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, 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, 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.

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