Serre's finiteness theorem says if $n$ is an odd integer, then $\pi\_{2n+1}(S^{n + 1})$ is the direct sum of $\mathbb{Z}$ and a finite group. By looking at the table of homotopy groups, say on [Wikipedia](https://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table_of_homotopy_groups), one empirically observes that if $n \equiv 3 \pmod 4$, then we in fact have
$$
 \pi\_{2n+1}(S^{n+1}) \cong \mathbb{Z} \oplus \pi\_{2n}(S^n).
$$
This holds for all the cases up to $n = 19$. On the other hand, for $n \equiv 1 \pmod 4$ (and $n \neq 1$), the order of the finite part drops by a factor of $2$ when passing from $\pi\_{2n}(S^n)$ to $\pi\_{2n+1}(S^{n+1})$.


From the EHP sequence, we know that these two are the only possible scenarios. Indeed, we have a long exact sequence
$$
 \pi\_{2n+2}(S^{2n+1}) \cong \mathbb{Z}/2\overset{P}{\to} \pi\_{2n}(S^n) \overset{E}{\to} \pi\_{2n+1}(S^{n+1}) \overset{H}{\to}\pi\_{2n+1}(S^{2n+1}) \cong \mathbb{Z}.
$$
Since $H$ kills of all torsion, one sees that the map $E$ necessarily surjects onto the finite part of $\pi\_{2n+1}(S^{n+1})$. So the two cases boil down to whether or not $P$ is the zero map. What we observed was that it is zero iff $n \equiv 3 \pmod 4$, up to $n = 19$.


Since $\pi\_{2n+2}(S^{2n+1}) \cong \mathbb{Z}/2\mathbb{Z}$ has only one non-zero element, which is the suspension of the Hopf map, it seems like perhaps one might be able to check what happens to this element directly. However, without a good grasp of what the map $P$ (or the preceeding $H\colon\pi\_{2n+2}(S^{n+1}) \to \pi\_{2n+2}(S^{2n+1})$) does, I'm unable to proceed.


Curiously, I can't seem to find any mention of this phenomenon anywhere. The closest I can find is [this MO question](https://mathoverflow.net/questions/50452/early-stabilization-in-the-homotopy-groups-of-spheres), but this phenomenon is not really about early stabilization, since for $n = 3, 7$, the group $\pi\_{2n-1}(S^n)$ is not the stable homotopy group, but something smaller. I'd imagine either this pattern no longer holds for larger $n$, or there is some straightforward proof I'm missing.


**Note:** Suggestions for a more descriptive title are welcome.



A refinement of Serre's finiteness theorem on unstable homotopy groups of spheres
$P(\eta) = [i\_n,i\_n] \circ \eta$, where $[i\_n,i\_n]: S^{2n-1} \rightarrow S^n$ is the Whitehead product of the identity map with itself. So you are asking if this composite is null.


I don't know if this is an easy problem. One sufficient condition is that $[i\_n,i\_n]$ be divisible by 2, but, ha, ha, this is now known to only rarely happen, thanks to the Hill-Hopkins-Ravenel theorem on the Kervaire invariant. (See [HHR, Thm 1.5].) But it does for $n=63$, so $P$ is zero in that case.


I'd search the old literature - papers of Mahowald, Barratt, James and their collaborators - for any general results, if they exist. Mahowald, in particular, has many papers with examples of families of elements on which $H$ acts nonzero.


Added 30 minutes later: [Mahowald, Some Whitehead products in $S^n$, Topology 4, 1965, Theorem 1.1.2(a)] answers your question. $P(\eta)$ is zero if $n \equiv 3 \mod 4$ and is nonzero in basically all other cases. 


