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?

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Simon Parker
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    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
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    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
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    @BarryCipra If you see [this wikipedia paragraph](https://en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results), 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
  • @barakmanos See the comment right above yours. – Arthur Nov 02 '16 at 14:59
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    @Arthur, excellent, thanks. Do you want to write up an answer, or shall I? – Barry Cipra Nov 02 '16 at 14:59
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    @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
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    Well, Parcly Taxel has relieved us both of the burden. – Barry Cipra Nov 02 '16 at 15:08
  • 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
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    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
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    I wonder if I am the only one who read the title and thought of Taxi numbers. – Aron Nov 03 '16 at 08:04
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    They become a lot less interesting when you realize that they are the only ones – Mike Vonn Nov 03 '16 at 12:44
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    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?](https://www.quora.com/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

1 Answers1


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.

Parcly Taxel
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    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
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    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