# $\omega$ of a highly composite number.

A number is highly composite if it is the smallest number that has more divisors that any number less than it.

Let $$(h_n)_{n\ge 1}$$ be the sequence of highly composite numbers and $$\omega(n)$$ denote the number of distinct prime divisors of $$n$$. I noticed that $$\omega(h_n)$$ is a nondecreasing sequence but I don't know why. Here's my attempt,

Assume not then we can find an $$n$$ such that $$\omega(h_n)>\omega(h_{n+1})$$ Write $$h_n=\prod_{i=1}^rp_i^{a_i}\text{ and } h_{n+1}=\prod_{i=1}^sp_i^{b_i}$$ with $$r>s$$ and $$p_i$$ is the ith prime number. Also the $$a_i$$'s and the $$b_i$$'s are both nonincreasing sequences (those are all well-known facts about highly composites). Therefore $$1>\frac{h_n}{h_{n+1}}=\frac{\prod_{i=1}^sp_i^{a_i}\prod_{i=s+1}^rp_i^{a_i}}{\prod_{i=1}^sp_i^{b_i}}=\prod_{i=1}^sp_i^{a_i-b_i}\prod_{i=s+1}^rp_i^{a_i}$$ Now I don't know what to do I think if we can somehow get a handle on $$a_i-b_i$$ then we may get a contradiction.

• How many values have you checked? Per wiki $h_{25}=2^3\cdot3^2\cdot5\cdot 7\cdot 11$ but $h_{26}=2^4\cdot 3^4\cdot 5\cdot 7$.
– Sil
May 30 at 23:14
• Consider the highly composite numbers $2=2^1$ and $4=2^2$ with divisors $\{1,2\}$ and $\{1,2,4\}$ respectively where $\omega(2)=\omega(4)=1$. Prime factors with larger exponents are sort of weighted heavier than prime factors with smaller exponents. May 30 at 23:49
• @StevenClark what do you mean.
– PNT
May 31 at 0:13
• The situation is more rigid for the subsequence called superior highly composite numbers. To get to the next larger SHC number, we multiply the current one by a specific prime. en.wikipedia.org/wiki/Superior_highly_composite_number May 31 at 0:48

Your conjecture is not true. OEIS sequence A002182 is the sequence of highly composite numbers, and it gives \begin{align*}h_{25} &= 27720 = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \\ h_{26} &= 45360 = 2^4 \cdot 3^4 \cdot 5 \cdot 7 \end{align*} where $$d(h_{25}) = 2^5 \cdot 3 = 96$$ and $$d(h_{26}) = 2^2 \cdot 5^2 = 100$$. And $$\omega(h_{25}) = 5 > \omega(h_{26}) = 4.$$