Even numbers of the form $\frac {n(n+1)}{2}$ and twin primes If we take some even number of the form $\frac{n(n+1)}{2}$ and add $1$ to it and also subtract $1$ from it then we have a mapping $\frac{n(n+1)}{2}\to\left\{\frac{n(n+1)}{2}-1,\frac{n(n+1)}{2}+1\right\}$.
From some even numbers of that form that I checked only number $6=\frac{4\cdot 3}{2}$ maps to a twin prime pair $6 \to \{5,7\}$.
Is there any simple explanation why $\left\{\frac{n(n+1)}{2}-1, \frac{n(n+1)}{2}+1\right\}$ is so rarely a twin prime pair?

Note that: $$\frac{n(n+1)}{2} - 1 = \frac{n^2 + n -2}{2} = \frac{(n+2)(n-1)}{2}$$
Now if we want this number to be prime the $2$ in the denominator must cancel out one of the factors in the numerator. Thus we must have $n+2 = 2$ or $n-1=2$. This doesn't leave too much possibilities. And you can see that the only such pair is $(5,7)$.
