What's so special about a homotopy $15$-sphere? I just saw a table which counts diffeomorphism classes of homotopy $n$-spheres (that is, spaces homotopy equivalent to $n$-dimensional spheres). Such a table can be seen on the first page of this paper. Most of these numbers are less than $10$, and all of them except $15$ is less than a thousand. But then $15$ has $16,256$ different classes.
I always thought of higher dimensional spheres as being fairly homogeneous as you went out, so this variance came as a surprise. My question is whether there is an easy way to explain such a large jump.  Also, is there a more current table of values available?
 A: 
I always thought of higher dimensional spheres as being fairly homogeneous as you went out

In fact topology in $n$ dimensions, even the topology of spheres, depends in some surprisingly delicate ways on the $2$-adic properties of $n$ (see also this MO question). 
Most basically, the Euler characteristic of the $n$-sphere $S^n$ is $1 + (-1)^n$; that is, it's $0$ when $n$ is odd and $2$ when $n$ is even. This hints at important differences between the behavior of the even-dimensional and odd-dimensional spheres; for example, by the Poincaré–Hopf theorem, only the odd-dimensional spheres can admit nonvanishing vector fields. 
More generally, the classical vector fields on spheres problem asks for the maximum number of linearly independent nonvanishing vector fields on $S^n$. The answer turns out to be the following: let $\nu_2(n + 1)$ denote the largest value of $k$ such that $2^k \mid (n + 1)$. Write $\nu_2(n + 1) = 4d + c$ where $0 \le c \le 3$. Then the maximum number is exactly
$$2^c + 8d - 1.$$
In particular:


*

*If $n$ is even then $k = c = d = 0$ and the answer is $0$, which we already knew.

*If $n$ is $1 \bmod 4$ then $k = c = 1, d = 0$ and the answer is $1$. 

*If $n$ is $3 \bmod 8$ then $k = c = 2, d = 0$ and the answer is $3$. 

*If $n$ is $7 \bmod 16$ then $k = c = 3, d = 0$ and the answer is $7$. 

*If $n$ is $15 \bmod 32$ then $k = 4, c = 0, d = 1$, and the answer is $7$ again.


The appearance of the numbers $1, 3, 7$ here is related to the spheres $S^1, S^3, S^7$ being the unit spheres in the complex numbers $\mathbb{C}$, the quaternions $\mathbb{H}$, and the octonions $\mathbb{O}$ respectively. In fact this (hard) theorem implies Hurwitz's theorem that the real numbers, complex numbers, the quaternions, and the octonions are the only finite-dimensional real nonassociative algebras equipped with a certain kind of norm. The appearance of the number $8$ here is related to Clifford algebras, which can be used to construct the required vector fields.
So that's some context to convince you that the behavior of spheres is not necessarily uniform, but in general can depend delicately on the dimension. This is no less true for the problem of counting exotic spheres (equivalent to counting homotopy spheres by the generalized topological Poincaré conjecture), although here I know even less than about the vector fields on spheres problem. Let me try to say some things anyway.
It turns out that most of what makes the group of exotic spheres large in dimensions $4k - 1$ is a subgroup of exotic spheres that bound parallelizable manifolds. The non-exotic sphere $S^{4k-1}$ itself bounds the $4k$-disk $D^{4k}$, but (as I understand it) the exotic spheres bound other weirder $4k$-dimensional parallelizable manifolds (or don't bound any parallelizable manifolds at all). 
Manifold topology in dimensions divisible by $4$ is well-known to be different from manifold topology in other dimensions: most basically, as Sam Lisi says in the comments, those are precisely the dimensions in which the signature of a manifold is well-defined, and as I understand it this is how you distinguish those weirder $4k$-dimensional parallelizable manifolds. 
