The three-prime floor-function set formula

About 5 years old · traced to 2021

Let $G(x)$ denote the number of primes in the set $\left\{\left\lfloor x/n\right\rfloor:1\le n\le x\right\}$ . Suppose that $x=p_1p_2p_3$ , where $p_1,p_2,p_3$ are primes satisfying $2<p_1<p_2<p_3$ . The three-prime floor-function set formula. Then

G(x)=\begin{cases} G(x-1)&\text{if }p_1p_2>p_3,\\ G(x-1)+1&\text{if }p_1p_2<p_3. \end{cases}

This statement is presented as a generalisation suggested by a limited computational investigation; unlike the preceding prime and semiprime cases, no proof or resolution is supplied in the source.

References

Primary source

Randell Heyman, “Primes in floor function sets”, arXiv:2111.00408 (2021).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2021 conjecture remains unconfirmed, although a complete proof has since been posted and not independently verified.

Randell Heyman proposed this three-prime formula in 2021 after a limited Maple investigation. It predicts whether $G(x)$ changes when $x=p_1p_2p_3$ according to the comparison between $p_3$ and $p_1p_2$ .

Known results

For prime $x\ne3$ , $G(x)=G(x-1)+1$ (Heyman, 2021; preceding work attributed to Adams–Watters).
For $x=pq$ with odd primes $p,q$ , $G(x)=G(x-1)+1$ (Heyman, 2021).

Posted attempt

A reader claims a stronger identity for every odd $N\ge5$ , namely that $G(N)-G(N-1)$ counts prime divisors $q$ of $N$ satisfying $q^2\ge N$ ; this would prove the three-prime formula completely. The argument has not been independently verified.

Current status (as of August 2026): The conjecture has an unverified complete-proof claim, but no independent verification or published corroboration is recorded.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

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.