Let $\pi_k(n)$ be the almost prime counting function, then $\pi_k(2^kn)$ reaches a max value, since $\pi_k(2^kn)=\pi_{k+1}(2^{k+1}n)$ for large enough $k$.
(eg, $\pi_{5}(272)=\pi_{6}(2\times272)=\pi_{7}(4\times272)=\pi_{8}(8\times272)\dots$)
Is it true that, for fixed $n$, that $\dfrac{\pi_k(2^kn)}{max.\pi_k(2^kn)}\sim 1-e^{-k^2/2\log(n)}$ for any $k$?
($n=10,n=10^3,n=10^6$ respectively.)
AlmostPrimePi[k_Integer, n_] :=
Module[{a, i}, a[0] = 1; If[k == 1, PrimePi[n], Sum[PrimePi[n/Times @@
Prime[Array[a, k - 1]]] - a[k - 1] + 1, Evaluate[Sequence @@ Table[{a[i],
a[i - 1], PrimePi[(n/Times @@ Prime[Array[a, i - 1]])^(1/(k - i + 1))]},
{i, k - 1}]]]]]
n = 100000; range1 = 15; range2 = 35;
m = AlmostPrimePi[range2, 2^range2 n];
Show[Plot[1 - 1/(E^((k^2)/(2 Log[n]))), {k, 0, range1}],
ListLinePlot[Join[{0}, Table[AlmostPrimePi[k, 2^k n], {k, 1, range1}]/m],
DataRange -> {0, range1}, PlotStyle -> Red]]
($k=35$ works up to $n=10^6$, since $2^{35}\times10^6<3^{35}$. range2 should be increased for $n<10^6$ to satisfy $2^{k}n<3^{k}$.)
If this is the case, what does this have to do with the birthday problem?