Set with $k$-subset unique sum Probably, I should call it a sequence, anyway, is there a sequence/set (Fibonacci is a valid answer for this question (minus the first three Fibonacci numbers including zero), but too big) where any $k$-subset (subset/aggregation of terms within sequence, with $k$ elements) will have a unique sum that can't be achieved by any other k-subset?
This PDF is somewhere along the same lines... Thanks all.
Edit I'm asking for the smallest such series, and it must satisfy (as one commenter asked) for all $k$ (well, if that's not possible, just odd or even $k$ will do)
 A: I'll assume you want your $a_i$ to be positive integers, and that you want the statement true for all $k$.
Let $a_1 <  a_2 < ... < a_n ...$ be such a sequence.  Then for any $n$, and $k\leq n$, the $k$-subsets of $\{a_1,...,a_n\}$ must be distinct.  But the sum of any $k$-subset of this set is at most $ka_n$. So there must be at least $n\choose k$ distinct numbers from $1$ to $ka_n$(*), and hence ${n\choose k} \leq ka_n$, or
$$a_n \geq \frac{1}{k} {n\choose k}$$
Now, if $n=2m$ and $k=m$ then you get:
$$a_{2m} \geq \frac{1}{m} {2m\choose m}$$
The right hand side is approximately $\frac{(m+1)2^{2m}}{m^{5/2}\sqrt{\pi}}$.  This can be seen via Catalan numbers, see: http://en.wikipedia.org/wiki/Catalan_number .
That would seem to imply that the Fibonacci sequence does not satisfy your original condition.  The Fibonacci sequence grows like $\frac{1}{\sqrt 5}\phi^n$, where $\phi = \frac{1+\sqrt{5}}{2} < 2$.
Indeed, if $26 = 21 + 3 + 2 = 13 + 8 + 5$, so this is not true for the Fibonacci numbers when $k=3$.
(*) Edit: Of course, the range is even stronger. $k$ distinct integers from $1$ to $a_n$ add up to a value between $\frac{k(k+1)}{2}$ and $ka_n-\frac{k(k-1)}{2}=\frac{k(2a_n-k+1)}{2}$, for a total of $1+\frac{k(2a_n-2k)}{2}$ distinct possible values. So we can slightly improve the above:
$$\binom{n}{k}\leq 1+k(a_n-k)\\
\frac{1}{k}\binom{n}{k}-\frac{1}{k} +k \leq a_n$$
