega^0(M)$ just includes the zero-forms. Note that indeed $\d^2=0$, since $\mathrm{im}(\iota)=\lbrace 0\rbrace$.
My question is, what is the cohomology of this complex $$\H^{(k)}(M) := \frac{\mathrm{ker}\big(\d_{(k)}\big)}{\mathrm{im}\big(\d_{(k-1)}\big)},$$ in terms of the cohomology of the original de Rham complex?
Is it just $$ \H^{(k)}(M) \overset{?}{=} \bigoplus_{n=0}^k \H^{n}(M)?$$ Or is it something more elaborate? A more elaborate thing which would be natural would be $$\H^{(k)}(M) \overset{?}{=} \bigoplus_{n=0}^k (-1)^n \H^{n}(M).$$ By this I mostly mean that I would expect $\dim \H^{(k)}(M) = \sum_{n=0}^k (-1)^n\dim\H^n(M)$. The reason to suspect this, is that from the physical interpetation of the aforementioned complex in term of ghost fields, fields of degree $n=k-[\mathrm{even}]$ have ghost number zero, whereas fields of degree $n=k-[\mathrm{odd}]$ have ghost number one. This even/odd grading would then reflect the bosonic/fermionic statistics of the harmonic forms at each degree.