There is a famous old theorem by Kronecker that for every positive real $\alpha$ and $\epsilon>0$ there exists a positive integer n such that $\alpha n$ is within $\epsilon$ of an integer.

Recently I found that the same result is true if we replace $\alpha n$ by $\alpha n^2$ or any polinomial p such that $p(0)=0$.

Could this result be generalised to other functions? Particularly I'm curious about sequences $\alpha 2^n$ and $\alpha F_n$ where by $F_n$ I denote n-th Fibonacci number.

Does anyone know anything about it?

  • $\begingroup$ $\alpha$ has to be irational. $\endgroup$
    – Nick S
    Mar 11, 2011 at 1:17
  • 1
    $\begingroup$ @Nick, why must $\alpha$ be irrational? If $\alpha$ is, say, 17, then there are certainly integers $n$ such that $\alpha n^2$ is within $\epsilon$ of an integer (ditto for $2^n\alpha$ or $F_n\alpha$). $\endgroup$ Mar 11, 2011 at 4:51
  • $\begingroup$ Presumably Nick meant 'irrational unless the (reduced) denominator is a power of two' to exclude trivial cases like 17. $\endgroup$
    – Charles
    Apr 4, 2011 at 15:41

3 Answers 3


Well, as Gerry has pointed out, this is certainly not true for all $\alpha$. On the other hand, this is true for a.e. $\alpha$. More precisely, the sequence $2^n\alpha$ is equidistributed mod 1 for a.e. $\alpha$.

I believe this result is due to H. Weyl and can be found in Cornfeld, Fomin and Sinai `Ergodic Theory'. (I don't have it with me.)

The same must be true for the Fibonacci sequence, I'm sure.

So, what you probably need is for this to be true for all $\alpha$, except some small (countable?) set. After all, $\|2^n\alpha\|<\varepsilon$ is indeed much weaker than equidistibution.

Update. Come to think about it, the answer is as follows: let $$ \alpha = \sum_{k=1}^\infty a_k2^{-k} $$ be the binary expansion of $\alpha$. Then the sequence $2^n\alpha\bmod 1$ gets arbitrarily close to 0 if and only if the sequence $(a_k)$ has unbounded strings of 0s. In particular, any rational $\alpha$ is out of the picture, apart from the binary rationals, of course.

All in all, your set of $\alpha$'s is indeed of full measure, but the exceptional set is of Hausdorff dimension 1, i.e., pretty big.

For the Fibonacci sequence you'll need to replace binary expansion with the $\beta$-expansion, where $\beta=(\sqrt5-1/)2$, with the same conclusion.

  • 1
    $\begingroup$ @Nikita, yes, Weyl proved that if $a_1,a_2,\dots$ is any sequence of distinct integers then the sequence $a_1x,a_2x,\dots$ is uniformly distributed modulo 1 for almost all real $x$. This of course includes the Fibonaccis. Hardy and Littlewood had already studied the case $a_n=b^n$, $b$ an integer, $b\ge2$. $\endgroup$ Mar 11, 2011 at 4:59
  • $\begingroup$ Thanks for an answer, though in the case of Fibonacci sequence this trick works not so good as in binary case. Just as β-expansion is not unique it is not so obvious that there exists reals with uniformly bounded number of consecutive 0s. At least typical example $0.10(10)_\beta=10_\beta$ shows it. Though it looks like looking on expansions $\sum\limits_{k=1}^{\infty} a_k \beta^k$ with maximum value of $\sum a_k 2^{-k}$ must work. And a.e. part is certainly true. Thanks! $\endgroup$ Mar 11, 2011 at 14:46
  • $\begingroup$ Ostap, to make them unique, one usually forbids two consecutive 1's. Then they exhibit pretty much the same ergodic and arithmetic properties as the binary expansions. (See, e.g., my survey paper maths.manchester.ac.uk/~nikita/ad.pdf). The Fibonacci numbers are closely related to the powers of the golden ratio, of course - you just need to divide them by $sqrt5$, I guess. $\endgroup$ Mar 11, 2011 at 15:00
  • $\begingroup$ Actually we must forbid not only pairs of consecutive 1s but also infinite sequences of the form 010101... e.g. Let β be a golden ratio, then $\sum \beta^{2k−1}={\beta\over{\beta^2−1}}=1=\beta^0$.As you see there are no consecutive 1s in any of two expansions, yet they are equal. Though now I see that nonuniqueness isn't very bad there, sorry for misconseption. Though for β-expansion there exist a number $\alpha=\sum\limits_{k=1}^{\infty}\beta^{3k-1} ={\beta\over{\beta^3- 1}}= {\beta\over 2}$ with no unbounded string of 0s but $\alpha\beta^{3k-1}$ gets close to integer($F_{3n}$ are even). $\endgroup$ Mar 11, 2011 at 16:19
  • $\begingroup$ I agree that the case of the Fibonacci sequence is different. What we have here is the condition $\|\xi\tau^{n_k}\|\to0$ as $k\to\infty$ for some subsequence $n_k$ (with $\tau=(1+\sqrt5)/2$). If we had just $n_k=k$, then we would know that $\xi\in\mathbf Q(\tau)$, whence $\xi$ would have an eventually periodic greedy $\tau$-expansion (combined Pisot + K. Schmidt). I believe we must have $n_k=bk$ for some $b\ge1$ but don't know how to prove it. Perhaps, re-reading Cassels `Introduction in Diophantine Approximations' (chapter on Pisot-Vijayaraghavan numbers) might help... $\endgroup$ Mar 13, 2011 at 1:17

$\alpha2^n$ is clearly not going to work, e.g., for $\alpha=1/3$. One place to look is the Kuipers and Niederreiter book on uniform distribution of sequences, although uniform distribution is a bit of overkill for the question you are asking about.


You've received two good answers, but I'll elaborate a bit. Usually equidistribution on the torus (or more general, compact groups) wrt the Haar measure is achieved by computing the Weyl sums and showing that there are some cancellations.

The question you are referring to is studied in the area of so-called "sparse equidistribution" (although it is more of sparse density if you would like).

The problem with the harmonic-analytic approach is by summing (or integrating) over very sparse part of your period. It is usually not stright forward to bound such exponential sums. For example, Vinogradov proved for example that $ \{p_{n}x\} $ is equidistributed mod 1 for all irrational x, to bound the Weyl sums, he used sieves with what is called now Vinogradov sums, and a result about the odd Goldbach conjecture.

Now if you are interested in a metric result (i.e. a.e. x), then it is a very classical result that for every increasing unbounded sequence \$ {a_{n}}$ and for a.e. x, one have that ${a_{n}x}$ is equi. mod 1, this is done by taking the Weyl sum, computing its L^2 norm, and then sub. limit and integration by the DCT.

Now the question if such a result follows for every x is very subtle, and not always amenable to harmonic-analytic approach, and the current state of the art actually lies in the ergodic approaches.

If you have a sequence which is contained inside a geometric progression, then there exists x's for which $\{a_{n}x\}$ is not equi. more generally, for $\{q^{n}\}$ say, you can find x's whose orbit closure is with any Hausdorff dimension you want (the reason here that as a dynamical system, this is isomorphic to Bernoulli shift on $q$ letters). More generally, a result due to Boshernitzan says that if you have a lacunary sequence (the limit of the ratios of consecutive elements is larger than 1), then the Hausdorff dimension of the set of exceptional x's (such that $\{a_{n}x\}$ is not dense/equidistributed) is 1. On the contrary, Boshernitzan shown that if the sequence is non-lacunary (the ratio tends to 1, you should think about it as having sub-exp. growth), then the Hausdorff dimension of the set for which $\{a_{n}x\}$ is equi. is 1. There were even some old results due to Erdos from the 1950's about it (he talked about convolution of Bernoulli measures, which can be interpreted in this sense as well).

A very peculiar discovery by Furstenberg (67) shown that if you have a non-lacunary semigroup, say $\{2^{n}3^{m}\}$ then for every irrational x you get that $\{2^{n}3^{m}x\}$ is dense mod 1 (certainly not equidistributed). This result is very interesting, because you have density for every x. Moreover, recently, Bourgain-Lindenstrauss-P. Michel and Venkatesh proved an effective version of that theorem (meaning that you fix some epsilon, you can estimate how far you need to go in-order to find an element which is epsilon-close to an integer). An even recent work (by myself, still preprint), generalizing the Bourgain-Lindenstrauss paper, and I shown that sets like $\{2^{n}3^{3^{m}}3^{3^{k^2}}x\}$ are dense for every x.

About Fibonacci sequences, it follows from my work (based on other work of D. Meiri whith Yuval Peres and Elon Lindnstrauss), that you can prove density of sequences such as $\{2^{n}3^{3^{m}}F_{k}x\}$ for every irrational x. For the general Fibonachi sequences, you are basically in the lacunary case, which Boshernitzan already covered in certain sense.

  • $\begingroup$ wow, thank you for the answer. The Bernouli shift flawely mentioned in your post helped me in completely unrelated area. Thank you very much! $\endgroup$ Jun 15, 2011 at 16:37

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