Finding positive integer $n>10$ that maximizes $\frac{\sigma_0(n)}{2^{\log n}}$

Question

Among all the positive integer, which one integer, $n$, can make the number below the largest?

$$f(n)=\frac{\sigma_0(n)}{2^t}$$where $t=\log_{10}n$ and $\sigma_0$ is the divisor function.

For example, $1000$ has the value of $t=3$, as it has $16$ factors i.e. $\sigma_0(1000)=16$ and $f(1000)=2$.

By the way, the number has to be bigger than $10$.

Thanks!

Answer 1

Consider the expression: $$ n := 2^{n_1} \cdot 3^{n_2} \cdot 5^{n_3} \cdot 7^{n_4} \cdot \ldots$$ where $ \sigma_0(n) $ represents the number of positive factors of $ n $, given by $(n_1+1)(n_2+1)(n_3+1)(n_4+1)\ldots$, according to the fundamental counting principle.

Therefore, we want to maximize the following expression: $$ \frac{\sigma_0(n)}{2^{\log_{10} n}} = \frac{(n_1+1)(n_2+1)(n_3+1)(n_4+1)\ldots}{2^{n_1\log_{10}2 + n_2\log_{10}3 + n_3\log_{10}5 + n_4\log_{10}7 + \ldots}} $$

Simplifying further, we have: $$ = \prod \frac{n_r+1}{\left(2^{\log_{10}p_r}\right)^{n_r}} \quad \text{where,} \quad p_r \text{ is the } r^{th} \text{ prime number} $$

To maximize each term individually, we use the fact that the expression $ \frac{x+1}{a^x} $ is maximized at $ x = \frac{1-\ln(a)}{\ln{a}} $.

Therefore, the maximum value for each term occurs when: $$n_r = \frac{1-\ln{\left(2^{\log_{10}p_r}\right)}}{\ln\left(2^{\log_{10}p_r}\right)} $$

Approximating these values, we find: $$\boxed{n_1\approx3.79\rightarrow4\\n_2\approx2.03\rightarrow2\\n_3\approx1.06\rightarrow1\\n_4\approx0.71\rightarrow1\\\\n_{r\ge5}<<1\rightarrow0}$$

$\rightarrow$ nearby integer.

Therefore, the solution is $$ n = 2^4 \cdot 3^2 \cdot 5^1 \cdot 7^1 = 5040 = 7!$$, the nearby integer that maximizes the given expression.

Answer 2

So you're looking to maximise $\sigma_0(n)\cdot n^{-c}$ where $c = \log_{10}(2)$.

Suppose $n = \prod p_k^{a_k}$. Then you're looking at

$$\prod (a_k +1) \cdot p_k^{-a_kc}$$

The partial product containing $p_k$ is maximised whenever $a_k$ is nearest to $\frac{\ln(10)}{\ln(2)\ln(p_k)} -1 $ (via derivative).

Since $$1> 2\cdot x^{-\log_{10}(2)}$$

is satisfied by all $x \geq 11$, it follows that we can restrict $p_k \in \{2,4,5,7\}$. (since the inequality implies for all $x \geq 11$, it is better to just have $x^0$ instead of $x^n$ where $n >0$.)

For each of these $p$, you need to solve for the closest integer to the greatest $x$ such that:

$$(x+1) > x\cdot p^{c}$$

And that integer is the desired value of $a_k$ (since we are getting the value of $a_k$ that maximizes the partial product). You can equivalently use derivatives.

Putting together all $a_k$s you get the desired answer $7!$.