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Find a sequence $(a_n)$ such that $(a_{n^2})$ converges but $(a_n)$ diverges

I am running out of examples.

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    Do you mean $\lim_{n \to \infty} a_{n^2}$ or $\lim_{n \to \infty}a_{n}^2$? – 2017-02-08
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    $lim_{n\to\infty}a_{n^2}$ – 2017-02-08
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    i don't think this converge, since it will become $(-1)^{n^2}$, can't say it is even or odd – 2017-02-08
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    @user284275 Yes you are right sorry I misread – 2017-02-08
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    Try $$a_n=(-1)^{d(n)}$$ where $d(n)$ denotes [the number of divisors](https://en.wikipedia.org/wiki/Divisor_function) of $n$, and check that this is a mere reformulation of @S.C.B.'s example below. – 2017-02-08

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Define the sequence $(a_m)$ as $$a_m=\begin{cases} 0 & \text{if }\;\ m=n^2 \text{ for some } n \in \mathbb{Z}\\ 1 & \text{otherwise}\end{cases}$$ Note that $$\lim_{n\to\infty}a_{n^2}=0$$ but that the sequence $(a_n)$ does not converge.

As @Did pointed out in his comment, if $d(n)$ denotes the number of divisors of $n$, note that $$a_{n}=\frac{(-1)^{d(n)}+1}{2}$$ Is equivalent to the sequence mentioned above.

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This sequence also works $a_n=\sin(\pi\sqrt n)$.

With $a_n=0$ only when $n$ is a perfect square, else it takes various values in $[-1,1]$.