Solution 1:

Though it is clear what would count as a positive answer, the question is not precise enough that to make clear what counts as a negative answer.

Nonetheless, I believe that no similar formula could possibly be equivalent to denseness. The problem is that there are sequences that are dense but that are wildly non-equidistributed. For example, your formula would have to detect sequences constructed in the following way. Let $\{q_1,q_2,\ldots\}$ be an enumeration of $\mathbb{Q} \cap [0,1]$, and choose an increasing function $f : \mathbb{N} \rightarrow \mathbb{N}$ that goes to infinity very rapidly. For instance, $f$ could be a busy beaver function. Now define a sequence $\{x_n\}$ by the formula $$x_n = \begin{cases} q_k & \text{if } f(k)=n \text{ for some }k,\\ 0 & \text{otherwise.} \end{cases}$$

As you can see, $x_n$ is extremely close to the constant sequence $0$. I can't imagine any formula would be able to detect the tiny blips where it jumps to the various rational numbers.

Solution 2:

The problem is that equidistribution is a property of sequences, but density is a property of sets. You may wish to prove as an exercise that a countable subset of $[0,1]$ is dense if and only if it can be ordered in such a way as to be equidistributed. So the set $\lbrace\,x_0,x_1,\dots\,\rbrace$ is dense if and only if there is a permutation $\sigma$ such that $$\lim_{n\to\infty}{1\over n}\sum_{j=0}^{n-1}e^{2\pi i\ell x_{\sigma(j)}}=0$$ for every nonzero integer $\ell$.

Solution 3:

Here are some necessary and sufficient conditions:

1.a. The set $\{n\mid a\leqslant x_n\leqslant b\}$ is nonempty for every $0\leqslant a<b\leqslant 1$.

1.b. The set $\{n\mid a\leqslant x_n\leqslant b\}$ is infinite for every $0\leqslant a<b\leqslant 1$.

2.a. The sum of the series $\sum\limits_nf(x_n)$ is positive for every continuous nonnegative function $f$ not identically zero.

2.b. The series $\sum\limits_nf(x_n)$ diverges for every continuous nonnegative function $f$ not identically zero.