I am investigating properties of the number of divisors of composites between and adjacent to twin primes. When running some numeric calculations in Python (which are hopefully correct) I get the following result:
I want to somehow prove that this tendency is always true, meaning that the number between two twin primes tend to have more divisors than the ones adjacent to the twins.
This is my first attempt on doing something, hehe. Please tell me if this is at all true and if it somehow can be used to prove the suggested tenedency:
Prime numbers are always of the form $6k\pm 1$ except for the case when the prime is either $2$ or $3$. Let $p$ and $q$ be twin primes, and let $n$ be the number that lies between $p$ and $q$. If $p = 6k-1$, then $q = 6k+1$. Furthermore, we have that $n-2 = 6k-2$, $n = 6k$, and $n+2 = 6k+2$. Now, considering $n, n-1$ and $n+2$ under the $\tau (n)$ function, which represents the sum of divisors function, we have:
$\tau(n)=\tau(6k)\geq 4$ since $6k$ has at least the divisors $1, 2, 3$ and $6k$. $\tau(6k−2)\geq 3$ since $6k-2$ has at least the divisors $1, 2$ and $6k-2$.
Similarly, $\tau(6k+2)\geq 3$.
Therefore, we have that $\tau(n) - \tau(n-2) > 0$ and $\tau(n) - \tau(n+2) > 0$.
Given that the above holds true we can also conclude that $$\dfrac {\tau (n)}{\tau (n) + \tau (n-2) + \tau (n+2)} \geq \dfrac{4}{10}.$$