# The divisor function $$\sigma_0(n)$$

The divisor function $$\sigma_0(n)$$ gives the number of divisors of a natural number $$n$$.

Thus, $$\sigma_0(1)=1,\sigma_0(2)=2,\sigma_0(4)=3$$

Consider the function $$n=\sigma_{0,\min}^{-1}(k)$$, where $$n$$ is the smallest natural number with exactly $$k$$ divisors.

Is the following statement true?

$0<k_1<k_2 \implies \ \sigma_{0,\min}^{-1}(k_1) < \sigma_{0,\min}^{-1}(k_2)$

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