What are the smallest numbers $n$ such that $\dfrac{d(n)}{\ln(n)} \geq k$ where $d(n) = \sigma_0(n)$ is the number-of-divisors function?

Question

I have calculated $\dfrac{d(n)}{\ln(n)}$ on a few highly composite numbers up to 5040.

Here is what I got:

$\dfrac{d(120)}{\ln(120)} = 3.3420423$

$\dfrac{d(360)}{\ln(360)} = 4.0773999$

$\dfrac{d(1260)}{\ln(1260)} = 5.0428170$

$\dfrac{d(2520)}{\ln(2520)} = 6.12869166$

$\dfrac{d(5040)}{\ln(5040)} = 7.03798995$

So we have:

$k=3 : n=120$

$k=4 : n=360$

$k=5 : n=1260$

$k=6 : n=2520$

$k=7 : n=5040$

Notice how, except for $1260$, all of them are superior highly composite numbers!

Could someone calculate the next few $n$ up to $k=12$ ?

Answer 1

Let $\{p_i\}_{i=1}^{\infty}$ be the primes in the usual order: $p_1 = 2, p_2 = 3, \dots$ Let $f:\mathrm{Z} \rightarrow \mathrm{Z}: n \mapsto \frac{d(n)}{\ln n}$ be the function of interest. Let $n$ be an integer with factorization $n = \prod_{i=1}^\infty p_i^{e_i}$. Then we compute directly that $d(n) = \prod_{i=1}^{\infty} (e_i+1)$ and $\ln(n) = \sum_{i=1}^\infty e_i \ln p_i$.

Imagine a sequence of exponents $\{e_i\}_{i=1}^\infty$ (where only a finite number of exponents are nonzero so that the corresponding $n$ is finite). We may permute these $e_i$ without changing the number of divisors of the corresponding $n$s. To maximize $f$ over these corresponding $n$s we must minimize $\ln n$, which means we must assign the largest of the $e_i$ to the first several primes, then the next largest $e_i$, and so on, so that the sequence of exponents is nonincreasing. Therefore, the only numbers that can appear in the desired list of record values of $f$ have nonincreasing sequences of exponents in their prime decompositions.

To organize the search, sequentially fix $s$ and for each value of $s$, enumerate all nonincreasing sequences of exponents summing to $s$, i.e., such that $s = \sum_{i=1}^\infty e_i$. Then evaluate $f$ applied to each of these values and accumulate the records for each unit interval. (That is, for each unit interval, track the least integer which produced a value in that interval.) Applying this method, we get the sequence continuation (with corresponding approximate values of $f$ shown in parentheses): $$..., 15120 (8.313), 27720 (9.384), 55440 (10.986), 83160 (11.299), 110880 (12.397), ... $$

$f(55440) = 10.986$ suggests, that there are larger numbers than those on our list with smaller values of $f$ in their unit intervals. This is true, but not interesting. The sequence of powers of $2$ produce ever decreasing values. For instance $1.503 \dot{=} f(2^{14}) < f(2^{13}) < f(2^{12}) < \dots$

Answer 2

If we add the restriction $k< \frac{d(n)}{ln(n)} < k+1$ then:

k=8 n=15120

k=9 n=27720

k=10 n=55440

k=11 n=83160

k=12 n=110880

k=13 n=166320

k=14 n=277200

k=15 n=332640

k=16 n=554400

k=17 n=720720

k=18 n=1081080

k=20 n= 1441440

It seems there are no highly composite number (HCN) with $19< \frac{d(n)}{ln(n)} < 20$

but $19<\frac{d(1801800)}{ln(1801800)}<20$,

so if we consider the restriction $k< \frac{d(n)}{ln(n)} < k+1$

it's false n must be a HCN.