a(n)=sum(x=2,n,x/log(x)) closely approximates the number of primes < n^2.
In fact, the sum is as good as Li(n^2) but summing a(n) is rather time consuming for large n.
For n = 10^9,
 a(n) = 24739954333817884,
 Pi(n^2) = 24739954287740860,
 Li(n^2) = 24739954309690415,
 R(n^2) = 24739954284239494.
where Li = Logarithmic integral approximation of Pi, and R = Riemann's approximation of Pi.
Now x/(log(x)1) is a much better approximation of Pi(x) than x/log(x):
 10^18/(log(10^18)1)=24723998785919976,
 10^18/log(10^18)=24127471216847323.
Ironically though, a(n) = sum(x=2,n,x/(log(x)1) is way off Pi(n^2).
