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