Browse Problems

At least one of Catalan constant and the Gompertz constant is transcendental.

$\Gamma(1/n)$ for `n ≥ 2` is transcendental.

$\Gamma(1/4)$ is transcendental. [Ch84] Chudnovsky, G. (1984). Contributions to the theory of transcendental numbers.

$\Gamma(1/6)$ is transcendental. [Ch84] Chudnovsky, G. (1984). Contributions to the theory of transcendental numbers.

$\Gamma(1/3)$ is transcendental. [Ch84] Chudnovsky, G. (1984). Contributions to the theory of transcendental numbers.

$\Gamma(1/2)$ is transcendental. [Ch84] Chudnovsky, G. (1984). Contributions to the theory of transcendental numbers.

The Gompertz constant $\delta$ is transcendental.

[PPVW2016] 8.2(c): the number of elliptic curves over ℚ with rank ≥ 21 and naïve height at most `H` is asymptotically at most `H ^ o(1)`.

Are there infinitely many primes p such that p + 2 is prime?

The four exponential conjecture would imply that for any irrational number $t$, at least one of the numbers $2^t$ and $3^t$ is transcendental.

From [PPVW2016], Section 3.1: "from the mid-1960s to the present, it seems that most experts conjectured unboundedness."

For every finite union-closed family of sets, other than the family containing only the empty set, there exists an element that belongs to at least half of the sets in the family.

If the UC conjecture is tight for some family `A` then $|A| = 2^k$ for some $k$. Reference: Conjecture 3 in https://www.nieuwarchief.nl/serie5/pdf/naw5-2023-24-4-225.pdf.

Roberts and Simpson [Ro10] showed that the union-closed sets conjecture holds for set families of size at most 46. Their method, however, combined with the result of [Vu17], further shows that it …

The union-closed sets conjecture is sharp in the sense that if we replace the constant `1/2` with any larger constant, then the conjecture fails.

We can show the union-closed sets conjecture is true for the case where the set family contains some singleton.

Vuckovic and Zivkovic [Vu17] showed that the union-closed sets conjecture holds for set families whose universal set has cardinality at most 12. [Vu17] Vuckovic, Bojan; Zivkovic, Miodrag (2017). "The 12-Element Case of …

We can show the union-closed sets conjecture is true for the case where the universal set has cardinality 2, by brute force.

Yu [Yu23] showed that the union-closed sets conjecture holds with a constant of approximately 0.38234 instead of 1/2. [Yu23] Yu, Lei (2023). "Dimension-free bounds for the union-closed sets conjecture". Entropy. 25 (5): …

$108$ is the only primitive term.

If $P$ is the Promislow group, then the group ring $\mathbb{F}_p[P]$ has a non-trivial unit.

If $P$ is the Promislow group, then the group ring $\mathbb{C}[P]$ has a non-trivial unit.

Is it true that $$N(b) \ll \log \log b$$?

Asymptotic upper bound (1.2) in [Aa19]. Named after Hardy and Littlewood [HaL28].

From [Aa19] p.577: the trivial upper bound is $n^2$ (non asymptotic)

Let \(T_N = \sum_{k=1}^{N} k = \frac{N(N+1)}{2}\). If \(T_N\) is even (equivalently \(N \equiv 0 \pmod 4\) or \(N \equiv 3 \pmod 4\)), then under optimal play the game `Catch-Up(\(\{1, \ldots, N\}\))` …

The conjecture of van der Waerden, which states that the permanent of a doubly stochastic matrix is at least $n^{-n} n!$. Proved by Gyires [Gy80], Egorychev [Eg81], and Falikman [Fa81]. [Gy80] Gyires, …

The Vaught conjecture states that for a countable language L and a complete L-Theory T the number of countable models of T (up to isomorphism) is finite, $\aleph_0$ or $2^{\aleph_0}$.

The boundary of the Mandelbrot set is conjectured to have zero area.

The boundary of any Multibrot set is conjectured to have zero area. Note that we don't need to exclude the trivial cases `n = 0` and `n = 1` because the conjecture …

Weak form of Hall's conjecture: relax the exponent from $1/2$ to $1/2 - \varepsilon$.

It is conjectured that there are infinitely many Wolstenholme primes. *Reference:* [Wikipedia](https://en.wikipedia.org/wiki/Wolstenholme_prime#Expected_number_of_Wolstenholme_primes)

The **Zariski Cancellation Problem**: every polynomial ring over a field `k` of characteristic `0` is cancellative.

The single variable polynomial ring `k[X]` is cancellative in any characteristic

The two variable polynomial ring `k[X]` is cancellative in any characteristic

The positive characteristic case of the Zariski Cancellation Problem is false in dimension `3`

**The zero-divisor conjecture** If `G` is torsion-free, then the group algebra `K[G]` has no non-trivial zero divisors.

From [Yufei Zhao]: Is there a subset of $\{1, \ldots, N\}$ of size $N^{1/3 - o(1)}$ with no nontrivial solutions to $x + 2y + 3z = x' + 2y' + 3z'$?