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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.

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  • $\begingroup$ How many values have you checked? Per wiki $h_{25}=2^3\cdot3^2\cdot5\cdot 7\cdot 11$ but $h_{26}=2^4\cdot 3^4\cdot 5\cdot 7$. $\endgroup$
    – Sil
    Commented May 30, 2023 at 23:14
  • $\begingroup$ 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. $\endgroup$ Commented May 30, 2023 at 23:49
  • $\begingroup$ @StevenClark what do you mean. $\endgroup$
    – PNT
    Commented May 31, 2023 at 0:13
  • $\begingroup$ 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 $\endgroup$
    – Will Jagy
    Commented May 31, 2023 at 0:48

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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 $$ \omega(h_{25}) = 5 > \omega(h_{26}) = 4. $$

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