The divisor function $\sigma_0(n)$

Number Theory Level 3

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?

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

False True Maybe!. No conclusive proof. No counterexample

1 solution

Caleb Townsend
Feb 10, 2015

I will prove this by counter example. Let $k_1 = 5$ and $k_2 = 6$ . The smallest number with $k_1$ divisors is $16$ , and the smallest number with $k_2$ divisors is $12$ , i.e.
$n(5) = 16$
$n(6) = 12$
Therefore the statement is false, as $n(k_1)$ is not necessarily less than $n(k_2)$ , even if $k_1 < k_2.$

The divisor function σ 0 ( n ) \sigma_0(n) σ 0 ​ ( n )

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The divisor function $\sigma_0(n)$