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

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? Nov 2 '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. Nov 2 '16 at 14:48
• @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. Nov 2 '16 at 14:57
• Well, Parcly Taxel has relieved us both of the burden. Nov 2 '16 at 15:08
• I think we should drop the requirement $p\ne q$ and accept $n=9$ as an interesting number as well. Nov 2 '16 at 23:30

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.