What is the lower limit of $f(n):= \frac{1}{\ln{(n+1)}}\max_\limits{0\leq{m}\leq{n},m\in\mathbb{N}\\0<{p}\leq{n},p\in\mathbb{P}}\{v_p\binom{n}{m}\}$

Question

Let $v_p(n) = k$ express that $p^k \mid{n}$ and $p^{k+1}\nmid{n}$ for every positive integers $p,n$ and nonnegative integer $k$, and $[x]$ denote the integer part of a real number $x$.

Define $f(n):= \frac{1}{\ln{(n+1)}}\max_\limits{0\leq{m}\leq{n},m\in\mathbb{N}\\0<{p}\leq{n},p\in\mathbb{P}}\{v_p(\binom{n}{m})\}$ , where $n$ is a positive integer.

Question: what is the lower limit of $f(n)$ ,or $\liminf_\limits{n\to\infty} f(n)$ ?

Background: The function $f(n)$ is related with a question in OEIS A048278 :

• Find all positive integers $n$ such that $\binom{n}{m}$ is squarefree for each $0\leq{m}\leq{n}$.

It can be solved and generalized by proving that $f(n)$ has a positive lower bound.

Related result: Suppose $p\leq{n}$ is a prime. A formula of $\max_\limits{0\leq{m}\leq{n}}\{v_p(\binom{n}{m})\}$ says that $$\max_\limits{0\leq{m}\leq{n}}\{v_p(\binom{n}{m})\} = \left[\frac{\ln(n+1)}{\ln{p}}\right]-v_p(n+1)$$ which is a consequence of Kummer's theorem. It can also be derived from the following identity related with $\text{lcm}(1, 2, ..., n+1)$, $$\displaystyle (n+1) \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... ,{n \choose n} \right) = \text{lcm}(1, 2, ..., n+1)$$ and the proof of this identity has been discussed here.

The formula implies that

• ${f(n)}$ has upper bound $\frac{1}{\ln{2}}$.
• $f(n)$ doesn't have a limit for $n\in\mathbb{N}$ , since that if $n$ is big enough, it follows from the formula that $$\max_\limits{0\leq{m}\leq{n}\\p\leq{n},p\in\mathbb{P}}\{v_p(\binom{n}{m})\}= \begin{cases} \left[\frac{\ln(n+1)}{\ln{2}}\right]&\text{if }n \text{ is even},\\ \left[\frac{\ln(n+1)}{\ln{3}}\right]&\text{if }n+1 \text{ is a power of 2}. \end{cases} $$
• $\limsup\limits_{n\to\infty}f(n)=\frac{1}{\ln{2}}=1.442…$, as a result of the first case above.

The following is to show that $f(n)$ has a positive lower bound.

A lower bound of $f(n)$: Suppose $n+1=2^r3^s5^tw$ , where $r,s,t$ are nonnegative integers and $w$ is a positive integer which is relatively prime to $30$.

From the formula $$f(n)\ln(n+1)\geq\max_\limits{0\leq{m}\leq{n}}\{v_p(\binom{n}{m})\}=\left[\frac{\ln(n+1)}{\ln{p}}\right]-v_p(n+1)$$ It follows that $$2^{f(n)\ln(n+1)+1} > \frac{n+1}{2^r}$$ $$3^{f(n)\ln(n+1)+1} > \frac{n+1}{3^s}$$ $$5^{f(n)\ln(n+1)+1} > \frac{n+1}{5^t}$$ Multiply these inequalities above , we have $$30^{f(n)\ln(n+1)+1} > \frac{(n+1)^3}{2^r3^s5^t}\geq(n+1)^2$$ As a result, $$f(n)>\frac{2}{\ln{30}}-\frac{1}{\ln(n+1)}$$ Therefore, the lower bound of $f(n)$ exists. More precisely, the lower limit of $f(n)$ is not less than $\frac{2}{\ln{30}}=0.588…$ .

Similar methods seems to be unable to provide a lower bound $>\frac{2}{\ln{30}}$, but I doubt whether the lower limit of $f(n)$ is really this constant. Perhaps it is just eventually $\frac{1}{\ln{3}}=0.910…$ ?

Answer 1

It seems that the lower limit of $f(n)$ is $\frac{2}{\ln{30}}$ indeed.

For large $k$, if we choose $n=2^{[k\alpha]}3^{[k\beta]}5^{[k\gamma]}-1$, where

$$\alpha=\frac{1}{\ln{2}}-\frac{2}{\ln{30}}$$ $$\beta=\frac{1}{\ln{3}}-\frac{2}{\ln{30}}$$ $$\gamma=\frac{1}{\ln5}-\frac{2}{\ln{30}}$$

Then $\ln{(n+1)} = [k\alpha]\ln2+[k\beta]\ln3+[k\gamma]\ln5 $, which implies that

$$k-\ln30<\ln{(n+1)}\leq{k(\alpha\ln2+\beta\ln3+\gamma\ln5)}=k$$

Therefore, we have

$$\max_\limits{0\leq{m}\leq{n}}\{v_2(\binom{n}{m})\} = \left[\frac{\ln{(n+1)}}{\ln2}\right]-\left[k\alpha\right] < {1+\frac{\ln{(n+1)}}{\ln2}-k\alpha} \leq 1+k\frac{2}{\ln{30}}$$

Since $k<\ln30+\ln{(n+1)}$, then $$\max_\limits{0\leq{m}\leq{n}}\{v_2(\binom{n}{m})\}<3+\frac{2\ln{(n+1)}}{\ln{30}}$$

And similarly, $$\max_\limits{0\leq{m}\leq{n}}\{v_3(\binom{n}{m})\}<3+\frac{2\ln{(n+1)}}{\ln{30}}$$ $$\max_\limits{0\leq{m}\leq{n}}\{v_5(\binom{n}{m})\}<3+\frac{2\ln{(n+1)}}{\ln{30}}$$

For prime $p>7$, $$\max_\limits{0\leq{m}\leq{n}}\{v_p(\binom{n}{m})\} = \left[\frac{\ln{(n+1)}}{\ln{p}}\right] \leq{\frac{k}{\ln{p}}} \leq{\frac{k}{\ln7}} <{\frac{\ln30+\ln{(n+1)}}{\ln7}}$$

Note that $\frac{2}{\ln{30}}>\frac{1}{\ln7}$, so we have $$\limsup_\limits{k\to\infty}f(2^{[k\alpha]}3^{[k\beta]}5^{[k\gamma]}-1) \leq {\frac{2}{\ln{30}}}$$

While it has been proved that $\liminf_\limits{n\to\infty}f(n)\geq{\frac{2}{\ln{30}}}$, thus,

$$\liminf_\limits{n\to\infty}f(n)=\lim_\limits{k\to\infty}f(2^{[k\alpha]}3^{[k\beta]}5^{[k\gamma]}-1)={\frac{2}{\ln{30}}}$$