# Claim: $a$ has $90 \%$ primes less than $n$ If $n!= 2^s \times a \times b$ and $\lfloor{\frac{a}{b}}\rfloor = 2^{s-2}$

Description: We can write, $n!= 2^s \times a \times b \cdots (1)$

where $gcd(a,b)=1$ and $2^{s+1} \nmid n!$ .It is given that, $\lfloor{\frac{a}{b}}\rfloor = 2^{s-2}$.

Claim: If $n!= 2^s \times a \times b$ and $\lfloor{\frac{a}{b}}\rfloor = 2^{s-2}$, then $a$ has $90 \%$ primes less than $n$ as factor.

Proof:
Here, $\nu_ p(n)$ denotes the p-adic valuation of $n$ and $s_p(n)$ denotes the sum of the standard base-p digits of $n$, so,
$\nu _{p}(n!)={\frac {n-s_{p}(n)}{p-1}}$ ($\textit{Legendre’s formula}$). For $p = 2$, we obtain $\nu _{2}(n!)= n-s_{2}(n)$, where $s_{2}(n)$ is the number of $1$’s in the binary representation of $n$. The number of primes less than $n$ is denoted by $\pi(n)$. By $\textit{Prime Number Theorem}$, $\pi(n)\approx \frac{n}{\log n}$.

Since, $\lfloor{\frac{a}{b}}\rfloor = 2^{s-2}$

$\implies \log_2(a )-\log_2(b)= \log_2 (2)^{s-2}$ [ignoring a quantity $<1$ on R.H.S]

$\implies \log_2(a )-\log_2(b )=n- s_2(n)-2$

$\implies \sum_{p \mid a}\log_2( p ^{\nu _{p}(n!)})-\sum_{q \mid b}\log_2( q ^{\nu _{q}(n!)})=n- s_2(n)-2 \dots (2)$

Equation $(2)$ implies that,

$\sum_{p \mid a}\log_2( p ^{\nu _{p}(n!)})-\sum_{q \mid b}\log_2( q ^{\nu _{q}(n!)}) >0 \cdots (3)$

Now, we consider the smallest prime factor, $p_0$ of $a$ and the biggest prime factor$q_0$ of $b$. Note, $a$ has $i$ distinct prime factors, $b$ has $j$ prime factors and $i+j = \pi(n)-1\approx \frac{n}{\log n}-1$. Since,

$\sum_{p \mid a}\log_2( p ^{\nu _{p}(n!)}) < (\log_2(p_0 ^{\nu _{p_0}(n!)}) ) \times i$
and,

$\sum_{q \mid b}\log_2( q ^{\nu _{q}(n!)}) >(\log_2( q_0^{\nu _{q_0}(n!)}) ) \times j$ (See 1, Theorem 3.9 on page 7), so-

$(\log_2( p_0 ^{\nu _{p_0}(n!)}) ) \times i-(\log_2(q_0^{\nu _{q_0}(n!)}) ) \times j>\sum_{p \mid a}\log_2( p ^{\nu _{p}(n!)})-\sum_{q \mid b}\log_2(q ^{\nu _{q}(n!)}) \cdots (4)$

$\implies (\log_2(p_0 ^{\nu _{p_0}(n!)}) ) \times i-(\log_2( q_0 ^{\nu _{q_0}(n!)}) ) \times j>0$

$\implies i \times \log_2( p_0 ^{\nu _{p_0}(n!)}) > j \times \log_2( q_0 ^{\nu _{q_0}(n!)} )$

$\implies i > j \times \frac{ \log_2( q_0 ^{\nu _{q_0}(n!)}) }{\log_2( p_0 ^{\nu _{p_0}(n!)}) }$

Let, $T=\frac{ \log_2( q_0 ^{\nu _{q_0}(n!)}) }{\log_2(p_0 ^{\nu _{p_0}(n!)}) }$.

$\therefore i > (\pi(n)-1-i) \times T$

$\implies i > T \times (\pi(n) -1)- T \times i \implies i> \frac{T}{(1+T)} \times (\pi(n) -1)\cdots (5)$

If $\frac{T}{(1+T)} \approx 0.9$ so, $i> 0.9 \times (\pi(n)-1)$.

Query: Is the above proof correct? Please, let me know if anything is inconsistent.

Reference :

1. Diophantine equations involving arithmetic functions of factorials, Daniel M. Baczkowski, A Thesis for Master’s of Arts, Miami University.

#### Solutions Collecting From Web of "Claim: $a$ has $90 \%$ primes less than $n$ If $n!= 2^s \times a \times b$ and $\lfloor{\frac{a}{b}}\rfloor = 2^{s-2}$"

Since, $\lfloor{\frac{a}{b}}\rfloor = 2^{s-2}$
$\implies \log_2(\lfloor a \rfloor)-\log_2(\lfloor b \rfloor)= \log_2 (2)^{s-2}$…
This is certainly not correct. What happens if (for example) $a=13,b=3$?