# Sequence of $n$ terms, generated for a given $k$-subset sum

Before I make myself clear, I wish to say that I do not want to know a sequence like the binary sequence. If possible, I'd want a sequence smaller than the binary sequence, of polynomial growth rate (please avoid exponential growth rates of terms) - though the latter option need not necessarily be satisfied if not possible.

In other words, I mean to say, given a certain $k$ and a length $n$ for the sequence, we have to generate a sequence of $n$ terms whose $k$ terms' sum is always unique.

An extremely trivial example, $1, 2, 3, 4, 5$ is a sequence of $5$ terms, where no $k$ terms' sum (here $k=1$) are equal in any way, and $6, 16, 30, 48$ for $n=4, k=2$ and the $a$th term $=ak(a+k)$.

-

For the case $k=2$, you may be interested in the Mian-Chowla sequence, seen at http://oeis.org/A005282. It starts 1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 290, 361, 401, 475, 565, 593, 662, 775, 822, 916, 970, 1016, 1159, 1312, 1395, 1523, 1572, 1821, 1896, 2029, 2254, 2379, 2510, 2780, 2925, 3155, 3354, 3591, 3797, 3998, 4297, 4433, 4779, 4851, and each term is as small as possible given the restriction of $a+b\ne c+d$. You might be interested in $B_2$ sequences more generally. There is a copious literature.
EDIT: Guy, 3rd edition, page 351 (problem E28) writes, "Let $a_1\lt a_2\lt\cdots$ be an infinite sequence of integers for which all the triple sums $a_i+a_j+a_k$ are distinct. Erdos offers \$500.00 for a proof or disproof of an old conjecture of his, that$\lim(a_n/n^3)=\infty$." - Fine, what about other$k$? – Mach9 Apr 16 '12 at 15:12 I don't know. I'm sure there's a literature, as it's a natural question. Two suggestions: 1) see if it's in Richard Guy's book, Unsolved Problems In Number Theory. 2) calculate some small examples and look them up in the OEIS - that's what I did for$k=2$. – Gerry Myerson Apr 17 '12 at 0:22 BTW How did you pull that off for$k=2$? – Mach9 Apr 17 '12 at 12:17 Just because the pairwise sums are distinct, doesn't that make it distinct for all$k$? – Mach9 Apr 17 '12 at 12:19 No, e.g., in Mian-Chowla,$1+4+13=2+8+8$,$2+4+81=8+13+66$, etc. For$k=2\$ it was easy to find the first five or so terms by hand, then search the OEIS. – Gerry Myerson Apr 17 '12 at 13:06