A formula that counts exactly the twin prime averages occuring in an interval $[a,b]$ is surprisingly succinct!

Too long to insert in a comment. I don't think the formula can have any practical use.

Also, for large values ​​of n it does not give an exact result.

Yet this exact formula which has a similar complexity but which has no practical use:

$\pi_{\text{twin}} (n) = 1 + \\ \sum \limits_{k=1}^{\left\lfloor{\frac{n-1}{6}}\right\rfloor} {\left\lfloor \left( \displaystyle 1 + \displaystyle\sum_{a=1}^{\left\lfloor\frac{1+\sqrt{-1+6\cdot k}}{6}\right\rfloor} {\left( \left\lfloor\frac{-1+6\cdot k}{-1+6\cdot a}\right\rfloor - \left\lfloor\frac{-2+6\cdot k}{-1+6\cdot a}\right\rfloor \right)} + \displaystyle\sum_{a=1}^{\left\lfloor\frac{-1+\sqrt{-1+6\cdot k}}{6}\right\rfloor} {\left( \left\lfloor\frac{-1+6\cdot k}{1+6\cdot a}\right\rfloor - \left\lfloor\frac{-2+6\cdot k}{1+6\cdot a}\right\rfloor \right)} \\ + \displaystyle\sum_{a=1}^{\left\lfloor\frac{1+\sqrt{1+6\cdot k}}{6}\right\rfloor} {\left( \left\lfloor\frac{1+6\cdot k}{-1+6\cdot a}\right\rfloor - \left\lfloor\frac{6\cdot k}{-1+6\cdot a}\right\rfloor \right)} + \displaystyle\sum_{a=1}^{\left\lfloor\frac{-1+\sqrt{1+6\cdot k}}{6}\right\rfloor} {\left( \left\lfloor\frac{1+6\cdot k}{1+6\cdot a}\right\rfloor - \left\lfloor\frac{6\cdot k}{1+6\cdot a}\right\rfloor \right)} \right)^{-1}\right\rfloor}$