Number of possible sets for given N How many possible valid collections are there for a given positive integer N given the following conditions:
All the sums from 1 to N should be possible to be made by selecting some of the integers. Also this has to be done in way such that if any integer from 1 to N can be made in more than one way by combining other selected integers then that set of integers is not valid.
For example, with N = 7,
The valid collections are:{1,1,1,1,1,1,1},{1,1,1,4},{1,2,2,2},{1,2,4}
Invalid collections are:
{1,1,1,2,2} because the sum adds up to 7 but 2 can be made by {1,1} and {2}, 3 can be made by {1,1,1} and {1,2}, 4 can be made by {1,1,2} and {2,2} and similarly 5, 6 and 7 can also be made in multiple ways using the same set.
{1,1,3,6} because all from 1 to 7 can be uniquely made but the sum is not 7 (its 11).
 A: These are perfect partitions given in OEIS.  a(n-1) = sum of all a(i-1) such that i divides n and i < n.
A: The term I would use is "multiset".  Note that your multiset must contain 1 (as this is the only way to get a sum of 1).   Suppose there are $r$ different values $a_1 = 1, \ldots, a_r$ in the multiset, with $k_j$ copies of $a_j$.  Then we must have $a_j = (k_{j-1}+1) a_{j-1}$ for $j = 2, \ldots, r$, and $N = (k_r + 1) a_r - 1$.  Working backwards, if $A(N)$ is the number of valid multisets summing to $N$, for each factorization $N+1 = ab$ where $a$ and $b$ are positive integers with $b > 1$ you can take $a_r = a$, $k_r = b - 1$, together with any valid multiset summing to $a-1$.  Thus $A(N) = \sum_{b | N+1, b > 1} A((N+1)/b - 1)$ for $N \ge 1$, with $A(0) = 1$.  We then have, if I programmed it right,  1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 8, 1, 3, 3, 8, 1, 8, 1, 8, 3 for $N$ from 1 to 20.  This matches OEIS sequence A002033, "Number of perfect partitions of n".
A: Suppose your collection contains a finite set of distinct numbers $\{n_1 ... n_k\}$ and that the collection contains the number $n_i$ $t_i$ times (you can also suppose that the $n_i$ are sorted)
Then your condition is that for every number $x$ between 0 and $N$, $x$ can be written as $x = \Sigma u_i n_i$ with $0 \leq u_i \leq t_i$ in exactly one way.
This is possible if and only if $n_1 = 1$, forall $i$, $n_i = \Sigma_{j\lt i} t_i n_i$, and $N = \Sigma t_i n_i$.
You can prove this by doing an induction on $k$, starting at $k=N=0$ for the base case.
The base case is easy. The induction case is not difficult :
If you have a valid collection containing $k$ distinct terms that can make all numbers up to $N_k$, then the only way to extend it into a bigger collection is by adding $n_{k+1} = N_k+1$.
If you pick a smaller number, then $n_{k+1}$ can be written in two ways, if you pick a bigger number, then you can not make $N_k+1$.
And then, if you have a valid collection made of $k+1$ terms, then the sub-collection containing the first $k$ distinct terms is also valid and has to write every number up to $n_{k+1}-1$ for the same kind of reasons.
Thus, a valid collection of $k$ distinct terms that makes every number up to $N$ is determined by the sequence $(n_1=1, \ldots n_k, n_{k+1}=N+1)$ where $n_i$ divides $n_{i+1}$ :
 $$n_i = \Sigma_{j \lt i} t_j n_j = n_{i-1} +  t_{i-1} n_{i-1} = (t_{i-1} + 1) n_{i-1}$$
So this shows why they are successives multiples of each other and how to recover the $t_i$ from the sequence.
The 4 collections you gave as an example correspond respectively to the sequences $(1,8), (1,2,8), (1,4,8), (1,2,4,8)$.
The number of valid sequences depends on the exponents in the prime decomposition of $N+1$ :
if $N+1$ is a prime power $p^a$, then there are exactly $2^{a-1}$ valid sequences, since you have to choose wether you pick $p^i$ or not for $0 \lt i \lt a$.
If $N+1$ has several prime divisors, I don't think there is a nice formula giving the number of valid collections from the multiset of exponents in the prime factorisation of $N+1$. 
