Let $A(X)$ denote the Waldhausen's algebraic K-theory of a space $X$, and let $n$ be **odd**. 


1. Are the rational homotopy groups of $A(S^n)$ known?
2. Is the group $\pi\_{2k}(A(S^n))$ finite for all positive $k\ll n$?


A reference (or proof sketch) would be appreciated.


EDIT: I found that the answers are stated (without reference or proof) on page 1 in "Homological stability of diffeomorphism groups" by Alexander Berglund and Ib Madsen, <http://arxiv.org/abs/1203.4161>. Namely, the answers to both questions is yes.



Algebraic K-theory of odd-dimensional spheres
Let $\tilde A(X)$ be the reduced functor, i.e., the homotopy fiber of the map $A(X) \to A(\ast)$. Since $A(\*)$ is rationally a product of $K(Q,4j+1)$ for $j \ge 1$, we may as well study $\tilde A(X)$ instead.


The rational homotopy of $\Omega \tilde A(\Sigma Y)$ was studied in 


G. Carlsson, R. Cohen, T. Goodwillie, and W. Hsiang, *The free loop space and the algebraic K-theory of spaces*.
K-Theory 1 (1987), no. **1**, 53–82. 


(The paper has some gaps but these were later corrected.) If we assume that $Y$ is connected, then the rational homotopy type of $\Omega \tilde A(\Sigma Y)$ coincides with that of the functor:
$$
\prod\_{n\ge 1} Q(Y^{[n]}\_{h\Bbb Z\_n})
$$
where $Y^{[n]}$ is the $n$-fold smash product of $Y$, $\Bbb Z\_n$ acts by cyclic permutation and ${({-})}\_{h\Bbb Z\_n}$ means homotopy orbits. Rationally, the homotopy groups of $Q(Y^{[n]}\_{h\Bbb Z\_n})$ coincide the homology groups of $Y^{[n]}\_{\Bbb Z\_n}$. 
When $Y$ is a $j$-sphere, the rational homology is not hard to compute.


Finally, if $X = S^1$, we know that $\tilde A(S^1)$ is rationally the same as $B A(\*)$ by a version of Bass-Heller-Swan. But by what I mentioned above, $B\tilde A(\*)$ is a rationally the product of $K(\Bbb Q,4j+2)$, $j \ge 1$.


