ON types of squarefree numbers and comparing their amounts < a given integer N.

Question

Let an m-prime be a square-free number with m prime divisors. Also let the number of t-primes < N be represented as #(t-prime){N} (t and N being positive elements of integers). Is the following true? Given a fixed integer N ; #(2-prime){N} >= #(3-prime){N} >= #(4-primes){N} >=....etc.??

Answer 1

The statement is false -- although it holds for quite some time.

If $c_t(n)$ denotes the number of $t$-primes smaller than or equal to $n$, we have $$c_2(1279789)<c_3(1279789)$$ and also $$c_1(58)<c_2(58)$$ (the latter wasn't part of the conjectured series of inequalities, though).

Based on a quick non-rigorous reasoning (which could be wrong, of course), it seems that neither of the the inequalities can hold indefinitely: apparently, the count of $(t+1)$-primes starts as smaller than the count of $t$-primes, but it eventually catches up and overtakes the other (and this "eventually" happens at different point for each inequality). In order to see why, consider the estimate given in Wikipedia: $$\pi_t(n)\sim \frac{n}{\log n} \frac{\left(\log \log n\right)^{t-1}}{(t-1)!}$$ Although the result in Wikipedia is stated for numbers with $t$ prime factors when counting multiplicity, it should also be true for numbers with $t$ distinct prime factors and also for square-free numbers with $t$ distinct prime factors; which is precisely our $c_t(n)$. Since the expression on right-hand side grows asymptotically faster when $t$ is greater (higher exponent at $\log \log n$), it eventually outgrows the one with smaller $t$. However, the speed of growth is proportional to double logarithm of $n$, so it might be a very slow process...