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This should be a comment, mostly because I have not checked...

Fix $n$ and consider only $m=1$. For a natural number $\ell$ let $[\ell]=\lbrace 1,\dots,\ell\rbrace$.

Let $X_k(n)$, for each $k\geq0$, be the vector space with basis the set of all functions surjective $[n]\to[k+1]$. Then $X_\flat(n)$ is a simplicial vector space (with simplicial operations acting on $[k+1]$). We can do better: for each $k\geq0$, there is an endomorphism $t_k:X_k(n)\to X_k(n)$ such that $t_k^{k+1}=\mathrm{id}$, given simply by rotation in $[k+1]$. The map $t_k$ plays nicely with the simplicial structure and I think that in this way $X_\flat(n)$ becomes a cyclic vector space (a cyclic object in the category of vector spaces) in the sense of Alain Connes; see Jean-Louis Loday's book on cyclic homology.

Now if the underlying field is of characteristic zero, the Connes' complex $C^\lambda$ corresponding to the cyclic vector space $X_\flat(n)$ has $C^\lambda_k=X_k(n)/(1-t_k)$; i other words, we identify things with their image under $t_k$. Elements of this vector space are functions $[n]\to[k+1]$ up to a rotation in $[k+1]$, which is the same thing as a partition of $[n]$ in $k+1$ labeled parts, up to a cyclic rotation of the labeling. There are $S(n,k+1)(k+1)!$ labeled partitions of $[n]$ into $k+1$ parts, and when we look at them up to rotation in $[k+1]$ we get $S(n,k+1) k!$. Now $c(k+1,1)=k!$, so in the end $$\dim C^\lambda_k=S(n,k+1)c(k+1,1).$$

It follows that you are almost trying to compute the Euler characeristic of the complex $C^\lambda$.

I don't have paper at hand (and this margin &c) so I cannot try now: but one knows that the homology of $C^\lambda$ can be obtained from a different complex, Connes' cyclic bicomplex or his triangular complex. Maybe the homology of those complexes can be readily computed, and that would do pretty much what you want.

1

This should be a comment, mostly because I have not checked...

Fix $n$ and consider only $m=1$. For a natural number $\ell$ let $[\ell]=\lbrace 1,\dots,\ell\rbrace$.

Let $X_k(n)$, for each $k\geq0$, be the vector space with basis the set of all functions $[n]\to[k+1]$. Then $X_\flat(n)$ is a simplicial vector space (with simplicial operations acting on $[k+1]$). We can do better: for each $k\geq0$, there is an endomorphism $t_k:X_k(n)\to X_k(n)$ such that $t_k^{k+1}=\mathrm{id}$, given simply by rotation in $[k+1]$. The map $t_k$ plays nicely with the simplicial structure and I think that in this way $X_\flat(n)$ becomes a cyclic vector space (a cyclic object in the category of vector spaces) in the sense of Alain Connes; see Jean-Louis Loday's book on cyclic homology.

Now if the underlying field is of characteristic zero, the Connes' complex $C^\lambda$ corresponding to the cyclic vector space $X_\flat(n)$ has $C^\lambda_k=X_k(n)/(1-t_k)$; i other words, we identify things with their image under $t_k$. Elements of this vector space are functions $[n]\to[k+1]$ up to a rotation in $[k+1]$, which is the same thing as a partition of $[n]$ in $k+1$ labeled parts, up to a cyclic rotation of the labeling. There are $S(n,k+1)(k+1)!$ labeled partitions of $[n]$ into $k+1$ parts, and when we look at them up to rotation in $[k+1]$ we get $S(n,k+1) k!$. Now $c(k+1,1)=k!$, so in the end $$\dim C^\lambda_k=S(n,k+1)c(k+1,1).$$

It follows that you are almost trying to compute the Euler characeristic of the complex $C^\lambda$.

I don't have paper at hand (and this margin &c) so I cannot try now: but one knows that the homology of $C^\lambda$ can be obtained from a different complex, Connes' cyclic bicomplex or his triangular complex. Maybe the homology of those complexes can be readily computed, and that would do pretty much what you want.