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[CoPh96] Coppersmith, Don and Phillips, Steven, On a question of Erdős on subsequence sums.
SIAM J. Discrete Math. (1996), 173-177.
[Fr93] Freud, R., Adding numbers - on a problem of P. Erdős. James Cook Mathematical
Notes (1993), 6199-6202.
@[expose]publicsectionopenFilternamespaceErdos867
A finite set of naturals $A={a_1<\cdots<a_t}$ is consecutive-sum-free if it has no
solutions to $a_i+a_{i+1}+\cdots+a_j\in A$ with $i<j$; equivalently, whenever an interval
$[m,n]$ contains at least two elements of $A$, the sum of the elements of $A$ lying in
$[m,n]$ is not itself an element of $A$.
Is it true that if $A={a_1<\cdots <a_t}\subseteq {1,\ldots,N}$ has no solutions to
$$a_i+a_{i+1}+\cdots+a_j\in A$$
then
$$\lvert A\rvert \leq \frac{N}{2}+O(1)?$$
In fact this problem is false. Freud [Fr93] constructed a sequence with density $\geq 19/36$.
The current best bounds are due to Coppersmith and Phillips [CoPh96], who prove that the
maximal size of such an $A$ satisfies
$$\frac{13}{24}N -O(1)\leq \lvert A\rvert \leq \left(\frac{2}{3}-\frac{1}{512}\right)N+\log N.$$
The current best bounds are due to Coppersmith and Phillips [CoPh96], who prove that the
maximal size of such an $A$ satisfies
$$\frac{13}{24}N -O(1)\leq \lvert A\rvert.$$
The current best bounds are due to Coppersmith and Phillips [CoPh96], who prove that the
maximal size of such an $A$ satisfies
$$\lvert A\rvert \leq \left(\frac{2}{3}-\frac{1}{512}\right)N+\log N.$$