I am not sure, if this is a research problem. If not I will move this question to ME: Let $\Omega(n) = \sum_{p|n} v_p(n)$, which we might view as a random variable. Let $E_n = \frac{1}{n} \sum_{k=1}^n\Omega(k)$ be the expected value and $V_n=\frac{1}{n} \sum_{k=1}^n(E_n-\Omega(k))^2$ be the variance. Then $$\pi(n) \approx \frac{n\gamma(\frac{V_n}{E_n},1.4854177\cdot \frac{V_n}{E_n^2})}{\Gamma(\frac{V_n}{E_n})}$$ where $\Gamma=$ Gamma function, $\gamma=$ lower incomplete gamma function.

Background: I was trying to fit the gamma distribution to the random variable $\Omega(k)$ ,$1 \le k \le n$. The value $1.4854177$ is fitted to some data. My question is, if there is any heuristic why this approximation should be good, if at all, or if this is just an overfitting problem?

Below you can find some sage code which implements this:

```
def Omega(n):
return sum([valuation(n,p) for p in prime_divisors(n)])
means = []
variances = []
xxs = []
omegas = [Omega(k) for k in range(1,10^4)]
for nn in range(10^4,10^4+3*10^3+1):
n = nn
omegas.append(Omega(n))
print "---"
m = mean(omegas[1:-1])
v = variance(omegas[1:-1])
shape,scale = m^2/v,v/m
xx = find_root(lambda xx : n*(lower_gamma(shape,xx*1/scale)/gamma(shape) ).N()-prime_pi(n),1,2)
xx = 1.4854177706344873
approxPrimePi2 = n*(lower_gamma(shape,xx*1/scale)/gamma(shape) ).N()
primepi = prime_pi(n)
print primepi, approxPrimePi2,shape.N(),scale.N(),xx
print "---"
print "err2 = %s" % (abs(primepi-approxPrimePi2)/primepi)
xxs.append(xx)
means.append(m.N())
variances.append(v.N())
```