---

David Saxton and Andrew Thomason, Hypergraph containers, Invent. Math. 201 (2015), 925–992; arXiv:1204.6595, whose first version of 2012-04-30 already carries the statement as its Theorem 1.10; library card. Theorem 2.11 of the journal version states that the number of Sidon subsets of $\{1,\ldots,N\}$ lies between $2^{(1.16+o(1))\sqrt N}$ and $2^{(55+o(1))\sqrt N}$. Since the largest Sidon subset has size $f(N)=(1+o(1))\sqrt N$ (Erdős–Turán, Chowla, Singer), the lower bound reads $A(N)\geq 2^{(1.16+o(1))f(N)}$. Hence $A(N)/2^{f(N)}\to\infty$, which answers the first question yes, and $A(N)=2^{(1+o(1))f(N)}$ fails, which answers the second question no; the authors state the second consequence explicitly. The problem is thereby answered, with the two questions answered in opposite directions.

The lower bound is an elementary construction, given in Section 11 of the arXiv first version as the proof of its Theorem 1.10: a set of $p-1$ residues that is Sidon modulo $p(p-1)$ (Ruzsa) supplies any four pairwise disjoint subsets, placed in the translates of $\{1,\ldots,p(p-1)\}$ by $0$, $p(p-1)$, $2p(p-1)$ and $3p(p-1)$; each element lies in one of the four translates or is omitted, so the five choices for each element give $5^{p-1}$ Sidon subsets of $\{1,\ldots,4p(p-1)\}$, and $\tfrac12\log_2 5=1.16\ldots$ gives the exponent. The upper bound applies the paper's container theorem to the $4$-uniform hypergraph of additive quadruples. The journal version states Theorem 2.11 and defers the details of both bounds to the companion paper Online containers for hypergraphs, with applications to linear equations, J. Combin. Theory Ser. B 121 (2016), 248–283 (arXiv:1611.01433), where they appear as Theorem 1.10 and Section 5; both are linked above. Before this, the trivial bounds were $2^{f(N)}\leq A(N)\leq N^{(1/2+o(1))\sqrt N}$, and Cameron and Erdős had shown only that $A(N)/2^{f(N)}$ is unbounded. Kohayakawa, Lee, Rödl and Samotij (card), independently and in Random Structures Algorithms 46 (2015), 1–25, prove $A(N)\leq 2^{cf(N)}$ for a constant $c$ (Theorem 1.1); they note that their method allows any $c>\log_2(32e)=6.442\ldots$, and they write the proof for $c>\log_2(33e)=6.487\ldots$. The site's remarks (last edited 2025-10-15) give $6.442$ as the best upper bound. That paper restates the lower bound proved here. The limit of $\log_2 A(N)/f(N)$, if it exists, is not determined.

Acceptance. The paper is refereed (Invent. Math.). The site's curator, T. F. Bloom, records both questions as settled on the problem page (last edited 2025-10-15), the first positively and the second negatively, with the lower bound credited to this paper; that is the reviewed evidence named here.

Depends on. Nothing beyond the cited paper.