According to Hilton-Milnor theorem for $n\geq 2$
$$
\pi\_k(\mathbb{S}^n\vee\mathbb{S}^n)=
\pi\_k(\mathbb{S}^n)\oplus
\pi\_k(\mathbb{S}^n)\oplus
\bigoplus\_{i=1}^\infty
\pi\_k(\mathbb{S}^{m\_i}),
$$
where $m\_i$ is a sequence of integers that tend to $\infty$. (Correct me if this statement is incorrect.)



> 
> Is there an explicit formula for the sequence $m\_i$?
> 
> 
> 



> 
> For what values of $k$ is $\pi\_k(\mathbb{S}^n\vee\mathbb{S}^n)$ infinite?
> 
> 
> 


If it is difficult to find an explicit formula for $m\_i$ for an arbitrary $n$, I would like to see some examples where the answers to both of the above questions can be explicit.



$\pi_k(\mathbb{S}^n\vee\mathbb{S}^n)$
Yes, there is an explicit formula. It even describes the torsion elements in the homotopy groups of $S^n \vee S^n$. One statement is


$$\Omega \Sigma(A \vee B) \simeq \Omega \Sigma A \times \Omega \Sigma B \times \Omega \Sigma(\bigvee\_{i,j \geq 1} A^{\wedge i} \wedge B^{\wedge j})$$


where $\wedge$ is the smash product, and $\wedge i$ means the $i$-fold smash, e.g. $A^{\wedge 3} = A \wedge A \wedge A$.


So if you want to know the homotopy groups of $\Sigma(A \vee B)$ through a range, you need to apply the above formula inductively until the wedge product on the right side is sufficiently highly connected.


For example, if you apply this to $\pi\_4 (S^2 \vee S^2)$ you get a group of the form


$$\Bbb Z\_2 \oplus \Bbb Z\_2 \oplus \Bbb Z\_2 \oplus \Bbb Z \oplus \Bbb Z.$$


The copies of $\Bbb Z\_2$ are instances of $\pi\_4 S^2$ and $\pi\_4 S^3$ respectively, while the copies of $\Bbb Z$ are instances of $\pi\_4 S^4$ via the above formula.


You get $\pi\_k(S^n \vee S^n)$ finite when the iterated Whitehead brackets of the non-trivial elements in rational homotopy of $S^n$ do not appear in dimension $k$. So those would be all dimensions not in the list $n, 2n-1, 3n-2, 4n-3, \dotsc$.


edit: The level of explicitness you can get will depend on some constraints. If you let $k$ be relatively small, this can be done. But if you want to know the sequence $m\_i$ that's valid for all $k$, given that the sequence is a solution to a type of partition problem, there likely isn't a cute closed-form expression available. I suppose it goes without saying that the sequence $m\_i$ does not depend on $k$ -- the terms when $m\_i > k$ just don't matter to the answer.


