# On quantities with no very small odd prime factors; a response to Wlodzimierz Holsztynski

In response to a comment posted under Powers of $2$ and the products of initial odd primes , I shall raise some questions about quantities near $O_n= P_{n+1}/2$, the product of the first $n$ odd primes. (Note the change in meaning of $O_n$; $P_n$ is the $n$th primorial.)

Before going into the questions, note that for integers $d$ of absolute value less than the largest prime factor of $O_n$ and coprime to $O_n$, such integers d have only 2 as a prime factor and $O_n + d$ has all its prime factors greater than the $n+1$st prime, unless $d$ has absolute value 1, in which case 2 is also a prime factor. Let this set of integers be called $C_n$.

1. (From the other question) How large a power of 2 divides $O_n^2 - 1$? As seen elsewhere, this quantity is not a power of 2 when $n>1$. More interestingly, the greater the power of 2 which divides this quantity, the further away one can demonstrate easily $O_n$ is from a power of 2.

2. Let $f(n)$ be that integer $e$ such that $2^e$ precisely divides $O_n^2 - 1$. So $f(1)=3, f(2)=5, f(3)=4.$ Is $f$ unbounded as a function of $n$? How fast does it grow? Is $f$ the sum of two nonconstant integer valued functions, one of them periodic and the other monotonic?

3. How many primes are in $C_n$?

4. Form $Q_n$, the product of the integers in $C_n$. This product is a multiple of 8, so is not squarefree. How far is it from being squarefree? In particular, is $Q_n$ the multiple of the square of any odd prime?

5. Given recent work on prime gaps (Zhang et al.), can we hope for a Generalized Dirichlet theorem? Specifically, translate the constellation $C_n$ by multiples of $O_n$: are there infinitely many translates which contain two or more primes in that translate?

I hope this makes my comment regarding related questions appear less vague and less broad.

• @TMA, the definition of $\ C_n\$ is not clear to me. Thank you for developing the topic of $\ 2^d+O\$ beyond its (very) modest start. – Włodzimierz Holsztyński Apr 13 '14 at 1:49
• I've been havi.g trouble su mitting edits. Cn is the set O_n + d where d ranges over plus/minus small powers of 2. – The Masked Avenger Apr 13 '14 at 1:54
• Thank you for the clarification. (I can see that you have problems with editing :-) – Włodzimierz Holsztyński Apr 13 '14 at 2:04
• Could anybody tell me about the "admissible-set" tag? It's new to me, and I am not able to find anything about it (my Internet skills are weak). – Włodzimierz Holsztyński Apr 13 '14 at 3:54
• All kinds of people could understand all kinds of things under "admissible set." Such a tag ought to at least have a tag wiki. If you care so much about this tag why not write one? I guess you did not mean en.wikipedia.org/wiki/Admissible_set for example, which one might consider as reason for using that tag for your purpose as no go. If you must creat hyperspecialized tags please at least make them so it is clear what is meant. – user9072 Apr 13 '14 at 17:26

Answer to question 2: obviously $f(n)\geq 3$ for all $n$, but I will show that $f(n)=3$ for infinitely many $n$, so we cannot give any other lower bound for the growth of $f$. To see this, notice that if $p_2...p_{n+1}\not \equiv \pm 1 \pmod 8$ (where $p_i$ is the $i$th prime), then the highest power of $2$ in $O_n^2-1$ is $2^3$. On the other hand, if $p_2...p_{n+1}\equiv\pm 1 \pmod 8$ and $n$ is chosen so that $p_{n+2}\equiv \pm 3 \pmod 8$ (by Dirichlet's theorem, there are infinitely many such $n$), we get $p_2...p_{n+2}\not \equiv \pm 1 \pmod 8$, so in any case there are infinitely many $n$ for which $f(n)=3$. We can even say something about the density of such $n$. By denoting $A=\{n:f(n)>3\}$, $B=\{n:f(n+1)=3\}$ and $C=\{n:p_{n+2}\equiv \pm 3 \pmod 8\}$, the natural denisties satisfy $d(A)+d(B)=1$ and we have $A\cap C \subset B$. Hence, using the PNT in aritmetic progressions, which tells that $d(C)=\frac{1}{2}$, we get $d(B)\geq d(A)-\frac{1}{2}=\frac{1}{2}-d(B)$, and from this we solve $d(B)\geq \frac{1}{4}$.
Answer to question 3: To clarify the definition, $C_n=\{O_n\pm d: d<p_{n+2} \hspace{0.1 cm}\text{is a power of two}\}.$ So the number of elements up to $x$ in $\bigcup_{n\geq 1} C_n$ grows like $\log^2 x$. As far as I know, no "natural" set that contains logarithmically few elements has been proved to contain infinitely many primes, so estimating the number of primes in $C_n$ seems hopeless. But one would expect that even $O_n+2$ is prime infinitely often.
Answer to question 4: first notice that the greatest common divisor of any two elements of $C_n$ is one. Indeed, if $q>2$ is a prime satisfying $q\mid O_n\pm 2^{\alpha}, q\mid O_n\pm 2^{\beta}$ and $2^{\alpha}\leq 2^{\beta}<p_{n+2}$, then $q\mid 2^{\beta-\alpha}-1,$ but this is a contradiction since $q\geq p_{n+2}$. So the question of $Q_n$ being divisible by an odd prime square reduces to an element of $C_n$ having this property. Again, this seems extremely hard since for example the following problem seems easier but is known to be open: Is $2^{p}-1$ ($p$ prime) always squarefree?
Answer to question 5: This would at least follow from Dickson's conjecture. According to it, there should even exist translates of $C_n$ that consist entirely (except $O_n\pm 1$) of primes. The constellation is admissible for this conjecture since if $q>2$ was a prime that divided one element of $kO_n+C_n$ for every $k$, then $q$ would be coprime to $O_n$ and bigger than $p_{n+1}$, but then $kO_n \pmod q$ would attain all the values $1,...,q-1$ as $k$ varies, which is a contradiction.
• In the definition of $C_n$, which you got mostly correct, I originally intended $|d|$ to be a power of 2 less than $p_{n+1}$, and now I want it less than $p_{n+2}$. This does not change the fundamental character of your answers though, and I appreciate your making precise the definition of $C_n$, as well as your answers. – The Masked Avenger Apr 18 '14 at 20:55
• True, $p_{n+2}$ is a better limit for $|d|$ as everything works in the same way. I fixed it now. – Joni Teräväinen Apr 18 '14 at 21:27