Do you want to know about this specific sequence, or more generally?

Specifically, this is essentially a W state, about which there have been several questions in the past. You just need one extra term (effectively $|32\rangle$) that you could later change into $|11\rangle$, leaving the others untouched.

Let's say I've got a state
$$
(|1\rangle+|2\rangle+|4\rangle+|8\rangle+|16\rangle+|32\rangle)/\sqrt{6}
$$
This is the same as
$$
(|000001\rangle+|000010\rangle+|000100\rangle+|001000\rangle+|010000\rangle+|100000\rangle)/\sqrt{6}.
$$
We do controlled-nots, controlled off the first qubit targetting qubits 3, 5 and 6. This changes $|100000\rangle$ into $|101011\rangle$. Now we do a controlled-controlled-not controlled by qubits 3 and 5, targetting qubit 1. This changes the $|101011\rangle$ into $|001011\rangle$ without changing anything else. Thus, we're left with
$$
|0\rangle(|10000\rangle+|01000\rangle+|00100\rangle+|00010\rangle+|00001\rangle+|01011\rangle)/\sqrt{6},
$$
which is the state you want.

More generally, meaning a state of the form
$$
\frac{1}{\sqrt{r}} \sum_{i=0}^{r-1}|a^i\text{ mod }N\rangle,
$$
I advise looking at the order finding algorithm for quantum computers. You're basically talking about preparing one of the eigenvectors of the order finding unitary. While you would like to be able to prepare such a state in order to find the order $r$, we don't know how to do that without first finding $r$. But, when you've run the order finding algorithm, and it randomly outputs a value of $s/r$, the state of the second register is in the eigenvector related to $s/r$.
$$
\frac{1}{\sqrt{r}} \sum_{k=0}^{r-1}e^{-2\pi i k s/r}|a^k\text{ mod }N\rangle
$$
Once you've got that, and you know the value $s/r$, you should easily be able to compensate for the phases.

1Do you mean $x_{i+1}=x_0x_i\text{ mod }21=x_0^{i+2}\text{ mod }21$ (i.e. without the $f$)? – DaftWullie – 2019-01-18T16:03:57.250

Yes, this function doesn't need to be presented recursively as I did and that does indeed help quite a bit here. But, by the way, you should've written $x_{i+1} = x_0 x_i = x_0^{i + 1}$ and not $x_0^{i+2}$. – R. Chopin – 2019-01-19T12:31:43.583

So what do you get for $i=-1$? – DaftWullie – 2019-01-19T12:51:06.413

I don't understand: $f$ is only defined for $i \geq 0$. Can you clarify? – R. Chopin – 2019-01-21T00:46:33.397

Well, that's not stated anywhere. Implicit in your notation is that you want a sequence $x_0,x_0^2,x_0^3,\ldots$ starting from $x_0$. So, if I want to use the stated formula, to get $x_0$, I need $i=-1$. Or, for $x_1=x_0^2$, I need $i=0$. – DaftWullie – 2019-01-21T08:08:35.947