Are $14$ and $21$ the only "interesting" numbers?

Question

The numbers $14$ and $21$ are quite interesting.

The prime factorisation of $14$ is $2\cdot 7$ and the prime factorisation of $14+1$ is $3\cdot 5$. Note that $3$ is the prime after $2$ and $5$ is the prime before $7$.

Similarly, the prime factorisation of $21$ is $7\cdot 3$ and the prime factorisation of $21+1$ is $11\cdot 2$. Again, $11$ is the prime after $7$ and $2$ is the prime before $3$.

In other words, they both satisfy the following definition:

Definition: A positive integer $n$ is called interesting if it has a prime factorisation $n=pq$ with $p\ne q$ such that the prime factorisation of $n+1$ is $p'q'$ where $p'$ is the prime after $p$ and $q'$ the prime before $q$.

Are there other interesting numbers?

If $p_n$ denotes the $n$-th prime, then we search examples for $p_{n+1}p_{m-1}-p_np_m=1$. I have seen this before... but where? — Dietrich Burde, Nov 02 '16 at 14:46
Since one of $n$ and $n+1$ is even, you either have $p=2$ and $p'=3$ or $q'=2$ and $q=3$, so that either $q={3q'-1\over2}$ (if $pp'=2\cdot3$) or $p'={3p+1\over2}$ (if $q'q=2\cdot3$). The Prime Number Theorem limits the number of possibilities, and it should be fairly easy to find the upper limit. In effect, you want to find when a "$3/2$" version of Bertrand's Postulate kicks in. — Barry Cipra, Nov 02 '16 at 14:48
Given that there is no known general relation between the prime factorization of $N$ and the prime factorization of $N+1$ (except for the fact that they are always disjoint multi-sets, i.e., share no common elements), the answer to your question is most likely also unknown. — barak manos, Nov 02 '16 at 14:57
@BarryCipra If you see this wikipedia paragraph, they say that the $\frac65$ version of Bertrand's postulate kicks in at $25$. The $\frac32$ version can't kick in later, so there aren't a whole lot of primes that needs to be checked. — Arthur, Nov 02 '16 at 14:57
@Arthur, excellent, thanks. Do you want to write up an answer, or shall I? — Barry Cipra, Nov 02 '16 at 14:59
@BarryCipra You do it. It's your result. Also, I got enough upvotes already from an answer basically referring to Bertrand's postulate (it's my most upvoted answer yet). — Arthur, Nov 02 '16 at 15:00
When I saw the title of this question I thought it was gonna be the parody about Ramanujan's $14=7\cdot 2=2\cdot 7$. — BigbearZzz, Nov 02 '16 at 22:21
I think we should drop the requirement $p\ne q$ and accept $n=9$ as an interesting number as well. — Jeppe Stig Nielsen, Nov 02 '16 at 23:30
I wonder if I am the only one who read the title and thought of Taxi numbers. — Aron, Nov 03 '16 at 08:04
They become a lot less interesting when you realize that they are the only ones — Mike Vonn, Nov 03 '16 at 12:44
This is not ideal terminology. 'Interesting numbers' are a standard meta-mathematical concept, and the proof that there are no uninteresting numbers is the example par excellence of a meta-mathematical theorem. — jwg, Nov 04 '16 at 10:50
Finding primes such as $p_{n+1}p_{m-1}-p_np_m = 2$ might be more interesting. Like $5 \cdot 7 - 3 \cdot 11$ — ypercubeᵀᴹ, Nov 04 '16 at 17:04
Coincidentally I asked a fairly similar question on Quora last week: Is there any special name for the sequence of smallest consecutive pairs of numbers (m,m+1) whose factorizations contain all the first n primes? — smci, Nov 05 '16 at 13:33

Parcly Taxel · Accepted Answer · 2017-02-23T08:12:55.880

Note that exactly one of $n$ and $n+1$ is even. It follows that for $n$ to be interesting, either $n=3p$ and $n+1=2N(p)$ or $n=2p$ and $n+1=3P(p)$, where $P(p)$ and $N(p)$ are the previous and next primes to $p$ respectively. Rearranging we get that $p$ must satisfy one of the following two equations: $$\frac{3p+1}2=N(p)\tag1$$ $$\frac{2p+1}3=P(p)\tag2$$ However, by a 1952 result of Jitsuro Nagura, for $p\ge25$ there is always a prime between $p$ and $\frac65p$. In particular, if $p\ge31$ is a prime: $$\frac56p<P(p)<p<N(p)<\frac65p$$ But when $p\ge31$ the following inequalities are also true: $$\frac{2p+1}3<\frac56p\qquad\frac65p<\frac{3p+1}2$$ Therefore, if $p$ is to satisfy $(1)$ or $(2)$ above, it must be less than 31. This leaves a handful of cases to check for $p$, and we find that the only interesting numbers are 14 and 21 as conjectured.

The Nagura paper is a reference in the Wikipedia article on Bertrand's postulate. While those in the comments had saw it, sketching out the approach I use here, I already knew what to do; I did not read those comments in detail until after posting my answer.

For the record, it is worth pointing out that this method was previously sketched in the comments to the question by Barry Cipra and Arthur, including a link to Wikipedia (which cites Nagura's paper). — Bill Dubuque, Nov 02 '16 at 22:48
Even as a non-mathematician, that paper was great to read, interesting approach (though probably interesting mainly because I'm not well versed in the field). — Etheryte, Nov 02 '16 at 23:54

Are $14$ and $21$ the only "interesting" numbers?

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