Permutations, Fractions, Primes

[Leroy Quet]

First, take a permutation {b(k)} of the first n positive integers, and 
then take the partial sums, for 1<=m<=n,

s(m) = sum{k=1 to m} 1/b(k).

After reducing the n fractions {s(m)} so their numerators and 
denominators are coprime,
we can get a "score" equal to the total number of primes in both the 
numerators and denominators of the s(m)'s.

For example, if, for n=6, we have the permutation
(3,4,1,6,5,2),

then we have the s(m)'s
1/3, 7/12, 19/12, 7/4, 39/20, 49/20.

So we have the primes: 3,7,19,7, for a score of 4.

If we make a game of this, a better player might play the permutation
(2,6,1,4,5,3),

which gives the s(m)'s
1/2, 2/3, 5/3, 23/12, 127/60, 49/20.


So we have the primes: 2,2,3,5,3,23,127, for a score of 7.

My question is, what is the maximum possible score for each given n?

I get the sequence beginning
0,3,4,4,...

Thanks,
Leroy Quet