Semiprime Covering of Residue Classes Modulo a Primorial

Semiprime Covering of Residue Classes Modulo a Primorial

Let $p$ and $q$ be consecutive primes, with $q$ the smallest prime greater than $p$, and let

$p\#=\prod_{\ell\le p,\ \ell\text{ prime}}\ell$

denote the primorial of $p$.

Let $\mathcal S$ denote the set of semiprimes,

$\mathcal S=\{rs:r,s\text{ are primes}\},$

where $r=s$ is allowed.

I am interested in the following statement.

Proposed Semiprime Primorial Covering Theorem. For every prime $p\ge 5$, with $q$ the next prime after $p$, and for every residue class $a\pmod{p\#}$, there exists a semiprime $s\le q^2$ such that

$\gcd(a-s,p\#)=1.$

Equivalently, for every integer $n$, there exists a semiprime $s\le q^2$ such that

$\gcd(n-s,p\#)=1.$

Since $p$ and $q$ are consecutive primes, this means that every prime divisor of $n-s$ is at least $q$.

Another equivalent formulation is the following. Define

$\mathcal S_q=\{s\in\mathcal S:s\le q^2\}.$

Then the translates of the reduced residue system modulo $p\#$ by the elements of $\mathcal S_q$ cover every residue class modulo $p\#$:

$\bigcup_{s\in\mathcal S_q}\left(s+(\mathbb Z/p\#\mathbb Z)^\times\right)=\mathbb Z/p\#\mathbb Z.$

In other words, for every $a\in\mathbb Z/p\#\mathbb Z$, there is some $s\in\mathcal S_q$ for which

$a-s\in(\mathbb Z/p\#\mathbb Z)^\times.$

The bound $q^2$ appears naturally. In particular, for the residue class $a\equiv0\pmod{p\#}$, every semiprime composed entirely of primes at most $p$ has a nontrivial common factor with $p\#$, whereas

$\gcd(q^2,p\#)=1.$

Thus $q^2$ always handles the zero residue class.

The statement has been checked computationally for the following cases:

$p=5,\ 7,\ 11,\ 13,\ 17,\ 19,$

corresponding to

$p\#=30,\ 210,\ 2310,\ 30030,\ 510510,\ 9699690.$

In each of these cases, every residue class modulo $p\#$ is covered by some semiprime $s\le q^2$.

Moreover, in each tested case the smallest universal upper bound required to cover all residue classes is exactly $q^2$:

$\begin{array}{c|c|c|c} p & p\# & q & \text{maximum required semiprime shift} \\\hline 5 & 30 & 7 & 49\\ 7 & 210 & 11 & 121\\ 11 & 2310 & 13 & 169\\ 13 & 30030 & 17 & 289\\ 17 & 510510 & 19 & 361\\ 19 & 9699690 & 23 & 529 \end{array}$

The motivation is additive number theory. If the proposed statement holds for every $p$, choose $p$ to be the largest prime satisfying

$p\le n^{1/3},$

and let $q$ be the next prime. Then

$q>n^{1/3}$

and hence

$q^3>n.$

Choose a semiprime $s\le q^2$ such that

$\gcd(n-s,p\#)=1.$

Every prime factor of $n-s$ must then be at least $q$. If $n-s$ had three or more prime factors, counted with multiplicity, we would have

$n-s\ge q^3>n,$

which is impossible because $s>0$.

Therefore $n-s$ has at most two prime factors. Hence $n-s$ is either prime or semiprime, giving

$n=\text{semiprime}+\text{prime}$

or

$n=\text{semiprime}+\text{semiprime}.$

So my question is:

Is the proposed semiprime primorial covering statement known? Can it be proved, disproved, or related to a standard covering or sieve theorem? In particular, is there a way to prove that the bound $q^2$ suffices for every pair of consecutive primes $p<q$?