An analogue of the equidistribution theorem? Suppose that $(n_k)_{k\in \mathbb{N}}$ is a given increasing sequence of positive integers. 
Does there exist an (irrational) number $a$ such that 
$\{an_k\}:=(a n_k)\text{mod }1 \rightarrow 1/2$ as $k \rightarrow \infty$? 
 A: The answer is no in general. For many increasing sequences $(n_k)_{k\in \mathbb{N}}$ of positive integers, it happens for every irrational number $a$ that $\{an_k\}$ is dense or even equidistributed in the unit circle. See this Wikipedia article for some examples.
A: The sequence $n_{2k}=k^2, n_{2k+1}=k^2+1$ is a counterexample.
Indeed, if $\{ n_k a \} \to \frac{1}{2}$ then $\{ n_{2k}a \} \to \frac{1}{2}$ and $\{ n_{2k+1} a\} \to \frac{1}{2}$.
This implies that $a= n_{2k+1}a-n_{2k} a= \lfloor n_{2k+1}a\rfloor+ \{ n_{2k+1}a \} - \lfloor n_{2k}a\rfloor- \{ n_{2k}a \} $ and hence
$$ a \pmod{1} \equiv  \{ n_{2k+1}a \} - \{ n_{2k}a \} \to 0 \pmod{1} $$
This shows that $a \in \mathbb Z$, which contradicts  $\{ n_k a \} \to \frac{1}{2}$.
The same is true about any subsequence containing infinitely many pairs of consecutive intgers.
P.S. On another hand, if $a$ is an irrational number, it follows from the denseness of $\{ na \}$ that there exists some $n_k$ such that $\{ n_ka \} \to \frac{1}{2}$.
This shows that there are many subsequences with this property, and I think one can argue that there are uncounatbly many such sequences.
