Primes which divide exactly one odd composite in a sequence of consecutive odd composites

Question

Consider two consecutive prime integers, and consider a sequence of consecutive odd composites which lie between those two primes. Does there always exist a prime which divides exactly one of the composites in the sequence?

Answer 1

Let $p$ and $p+2k+2$ be two consecutive primes. If $k$ is large enough, then we will show that there is a prime $q$ that divides exactly one of the numbers $p+2,p+4,\dots,p+2k$ under two assumptions:

i) the ABC conjecture;

ii) $k\le p^{1/(7\log\log p)}$.

Both of these assumptions are believed to be true, the second one following from Cramer's conjecture which predicts that $k=O(\log^2p)$.

As the Masked Avenger observes, it suffices to show that the largest prime factor of $P=\prod_{i=1}^k(p+2i)$ is $>k$. Assume on the contrary that all prime factors of $P$ are $\le k$. Note that if $$ \omega(p+2i)+\omega(p+2i+2)\le \frac{\log p}{3\log k}=:H $$ for some $i\in\{1,\dots,k-1\}$, then the ABC conjecture, applied to the equation $p+2i+2=(p+2i)+2$, would imply that $$ p \le p+2i+2 \ll {\text rad}(2(p+2)(p+4))^2 \le 4k^{2H}\le 4k^2p^{2/3},$$ which is a contradiction. So we may assume from now on that $$ \omega(p+2i)+\omega(p+2i+2)>H. $$ Then $$ \sum_{i=1}^k \sum_{q|(p+2i)} 1\ge \frac{H(k-2)}{2} \sim \frac{k\log p}{6\log k}. $$ On the other hand, $$ \sum_{i=1}^k \sum_{q|(p+2i)} 1=\sum_{q\le k} \sum_{\substack{1\le i\le k \\ p+2i\equiv 0\pmod q}}1 =\sum_{2<q\le k} \left(\frac{k}{q}+O(1)\right) =k\log\log k+O(k), $$ by Mertens' estimate. Comparing the two estimates, we deduce that $$ 6(\log k)\log\log k\gtrsim \log p, $$ which contradicts our assumption that $k\le p^{1/(7\log\log p)}$.

Of course, the two conjectures assumed are not known to be true yet (and we might be far from having a proof that is accepted to be true for the first one). It is possible to show some partial results though. Jutila [On numbers with a large prime factor. II. J. Indian Math. Soc. (N.S.) 38 (1974), no. 1, 2, 3, 4, 125–130 (1975)] has shown that the greatest prime factor of $(u+1)(u+2)\cdots (u+k)$ is $>k$ if $\exp\{c(\log x)^{2/3}(\log\log x)^{2/3}\} \le h \le x^{2/3}$. Of course, if $p$ and $p'$ are two consecutive primes, then we expect that $p'-p=O(\log^2p)$, so the result proven does not handle the original question. (On the other hand, we do know that $p'\le p + p^{0.525}$, by a result of Baker, Harman and Pintz.)

Answer 2

Let $k$ be the number of odd composites in the interval. I temporarily extend the notion of composite to include nonprimes.

Suppose the odd numbers have no prime factors. Then they are 1 and -1, the consecutive primes are 2 and -2, and the answer is no for an uninteresting reason, because I am looking at units as well.

Suppose $k=0$, so the surrounding primes are twin primes. Then the answer again is no, but for a different uninteresting reason.

Otherwise $k \gt 0$, and there is a prime $q$ which divides at least one of the odd composites in that interval. If $q \geq k$, then $q$ cannot divide any other odd composite in that interval, and the answer is yes.

Now that the underbrush has been cleared, we get to the interesting heart of the matter. Are there two odd positive consecutive primes $p$ and $p + 2k + 2$ which have $k$ odd composites between them, and such that every prime factor of those odd composites is less than $k$, and also divides at least two of the composites?

Just by counting, one can show that $k$ will not be small and that any such bracketing primes also will not be small. Note first that for every odd prime factor $q$, at most one prime power of $q$ can occur in the interval, by using Bertrand's postulate or something slightly weaker. Then there must be at least two odd prime factors and so $k \gt 5$.

Further one must have $\sum_i \lceil k/q_i \rceil \geq 2k - l$, where the sum is over the $l$-many distinct prime factors of (the product of) the odd composites. Also $l \lt \pi(k)$, so one will need quite a few primes to acheive this inequality. Using a rough approximation of sum of reciprocals, I guess $k \gt 47$. There may be literature on $k$-smooth numbers which will settle this, but I have no references for you.

Expanding on the remarks and inequality above, let the $l$-many distinct prime factors be denoted by $q_i.$ Each such will be a factor of at most $\lceil k/q_i \rceil$ of the odd composites in the interval, and at most one composite can have the form $q_i^t$ for that $i$ and some integer $t$. So each of the $k$ odd composites have at least one of the $q_i$ as a factor, and at most $l$ of them have exactly one of them as a factor. This is the basis for the somewhat weak inequality above.

Writing $M= \sum_i 1/q_i$, I weaken the inequality further to $M(k-1) + l \geq 2k - l$ and rearrange to get the sum of prime reciprocals $M \geq 2 - \frac{2(l-1)}{k-1}$. Since $l \lt \pi(k)\leq k/2$ when $k \geq 4$, one has $M \geq 1$ which implies $k\geq 30$ and $l \geq 9$. This bootstraps to (since now $l/k \leq 9/30$ by considering integers coprime to $30$) $M\geq 7/5$ which cranks up the allowed values of $k$ and $l$ to $164$ and $37$. Doing this twice more, followed by looking closely at $(l- 1)/(k- 1)$ when $k$ is one more than a prime and $l$ is the count of odd primes takes us to $k \geq 542$ and $p \gt 10^{15}$, since we are now talking about prime kilogaps.