Why is the number of exotic spheres equivalent to $S^7,S^{11},S^{15},S^{27}$ equal to perfect numbers?

Question

https://oeis.org/A001676 Number of exotic spheres

https://oeis.org/A000396 Perfect numbers

$S^7 \to 28$ = 2nd perfect number (28)

$S^{11} \to 992$ = 2 times 3rd perfect number (496)

$S^{15} \to 16256$ = 2 times 4th perfect number (8128)

$S^{27} \to 69524373504$ = 2 times 5th perfect number (33550336)

These are the only cases. It doesn’t seem to exist any other correspondance between those 2 sets.

Answer 1

According to this blog post (which summarizes the results of Kervaire and Milnor), the number of exotic spheres of dimension $4k-1$ is $$\# \Theta_{4k-1} = R(k) \cdot \# H_{4k-1} \cdot B_{2k}/2k,$$ where:

• $R(k) = 2^{2k-2} (2^{2k-1} - 1)$ for odd $k$, twice that for even $k$;
• $H_{4k-1} = \pi^{\mathrm{st}}_{4k-1}$ is the $(4k-1)$th stable homotopy group of spheres;
• $B_{2k}$ is the $2k$-th Bernoulli number.

Note that the image of the $J$-homomorphism, a subgroup of $H_{4k-1}$, is cyclic, and its order is precisely the denominator of $B_{2n}/4n$, so the formula always gives an integer. Moreover for small $k$, the numerator of the Bernoulli number is $1$, and the J-homomorphism is surjective or has a small index (e.g. 2) so you really get an equality $\# \Theta_{4k-1} = R(k)$, or up to a factor $2$. For bigger $k$ this doesn't work.

Now it turns out that the 45 first perfect numbers are of the form $$P_p = 2^{p-1}(2^p-1)$$ where $p$ is prime, $p = 2, 3, 5, 7, 13, 17, 19, 31, 61$... This is of course the number $2 R(p)$ above. So this is how the two are related. Note that if $k$ isn't prime then you don't actually get a perfect number. (It is known that all even perfect numbers are of this form. Now don't ask me why perfect numbers are even... Because nobody knows!)

The example you found ($S^7$, $S^{11}$, $S^{15}$ and $S^{27}$) are all of the form $4k-1$, with $k = 2,3,4,7$. For these, the formula above gives for example with $k=2$: $$\# \Theta_7 = R(2) \cdot \# H_7 \cdot B_4 / 4 = 2^{2 \cdot 2 - 1} (2^{2 \cdot 2 - 1} - 1) \cdot 240 \cdot 1/30 \cdot 1/4 = P_2.$$ For the other $k$ you can do similar computations.