An explanation by Pharap as per @Odomontois' request:
This answer is abusing inheritance and type parameters to create recursion. To understand what's happening, it's easier to first simplify the problem. Consider class X<A> { class Y : X<Y> { Y y; } }, which generates the generic class X<A>, which has an inner class Y. X<A>.Y inherits X<Y>, hence X<A>.Y also has an inner class Y, which is then X<A>.Y.Y. This then also has an inner class Y, and that inner class Y has an inner class Y etc. This means that you can use scope resolution (.) ad infinitum, and every time you use it, the compiler has to deduce another level of inheritance and type parameterisation.
By adding additional type parameters, the work the compiler has to do at each stage is further increased.
Consider the following cases:
In class X<A> { class Y : X<Y> { Y y;} } type param A has a type of X<A>.Y.
In class X<A> { class Y : X<Y> { Y.Y y;} } type param A has a type of X<X<A>.Y>.Y.
In class X<A> { class Y : X<Y> { Y.Y.Y y;} } type param A has a type of X<X<X<A>.Y>.Y>.Y.
In class X<A,B> { class Y : X<Y,Y> { Y y;} } type param A is X<A,B>.Y and B is X<A,B>.Y.
In class X<A> { class Y : X<Y> { Y.Y y;} } type param A is X<X<A,B>.Y, X<A,B>.Y>.Y and B is X<X<A,B>.Y, X<A,B>.Y>.Y.
In class X<A> { class Y : X<Y> { Y.Y.Y y;} } type param A is X<X<X<A,B>.Y, X<A,B>.Y>.Y, X<X<A,B>.Y, X<A,B>.Y>.Y>.Y and B is X<X<X<A,B>.Y, X<A,B>.Y>.Y, X<X<A,B>.Y, X<A,B>.Y>.Y>.Y.
Following this pattern, one can only imagine1 the work the compiler would have to do to to deduce what A to E are in Y.Y.Y.Y.Y.Y.Y.Y.Y in the definition class X<A,B,C,D,E>{class Y:X<Y,Y,Y,Y,Y>{Y.Y.Y.Y.Y.Y.Y.Y.Y y;}}.
1 You could figure it out, but you'd need a lot of patience, and intellisense won't help you out here.
