Erdős Problem #861 — How many Sidon subsets does an interval have?

About 36 years old · traced to

Let f(N)f(N) be the maximum size of a Sidon subset of [N]={1,…,N}[N]=\{1,\ldots,N\}, and let A(N)A(N) be the number of Sidon subsets of [N][N]. Does A(N)/2f(N)→∞A(N)/2^{f(N)}\to\infty? Is A(N)=2(1+o(1))f(N)A(N)=2^{(1+o(1))f(N)}?

References

Additional references

D. Saxton and A. Thomason, Hypergraph containers, Inventiones Mathematicae 201 (2015), 925–992.

Progress summary

Refreshed
Claimed progress

A 2018 result shows the count is sometimes larger than the obvious lower bound by an unbounded factor, but neither question is settled.

Cameron and Erdős asked whether the number of Sidon subsets of an interval substantially exceeds the contribution from largest Sidon sets, and whether it has the conjectured exponential scale.

Known results

  • The recorded estimates are 2(1.16+o(1))N≤A(N)≤2(6.442+o(1))N2^{(1.16+o(1))\sqrt N}\le A(N)\le 2^{(6.442+o(1))\sqrt N}.
  • The maximum Sidon-set size satisfies the scale f(N)∼Nf(N)\sim\sqrt N used in the comparison.

2018 enumeration result

The paper On the number of generalized Sidon sets reports lim sup⁡N→∞A(N)2−f(N)=∞\limsup_{N\to\infty}A(N)2^{-f(N)}=\infty. Thus the ratio is unbounded along a subsequence, but this does not prove A(N)/2f(N)→∞A(N)/2^{f(N)}\to\infty or A(N)=2(1+o(1))f(N)A(N)=2^{(1+o(1))f(N)}. Its 2Θ(N)2^{\Theta(\sqrt N)} result concerns generalized, not ordinary, Sidon sets.

Current status (as of September 2026): The subsequential unboundedness result is reported, while both stated asymptotic questions remain open.

Sources

Solutions 0

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