I am trying to prove that for all $N>1$, the number of divisors of the lowest common multiple of the first $N$ natural numbers is divisible by twice the number of distinct prime factors of $N$.
$$2\,\omega(N)\, | \,\tau(\operatorname{lcm}(1,2,3,…,N))\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(0)$$
For the greater proportion of naturals, this can be seen as a consequence of the number of divisors of $N$ being divisible by the number of it's unique prime factors, the density of these values when compared to it's complement (the number of divisors of $N$ being not being divisible by the number of it's unique prime factors) being relatively high. OEIS entry relevant
Then the factor of 2 can be seen as evident in observing that all values of $\tau(\operatorname{lcm}(1,2,3,…,N))$ appear to be even, so a preliminary lemma requiring proof is to show that $\operatorname{lcm}(1,2,3,…,N)$ must always have an even number of divisors, therefore also an even number of proper divisors greater than $1$ as well.
Any natural number $N$ has a unique prime factorization product:
$$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}\cdot\cdot\cdot p_{\omega(N),N}^{v_{\omega(N),N}}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)$$
$$\mathfrak L=\operatorname{lcm}(1,2,3,…,N)=p_{1,\mathfrak L}^{v_{1,\mathfrak L}}p_{2,\mathfrak L}^{v_{2,\mathfrak L}}p_{3,\mathfrak L}^{v_{3,\mathfrak L}}\cdot\cdot\cdot p_{\omega(\mathfrak L),\mathfrak L}^{v_{\omega(\mathfrak L),\mathfrak L}}$$
So in light of the number of divisors given by:
$$\tau(\mathfrak L)=\prod_{j=1}^{\omega(\mathfrak L)}(v_{j,\mathfrak L}+1)$$
We know that the lowest common multiple of the first $N$ natural numbers must have at least one prime factor with an odd multiplicity in order for it's number of divisors to be even.
So that is as far as ive got as well as starting to making considerations of previous study on this topic which I will link here
I'm really just looking for a hint please.
Best Answer
$2\cdot 3\cdot 5\cdot 7\cdot 13\cdot 17\cdot 19 = 881790$ doesn't appear to qualify - there are no $6$th, $13$th or $20$th powers in the LCM.