Given 'n' number of unique positive integers, how many unique sum values can I make? Suppose I have three integer values
[1,2,3]

With above three values, I can form total 8 values ($2^3$).
 0                        (Pick 0 value)
 1, 2, 3,                 (Pick only 1 value)
 3[1+2], 4[1+3], 5[2+3],  (Pick 2 values)
 6[1+2+3]                 (Pick 3 values)

However, Notice how there are duplicate outcomes.
3 and 3[1+2] ?
So I would say that using three numbers 1,2,3, I can form total 7 unique numbers.
But is there any formula I can apply to find out the cardinality of only the unique outcomes ?
Some more example set of numbers to consider.
[1,2,3,4]
[1,2,3,4,5]
[1,3,5,7]


In the following list I wrote $(n,values)$ where $n$ is the number of different elements in the set and $values$ is the number of different values we can get taking $0$ elements, $1$ element, $2$ element, and so on.
$$(0,1),(1,2),(2,4),(3,7),(4,11),(5,16),(6,22),\ldots$$
$$values=\frac{1}{2} \left(n^2+n+2\right)$$
Hope this can help

Edit
Only works for lists of consecutive integers, like $\{1,2,\ldots,n\}$
