p_n.dat contains

f
a
b
j
k

such that #E(GF(p^n)) = #E(GF(p)[X]/(f)) = k (a prime number). 

(E: y^2 = x^3 + ax + b, j(E) = j)

Example:

[1 4 0 0 0 1]
[4 2 3 4 3]
[1 3 2 1 2]
[0 3 0 1 3]
3119

(in 5_5.dat)

=> #E(GF(5)[X]/(1 + 4X + X^5)) = 3119