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?


1 Answer 1


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.

  • $\begingroup$ Thanks. Since it is for all alpha, beta, isn't mine a counterexample? $\endgroup$
    – joro
    Dec 18, 2014 at 13:18
  • 2
    $\begingroup$ @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 $\lim inf=0$. $\endgroup$ Dec 18, 2014 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, 2014 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. $\endgroup$ Dec 18, 2014 at 14:21
  • $\begingroup$ Indeed. Dumb me... $\endgroup$
    – joro
    Dec 18, 2014 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.