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Open problemEditorial · Combinatorics

Hadamard Matrix Conjecture

For every positive integer nn divisible by 44, there is an n×nn\times n matrix HH with entries in {±1}\{\pm1\} and HHT=nIHH^{\mathsf T}=nI.

Source checked Jul 24, 20261 pinned Lean statementInspect problem
Source labels openarXiv · Combinatorics

Independent Domination of Regular Graphs, Conjecture 1.6

Conjecture 1.6 (Even case).* For a nonempty isolate-free graph GG on nn vertices, if DD is even, then (D+2)2i(G)(D2+4)n(D + 2)^2 \cdot i(G) \leq (D^2 + 4) \cdot n.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openarXiv · Combinatorics

Independent Domination of Regular Graphs, Conjecture 1.6

Conjecture 1.6 (Odd case).* For a nonempty isolate-free graph GG on nn vertices, if DD is odd, then (D+1)(D+3)i(G)(D2+3)n(D + 1)(D + 3) \cdot i(G) \leq (D^2 + 3) \cdot n.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openarXiv · Combinatorics

Digit $2$ in base $3$ representation of $2^n$

For n>8n > 8, 2n2^n is not the the sum of distinct powers of 33. Expressed here in terms of the base 33 digits of nn.

This conjecture is equivalent to the halting of a 1515-state 22-symbol Turing Machine.

TODO(lezeau): Formalize the Turing Machine version of this problem.

Source: Hardness of Busy Beaver Value BB(15): https://link.springer.com/chapter/10.1007/978-3-031-72621-7_9 This is also https://arxiv.org/abs/2107.12475.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1

If A{1,...,N}A\subseteq\{1, ..., N\} with A=n|A| = n is such that the subset sums aSa\sum_{a\in S}a are distinct for all SAS\subseteq A then

N2n. N \gg 2 ^ n.
Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1: Real

A generalisation of the problem to sets A(0,N]A \subseteq (0, N] of real numbers, such that the subset sums all differ by at least 11 is proposed in [Er73] and [ErGr80].

[Er73] Erdős, P., Problems and results on combinatorial number theory. A survey of combinatorial theory (Proc. Internat. Sympos., Colorado State Univ., Fort Collins, Colo., 1971) (1973), 117-138.

[ErGr80] Erdős, P. and Graham, R., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathematique (1980).

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 10

Is there some kk such that every integer is the sum of a prime and at most kk powers of 22?

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 10: Granville Soundararajan Odd

Granville and Soundararajan [GrSo98] have conjectured that at most 33 powers of 22 suffice for all odd integers, and hence at most 44 powers of 22 suffice for all even integers.

Ref: Granville, A. and Soundararajan, K., A Binary Additive Problem of Erdős and the Order of 22 mod p2p^2

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 10: Grechuk

Bogdan Grechuk has observed that 11171751461117175146 is not the sum of a prime and at most 33 powers of 22, and pointed out that parity considerations, coupled with the fact that there are many integers not the sum of a prime and 22 powers of 22 suggest that there exist infinitely many even integers which are not the sum of a prime and at most 33 powers of 22).

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1068

Does every graph with chromatic number 1\aleph_1 contain a countable subgraph which is infinitely connected?

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 108

For every r ≥ 4 and k ≥ 2 is there some finite f(k,r) such that every graph of chromatic number ≥ f(k,r) contains a subgraph of girth ≥ r and chromatic number ≥ k?

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1093: I

Are there infinitely many binomial coefficients with deficiency 1?

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1093: Ii

Are there only finitely many binomial coefficients with deficiency > 1?

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1145

Let A={1a1<a2<}A=\{1\leq a_1 < a_2 < \cdots\} and B={1b1<b2<}B=\{1\leq b_1 < b_2 < \cdots\} be sets of integers with an/bn1a_n/b_n\to 1.

If A+BA+B contains all sufficiently large positive integers then is it true that lim sup1A1B(n)=\limsup 1_A\ast 1_B(n)=\infty?

A conjecture of Erdős and Sárközy.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1167

Erdős Problem 1167.* Let r2r \geq 2 be finite, γ2\gamma \geq 2, and λ\lambda be an infinite cardinal. Let κα\kappa_\alpha be cardinals for all α<γ\alpha < \gamma. Is it true that 2λ(κα+1)α<γr+12^\lambda \to (\kappa_\alpha + 1)_{\alpha < \gamma}^{r+1} implies λ(κα)α<γr?\lambda \to (\kappa_\alpha)_{\alpha < \gamma}^r? Here ++ means cardinal addition, so that κα+1=κα\kappa_\alpha + 1 = \kappa_\alpha if κα\kappa_\alpha is infinite.

A problem of Erdős, Hajnal, and Rado.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1167

Finite-target case.* When all κα\kappa_\alpha are finite, κα+1\kappa_\alpha + 1 is the ordinary natural-number successor. Special case of erdos_1167.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1167

Binary-color case.* The γ=2\gamma = 2 specialization (two color classes).

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1167

Infinite-target case.* When all κα0\kappa_\alpha \geq \aleph_0 are infinite and bounded by λ\lambda, κα+1=κα\kappa_\alpha + 1 = \kappa_\alpha, so the hypothesis simplifies to a "pure" stepping-down lemma:

\lambda \to (\kappa_\alpha)_{\alpha<\gamma}^r.$$ The condition $\kappa_\alpha \leq \lambda$ is needed to avoid a size obstruction: without it, the conclusion would require a subset of $\lambda$ of size $\kappa_\alpha > \lambda$, which is impossible (see `infinite_targets_needs_bound`).
Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openErdős Problems · Combinatorics

Erdős Problem 1167

r=2r = 2 case.* The stepping-down from 3-uniform to 2-uniform partition relations: 2λ(κα+1)α<γ32^\lambda \to (\kappa_\alpha + 1)_{\alpha<\gamma}^3 implies λ(κα)α<γ2\lambda \to (\kappa_\alpha)_{\alpha<\gamma}^2. Generalises the classical Erdős–Rado stepping-up/down theorem for pairs.

Source checked Jul 26, 20261 pinned Lean statementInspect problem