The three-prime floor-function set formula

In fact, a stronger statement holds for every odd integer $N\ge5$:

$$G(N)-G(N-1) = \#\{q\text{ prime}:q\mid N,\ q^2\ge N\},$$

where $G(M)$ counts the distinct primes in

$$S(M)=\left\{\left\lfloor\frac Mj\right\rfloor: 1\le j\le M\right\}.$$

For every integer $q\ge2$, the number of indices $j$ for which $\lfloor M/j\rfloor=q$ is

$$D_M(q) = \left\lfloor\frac Mq\right\rfloor - \left\lfloor\frac M{q+1}\right\rfloor.$$

Thus $q\in S(M)$ if and only if $D_M(q)>0$.

If $N$ and $q$ are both odd, then $q+1$ is even and cannot divide $N$. Consequently

$$D_N(q)-D_{N-1}(q)=\mathbf1_{\{q\mid N\}}.$$

Therefore no odd prime disappears when passing from $N-1$ to $N$, and a new odd prime must divide $N$.

Write $N=qm$ for such a prime divisor. Then

$$D_{N-1}(q) = m-1-\left\lfloor\frac{qm}{q+1}\right\rfloor = \left\lceil\frac m{q+1}\right\rceil-1.$$

Hence $q$ appears for the first time precisely when

$$m\le q+1.$$

Since $m$ and $q$ are odd, this is equivalent to $m\le q$, or

$$q^2\ge N.$$

Finally, $2\in S(M)$ for every $M\ge4$, so the prime $2$ contributes no change when $N\ge5$. This proves the general formula.

Now suppose

$$N=p_1p_2p_3,\qquad 2<p_1<p_2<p_3.$$

Neither $p_1$ nor $p_2$ can satisfy $q^2\ge N$. The only possible new prime is $p_3$, and

$$p_3^2\ge N \quad\Longleftrightarrow\quad p_3>p_1p_2.$$

Equality is impossible because $p_3$ is prime. Therefore

$$G(N)-G(N-1) = \begin{cases} 1,&p_3>p_1p_2,\\ 0,&p_3<p_1p_2, \end{cases}$$

which proves both asserted cases.