$\pi_k(\mathbb{S}^n\vee\mathbb{S}^n)$ 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.
 A: 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.
