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):
Ironically though, a(n) = sum(x=2,n,x/(log(x)-1) is way off Pi(n^2).