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which would correspond to z = − 2. Let us also work under the assumption that p has the expansion 

and determine what values of y$_{1}$, y$_{2}$, . . . we should have. First, according to (30), we should have n/ 2 n/ 2 

which we rewrite as 

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Setting p = 1 + y$_{1}$/n + y$_{2}$/n $^{2}$+ · · · and q = 1 + z/n and performing a series expansion using (35), we find the right hand side to be 

and the first two terms of the left hand side are 

We solve this term by term. First 

has the solution y$_{1}$ = 2 w where 

Here s = W ( x ) is the Lambert function solving s e s = x . Note that for z ≥ − 2 we have y$_{1}$ = z but this is not the case for z < − 2 where we have y$_{1}$ > z . The second coefficient is 

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with w as before. We could go on and solve for y$_{3}$, y$_{4}$, . . . but, for the case in hand we actually only need y$_{1}$ . We will henceforth drop the subscript and refer to it as simply y . 

What shape does the distribution have at this particular p and q ? What we are seeking is an expression for R$_{p,q}$ ( n, n/ 2 , ℓ ) and this is where we start using (49), (50) and (51). We define ℓ = x n 3 / $^{4}$, just as we did in the case of p = q . First we need an expression for q/p and we let z 

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