# Are the twin primes the only positive double zeros of this real function?

For $x \in \mathbb{R}, x \ge 1$ define $$f(x) = \sin\frac{\pi(\Gamma(x)+1)}{\lfloor x \rfloor}$$

By Wilson's theorem the positive integer zeros of $f(x)$ are exactly the primes and the positive real zeros are when $\Gamma(x)+1$ is an integer multiple of $\lfloor x \rfloor$.

One can replace $\lfloor x \rfloor$ by $x - \frac{i \, \log\left(-e^{\left(-2 i \, \pi x\right)}\right)}{2 \, \pi} - \frac{1}{2}$ or $x + \frac{\arctan\left(\cot\left(\pi x\right)\right)}{\pi} - \frac{1}{2}$.

Consider $$t(x) = f(x) f(x+2)$$

Double positive integer zeros of $t(x)$ appear to be the smaller twin of twin primes, though the derivative of floor at integers complicate things.

Q1 Is it true that the smaller twin is a double zero of $t(x)$ (possibly with some relaxations of the definition)?

Q2 Does $t(x)$ have positive double real zeros which are not integers?

This might happen when $\Gamma(x)+1$ is an integer multiple of $\lfloor x \rfloor$,

$\Gamma(x+2)+1$ is an integer multiple of $\lfloor x + 2\rfloor$ and by the functional equation $x (x + 1) \in \mathbb{Q}$, i.e. $\Gamma$ must be integer at algebraic numbers which are not integers.

This approach might be extended to multiple zeros of prime tuples or both $x,x^2+1$ are primes.

Added A plot suggest counterexamples might exist, though couldn't find a counterexample so far neither by root finding nor by inverting gamma (modulo errors).

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Added a question if the twins are really double zeros because the derivative of floor at integers complicates things. –  joro Sep 13 '13 at 10:55