I'm trying to create a sequence of elements which are dependent on their position in the sequence. What I mean is that I want the value $r$ in position 1 equal to 1, on position 2 equal to 2 and so on. E.g.
$$\langle r, r, r, r, r, r\rangle=\langle 1, 2, 3, 4, 5, 6\rangle$$
$$\langle 2^r, 2^r, 2^r, 2^r, 2^r, 2^r\rangle=\langle 2, 4, 8, 16, 32, 64\rangle$$
My first thought is having 2 sequences of elements. One of $r$ being all natural numbers and the second being the sequence I'm using $r$ in (The sequence called $\Bbb X$.) next to each other like so:
$$r=\langle 0, 1, 2, 3,\dots\rangle=\Bbb N$$
$$\Bbb X=\langle 2^{r+1}, 2^{r+1}, 2^{r+1}, 2^{r+1}, 2^{r+1}, 2^{r+1}\rangle$$
Both combining to make:
$$\langle 2, 4, 8, 16, 32, 64\rangle$$
Is there some sort of pre-existing symbol or notation to represent the value of whatever position an element is in a sequence, changing the value of an expression like $2^{r+1}$ based on its position?
Best Answer
I will use curly brackets $\{\}$ to denote sets and angle brackets $\langle\rangle$ to denote sequences.
A set is something fundamentally different from a sequence, since sets are not ordered. That is, the sets $\{1,1,1,2,4,3\}$ and $\{4,2,3,1\}$ are equal: multiplicity and order of elements does not matter. For sequences the order does matter, so $\langle 1,1,2\rangle$, $\langle 1,2\rangle$ and $\langle 2,1\rangle$ are all distinct from each other.
If you want to create a sequence $\langle a_1,a_2,\dots,a_i,\dots\rangle$ (finite or infinite) with each element having a value dependent on its position, I would write it as $\langle f(i)\mid i\in I\rangle$, where $I$ is an ordered list of indices, and $f$ is a function with the index as input.
For example:
$\mathbb X=\langle 2^r\mid 1\leq r\leq 6\rangle=\langle2,4,8,16,32,64\rangle$
$\langle n\mid n\in\mathbb Z_{>0}\rangle=\langle1,2,3,4,\dots\rangle$
In both of these examples it is left implicit that the indices $r$ and $n$ are ordered from small to large in the usual way.
For creating sets you could use the same notation, but here the order doesn't matter. So as a set we have $\{2^r\mid 1\leq r\leq 6\}=\{2,4,8,16,32,64\}=\{64,2,8, 16,32,4\}$.