Sidon sets are sets $A \subset \mathbb{N}$ such that for all $a_j,b_j \in A$ holds $$a_1+a_2=b_1+b_2 \iff \{a_1,a_2\}=\{b_1,b_2\}.$$ Thus if you know the sum of two elements, you know which elements were added in the first place.

The general case is known in literature as $B_k[g]$ sets. Every result of adding up at most $k$ elements will occur at most $g$ times in all possible combinations. Sidon sets are therefore called $B_2[1]$ sets.

An example of a $B_3[1]$ set is $\{1,7,11\}$ with $k = 3$, $n = \lvert B_k[1]\rvert=3$ elements, $m = \max B_k[1] = 11$ and ${n+k \choose k}={6 \choose 3}=20$ distinct sums:

0=0, 1=1, 1+1=2, 1+1+1=3, 7=7, 7+1=8, 7+1+1=9, 11=11, 11+1=12, 11+1+1=13, 7+7=14, 7+7+1=15, 11+7=18, 11+7+1=19, 7+7+7=21, 11+11=22, 11+11+1=23, 7+7+11=25, 11+11+7=29, 11+11+11=33.

I didn't find a $B_3[1]$ set with $n=3$ and a smaller $m$ than 11. However $\{1, 8, 11\}$ has similar properties.

My Question

Considering $B_k[1]$ sets (only one solution for any sum of at most $k$ elements). What is the smallest $m$ for given $k$ and $n$ and how to construct a set with this property?

If the general case for any given $k$ isn't known, we may set $k=3$.

Note: In a former question ("Set of small numbers with distinct $k$-sums") a construction by Javier Cilleruelo was given. However it doesn't manage to find sets with the smallest $m$ possible.

  • 1
    $\begingroup$ You might consider choosing a more informative title in order to make more people see this question. $\endgroup$
    – Stefan Kohl
    Oct 16, 2015 at 12:41

1 Answer 1


Since the question asks for exact smallest values, not bounds or asymptotics, it is likely to be a difficult problem.

First note that instead of taking sums of at most $k$ elements, we can take sums of exactly $k$ elements, if we include a zero element. So the three-element solution $\{1,7,11\}$ corresponds to a four-element solution $\{0,1,7,11\}$ with this interpretation. The result (minimum possible largest element) is not affected by this change, but the number of elements changes by 1.

For $k=3$ this is the OEIS sequence A227358, "Length of shortest Golomb-like (for sums of triples) ruler with n marks". Currently (last updated 2013) the values listed are:

$$ \begin{array}{lllllllll} n& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ \hline m& 0& 1& 4& 11& 23& 45& 82& 129& 208& 309 \end{array} $$

using the "exactly $k$ elements" definition (so the $n$ are off by one with respect to the current question). The OIES entry links to a C program that computes the solutions, answering the "how to construct" part.

For $k=3$ and $n=4=3+1$ the results of the OP are confirmed: There are exactly four minimal solutions with $m=11$, namely:

 0     1     7    11
 0     1     8    11
 0     3    10    11
 0     4    10    11

(The C program lists only the first two, because the other two are obtained by reversal symmetry, i.e. by the mapping $x \mapsto 11-x$.)

See also A227588 for a table where both $k$ and $n$ are varying.


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