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If $i+2<2n-1$, then $[S^2\times S^i, S^n]$ is a group; how can we "compute" it?

The choice of the map on the $2$-cell is trivial (assuming $n>2$) and the choice on the $i$-cell is parametrized by $\pi_i(S^n)$. After choosing the map on the $i$-skeleton, it can always be extended to the $i+2$-cell (the Whitehead product of $\pi_i(S^n)$ and $\pi_2(S^n)$ is trivial). All extensions to the $i+2$ cell are parametrized by $\pi_{i+2}(S^n)$, which yields a conjecture that the cohomotopy group might be isomorphic to $\pi_i(S^n)\times \pi_{i+2}(S^n)$. However, it is not clear whether two maps represented via different extensions to the $i+2$-cells, cannot be homotopic in $[S^2\times S^i, S^n]$. Maybe the group $\pi_{i+1}(S^n)$ is involved?

Thanks for possible help.

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    You are probably already aware of this, but if $i + 2 < n$ every map $S^2\times S^i \to S^n$ is nullhomotopic by cellular approximation, so the group is trivial. – 2017-01-19
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    Ah, yes, I mean $i>n$, thanks. – 2017-01-19

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