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It's a hard calculation, and the hardest case is when n = 1 mod 8: https://arxiv.org/abs/1506.00952 Sergei O. Ivanov, Roman Mikhailov and Jie Wu wrote this in 2015 - and then realized that Brayton Gray could have known this result back in 1984. (2/n)
It instantly follows that πₙ(S³) has more than one element except for n=0,1,2. (Oh, duh - πₙ(S²) has just one element when n = 0. I ignored that case in my first tweet.) Curtis showed earlier that πₙ(S⁴) has more than one element for all n > 3. (3/n)
Mahowald and M. Mori showed that πₙ(S⁵) has more than one element for all n > 5. But that's where this business stops! If k > 5, πₙ of the k-sphere has just one element when n=k+4. I don't understand these computations: I merely admire them. (4/n, n = 4)
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