Proving that there exists $w$ such that $4x < 6w < 6x$ and $\gcd(w,\frac{x\#}{6})=1$ where $x \ge 7$ and $x\#$ is the primorial

Question

I am trying to show that for any integer $x \ge 7$, there exists $w$ with the following properties:

• $4x < 6w < 6x$
• $\gcd\left(6w,\frac{x\#}{6}\right)=1$

I thought that this would be pretty easy to do.

Since:

• $4\cdot7 < 36 < 6\cdot7$
• $4\cdot8 < 36 < 6\cdot8$
• $4\cdot9 < 48 < 6\cdot9$
• $4\cdot10 < 48 < 6\cdot10$
• $4\cdot11 < 48 < 6\cdot11$
• $4\cdot12 < 54 < 6\cdot12$
• $4\cdot13 < 54 < 6\cdot13$
• $4\cdot14 < 72 < 6\cdot14$
• $4\cdot15 < 72 < 6\cdot15$
• $4\cdot16 < 72 < 6\cdot16$
• $4\cdot17 < 72 < 6\cdot17$
• $4\cdot18 < 96 < 6\cdot18$
• $4\cdot19 < 96 < 6\cdot19$
• $4\cdot20 < 96 < 6\cdot20$

So, it can be assumed that it is true up to some $x \ge 20$. If $x$ is odd, then it is true since there exists $6w$ by assumption for:

$$4\left(\frac{x+1}{2}\right) < 6w < 6\left(\frac{x+1}{2}\right)$$

So, it follows that:

$$4\cdot(x+1) < 12w < 6\cdot(x+1)$$

If $x$ is even and $x \equiv 2 \pmod 3$, it is true since there exists $6w$ by assumption for:

$$4\left(\frac{x+1}{3}\right) < 6w < 6\left(\frac{x+1}{3}\right)$$

So, it follows that:

$$4(x+1) < 18w < 6(x+1)$$

For all other cases where $x$ is even, we know it is true since there exists $6w$ by assumption for:

$$4\cdot\left(\frac{x}{2}\right) < 6w < 6\cdot\left(\frac{x}{2}\right)$$

So, it follows that:

$$4x < 12w < 6x$$

We know that:

• $12w \ne 4x+1$
• $12w \ne 4x+2$ since $4 \nmid 4x+2$
• $12w \ne 4x+3$
• $12w \ne 4x+4$ since $x \not\equiv 2 \pmod 3$

So it follows that $12w > 4x+4$ as required.

This seems like a very long way to prove this? Is there an easier way? Did I make a mistake in my reasoning?

Answer 1

Since $w < x$, the condition $(6w, \#x/6)=1$ amounts to $w$ being $3$-friable (also called $3$-smooth), that is $w = 2^r 3^s$ for some $r,s \ge 0$. Notice that we just need $\frac{2x}{3} < w < x$, so:

• For $k\ge 0$, $w = 2^k$ works whenever $x$ lies in the dilated interval $ 2^k\cdot (1,\tfrac32)$.
• For $k\ge 3$, $w = \tfrac98\cdot 2^k$ works whenever $x \in 2^k \cdot(\tfrac98, \tfrac{27}{16})$.
• For $k\ge 1$, $w = \tfrac32\cdot 2^k$ works whenever $x \in 2^k \cdot(\tfrac32, \tfrac94)$.

Now just observe that the union of the three open intervals above contains the half-open interval $(2^k,2^{k+1}]$. Adjusting $k$ will then cover all sufficiently large $x$ (we need the restrictions on $k$ in order for the chosen value of $w$ to be an integer). In particular, starting at $w = \tfrac32\cdot 2^2 = 6$ (and then cycling through to $w = 2^3$ and $w = \tfrac98 \cdot 2^3$) this covers all $x > 6$.

If it wasn't for the strict inequalities in the subject line, this construction would've worked with just $2^k$ and $\tfrac32\cdot 2^k$, but we needed the $\tfrac98$ case in order to have a positive overlap between the $x$-intervals.