This is joint work with someone. We got numerical evidence and argument against Littlewood conjecture, though mistakes are certainly possible.

Littlewood conjecture states that for any two real numbers $\alpha$ and $\beta$,

$$ \liminf_{n\to\infty} \ n\,\Vert n\alpha\Vert \,\Vert n\beta\Vert = 0$$

where $\Vert \,\Vert$ is here the distance to the nearest integer.

Let $f(n,\alpha,\beta)=n\,\Vert n\alpha\Vert \,\Vert n\beta\Vert$.

$L_n$ are Lucas numbers, $\phi=(1+\sqrt{5})/2=1.618033988\ldots,\psi=1-\phi=(1-\sqrt{5})/2=-0.618033988\ldots,L_n=\phi^n+\psi^n$, and $\{\,\}$ is the fractional part.

**Claim 1** $f(L_{6n+1},\phi/2,\phi/2) \sim L_{6n+1}/4$

Plot of $\log\log f(L_{6n+1},\phi/2,\phi/2)$ and $f(L_{6n+1},\phi/2,\phi/2)/(L_{6_n+1}/4) $:

Since $|\psi|<1$, for $n$ large enough $\psi^n$ tends to $+0$ for even $n$ and to $-0$ for odd $n$. This make $\{\phi^{2n}\}$ tend to $1$.

$\Vert x \Vert= \min(\{x\},1-\{x\})$.

Since $L_{6n+1}$ is odd, the $2$ in $\phi/2$ remains.

$\{L_{6n+1}\phi/2\}= \{\phi^{6n+2}/2+\psi^{6n+1} \phi/2\}$.

As $n$ tends to infinity $\psi^{6n+1} \phi/2$ tends to $-0$ and $\{\phi^{6n+2}/2\}$ tends to $1/2$. One can get explicit bounds.

This makes $\Vert \{L_{6n+1}\phi/2 \Vert \sim \frac12$ and $f(L_{6n+1},\phi/2,\phi/2) \sim L_{6n+1}/4$.

Q1 How to explain the experimental data in the plot?

Q2 Can the argument be made rigorous if it is correct?