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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) $:

enter image description here

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?

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In order to prove the Littlewood conjecture it is enough to prove that $$\lim_{k\to \infty} f(n_k,\alpha,\beta)=0$$ for some subsequence $n_k$. In your case, $\alpha=\beta=\frac{\phi}{2}$. And you can show that $$\lim_{k\to \infty}f(L_{6k+1},\phi/2, \phi/2)=\infty,$$ however by choosing a different subsequence, one can obtain $$\lim_{k\to \infty} f(2L_k,\phi/2,\phi/2)=0,$$ and therefore Littlewood's conjecture is true for these particular constants. The proof is basically the same as your analysis above: $||2L_k\cdot\frac{\phi}{2}||=||\phi^{k+1}+\phi \psi^k||=|\sqrt{5}\psi^k|$ and the result follows.

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  • $\begingroup$ Thanks. Since it is for all alpha, beta, isn't mine a counterexample? $\endgroup$ – joro Dec 18 '14 at 13:18
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    $\begingroup$ @joro, not quite, because although $f(n)$ is very large for some values of $n$ which you considered, it will be very small for the values I mentioned above. That's enough to imply $\lim inf=0$. $\endgroup$ – Gjergji Zaimi Dec 18 '14 at 13:21
  • $\begingroup$ Might be wrong, but you contradict several papers and wikipedia (might not be reliable source). "Only a finite number of elements of the sequence are less than $b-\varepsilon$". en.wikipedia.org/wiki/… $\endgroup$ – joro Dec 18 '14 at 14:11
  • $\begingroup$ I don't see any contradiction with that article. Only finitely many elements of a sequence can be below the $\liminf$, however, it is possible for infinitely many to be above it. $\endgroup$ – Gjergji Zaimi Dec 18 '14 at 14:21
  • $\begingroup$ Indeed. Dumb me... $\endgroup$ – joro Dec 18 '14 at 15:27

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