Browse Problems
Search and filter the ASI Benchmark problem set. Click any problem to see its Lean statement and recent submissions.
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| Prize | Problem | Status | ID |
|---|---|---|---|
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artin_primitive_roots.parts.i
**Artin's Conjecture on Primitive Roots**, first half. Let $a$ be an integer that is not a square number and not $−1$. Then the set $S(a)$ of primes $p$ such that $a$ is …
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Open | Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.parts.i |
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artin_primitive_roots.parts.ii
**Artin's Conjecture on Primitive Roots**, second half. Write $a = a_0 b^2$ where $a_0$ is squarefree. Under the conditions that $a$ is not a perfect power and $a_0\not\equiv 1\pmod{4}$ (sequence A85397 in …
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Open | Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.parts.ii |
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artin_primitive_roots.variants.part_ii_power_squarefreePart_modeq_one
**Artin's Conjecture on Primitive Roots**, second half, power version If $a = b^m$ is a perfect power of a number $b$ whose squarefree part $b_0\equiv 1 \pmod{4}$, then the density of the …
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Open | Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.variants.part_ii_power_squarefreePart_modeq_one |
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artin_primitive_roots.variants.part_ii_power_squarefreePart_not_modeq_one
**Artin's Conjecture on Primitive Roots**, second half, power version If $a = b^m$ is a perfect odd power of a number $b$ whose squarefree part $b_0\not\equiv 1 \pmod{4}$, then the density of …
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Open | Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.variants.part_ii_power_squarefreePart_not_modeq_one |
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artin_primitive_roots.variants.part_ii_square_or_minus_one
**Artin's Conjecture on Primitive Roots**, second half, different residue version If $a$ is a square number or $a = −1$, then the density of the set $S(a)$ of primes $p$ such that …
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Solved | Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.variants.part_ii_square_or_minus_one |
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arxiv.id0911_2077.conjecture6_3
Empirical evidence seems to suggest that Slud's bound does not hold for all $p$, and in fact, as $n\to\infty$, the maximal permissible $p$ shrinks to $\frac{1}{2}$. Also, the following appears to be …
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Solved | Arxiv__Conjecture6_3__arxiv.id0911_2077.conjecture6_3 |
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a_unbounded
Counter-conjecture to `a_isBigO`: $a(n) / (\log n \log \log n)$ is unbounded.
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Open | OEIS__34693__a_unbounded |
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avg_rank_lt_0885
Theorem 3 of [BS2013]: when elliptic curves over ℚ are ordered by height, their average rank is < .885.
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Solved | Wikipedia__EllipticCurveRank__avg_rank_lt_0885 |
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balanced_primes
Let $p_k$ be the $k$-th prime number. Are there infinitely many $n$ such that $(p_n + p_{n+2}) / 2$ is prime?
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Open | Wikipedia__BalancedPrimes__balanced_primes |
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balanced_primes_order
Let $p_k$ be the $k$-th prime number. Are there infinitely many $n$ such that $p_n = \dfrac{\sum_{i = 1} ^ k p_{n - i} + p_{n + i}}{2*k}$?
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Open | Wikipedia__BalancedPrimes__balanced_primes_order |
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bateman_horn_conjecture
**The Bateman-Horn Conjecture** Given a finite collection of distinct irreducible polynomials non-constant $f_1, f_2, \dots, f_k \in \mathbb{Z}[x]$ with positive leading coefficients that satisfy the Schinzel condition, the number of positive integers …
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Open | Wikipedia__BatemanHornConjecture__bateman_horn_conjecture |
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baumgartner_headline
The statement for which Baumgartner actually writes a proof.
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Solved | ErdosProblems__198__baumgartner_headline |
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baumgartner_strong
Let $V$ be a vector space over the rationals and let $k$ be a fixed positive integer. Then there is a set $X_k \subseteq V$ such that $X_k$ meets every infinite arithmetic …
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Solved | ErdosProblems__198__baumgartner_strong |
| - | BB_5 | Solved | Wikipedia__BusyBeaver__BB_5 |
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BB_6
Determine the value of the Busy Beaver function at n = 6.
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Open | Wikipedia__BusyBeaver__BB_6 |
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beaver_math_olympiad_problem_2_antihydra
[BMO#2](https://wiki.bbchallenge.org/wiki/Beaver_Math_Olympiad#2._Hydra_and_Antihydra) Antihydra is a sequence starting at 8, and iterating the function $$H(n) = \left\lfloor \frac {3n}2 \right\rfloor.$$ The conjecture states that the cumulative number of odd values in this sequence is …
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Open | Other__BeaverMathOlympiad__beaver_math_olympiad_problem_2_antihydra |
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beaver_math_olympiad_problem_2_antihydra.variants.set
[BMO#2](https://wiki.bbchallenge.org/wiki/Beaver_Math_Olympiad#2._Hydra_and_Antihydra) formulation variant Alternative statement of beaver_math_olympiad_problem_2_antihydra using set size comparison instead of a recurrent sequence b.
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Open | Other__BeaverMathOlympiad__beaver_math_olympiad_problem_2_antihydra.variants.set |
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beaver_math_olympiad_problem_3
[BMO#3][https://wiki.bbchallenge.org/wiki/Beaver_Math_Olympiad#3._1RB0RB3LA4LA2RA_2LB3RA---3RA4RB_(bbch)_and_1RB1RB3LA4LA2RA_2LB3RA---3RA4RB_(bbch)] Let $v_2(n)$ be the largest integer $k$ such that $2^k$ divides $n$. Let $(a_n)_{n \ge 0}$ be a sequence such that $$a_n = \begin{cases} 2 & \text{if } n=0 \\ a_{n-1}+2^{v_2(a_{n-1})+2}-1 …
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Solved | Other__BeaverMathOlympiad__beaver_math_olympiad_problem_3 |
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beaver_math_olympiad_problem_4
[BMO#4](https://wiki.bbchallenge.org/wiki/Beaver_Math_Olympiad#4._1RB3RB---1LB0LA_2LA4RA3LA4RB1LB_(bbch)) Bonnie the beaver was bored, so she tried to construct a sequence of integers $\{a_n\}_{n \ge 0}$. She first defined $a_0=2$, then defined $a_{n+1}$ depending on $a_n$ and $n$ using the …
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Solved | Other__BeaverMathOlympiad__beaver_math_olympiad_problem_4 |
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beaver_math_olympiad_problem_5
[BMO#5](https://wiki.bbchallenge.org/wiki/Beaver_Math_Olympiad#5._1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE_(bbch)) Let $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ be two sequences such that $(a_1, b_1) = (0, 5)$ and $$(a_{n+1}, b_{n+1}) = \begin{cases} (a_n+1, b_n-f(a_n)) & \text{if } b_n \ge f(a_n) …
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Open | Other__BeaverMathOlympiad__beaver_math_olympiad_problem_5 |
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bounded_burnside_problem
Let $G$ be a finitely generated group, and assume there exists $n$ such that for every $g$ in $G$, $g^n = 1$. Is $G$ necessarily finite?
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Open | Wikipedia__BoundedBurnsideProblem__bounded_burnside_problem |
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bounded_gap_legendre
If there exists a constant `c > 0` such that `(n + 1).nth Nat.Prime - n.nth Nat.Prime < (n.nth Nat.Prime) ^ (1 / 2 - c)` for all large `n`, then Legendre's …
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Solved | Wikipedia__LegendreConjecture__bounded_gap_legendre |
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BoxdotConjecture
Boxdot Conjecture: every normal modal logic that faithfully interprets KT by the boxdot translation is included in KT.
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Solved | Arxiv__BoxdotConjecture__BoxdotConjecture |
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brocard_conjecture
**Brocard's Conjecture** For every `n ≥ 2`, between the squares of the `n`-th and `(n+1)`-th primes, there are at least four prime numbers.
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Open | Wikipedia__BrocardConjecture__brocard_conjecture |
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brocard_conjecture.ferreira_large_n
Ferreira proved that Brocard's conjecture is true for sufficiently large n.
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Solved | Wikipedia__BrocardConjecture__brocard_conjecture.ferreira_large_n |
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bruck_ryser
Bruck and Ryser have proved that if $n \equiv 1 (\mod 4)$ or $n \equiv 2 (\mod 4)$ then $n$ must be the sum of two squares.
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Solved | ErdosProblems__723__bruck_ryser |
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bunyakovsky_conjecture
**Bunyakovsky conjecture** If a polynomial $f$ over integers satisfies both Schinzel and Bunyakovsky conditions, there exist infinitely many natural numbers $m$ such that $f(m)$ is prime.
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Open | Wikipedia__Bunyakovsky__bunyakovsky_conjecture |
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busy_beaver_math_olympiad_problem_1
[BMO#1](https://wiki.bbchallenge.org/wiki/Beaver_Math_Olympiad#1._1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE_(bbch)) Let $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ be two sequences such that $(a_1, b_1) = (1, 2)$ and $$(a_{n+1}, b_{n+1}) = \begin{cases} (a_n-b_n, 4b_n+2) & \text{if }a_n \ge b_n \\ …
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Open | Other__BeaverMathOlympiad__busy_beaver_math_olympiad_problem_1 |
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c_2_lower
Lower bound for $c(2)$ from Green's first paper ([Gr01]); the constant is `sqrt(4/7)` (about 0.7559).
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Solved | GreensOpenProblems__35__c_2_lower |
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carchimaelTotient_bound
In Theorem 6 in [F1998], Kevin Ford proves that the smallest counterexample to Carmichael's totient function conjecture must be $≥ 10 ^ (10 ^ 10)$
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Solved | Wikipedia__CarmichaelTotient__carchimaelTotient_bound |
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casas_alvero_conjecture
The Casas-Alvero conjecture states that in characteristic zero, if a monic polynomial `P` has the Casas-Alvero property, then `P = (X - α)ᵈ` for some `α`.
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Open | Paper__CasasAlvero__casas_alvero_conjecture |
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casas_alvero.double_prime_power
The Casas-Alvero conjecture holds for polynomials of degree `2p^k` where `p` is prime. This was proved by Graf von Bothmer, Labs, Schicho, and van de Woestijne. Reference: [The Casas-Alvero conjecture for infinitely …
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Solved | Paper__CasasAlvero__casas_alvero.double_prime_power |
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casas_alvero.positive_char_counterexample
The Casas-Alvero conjecture fails in positive characteristic `p` for polynomials of degree `p + 1`. This was shown by Graf von Bothmer, Labs, Schicho, and van de Woestijne. Reference: [The Casas-Alvero conjecture …
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Solved | Paper__CasasAlvero__casas_alvero.positive_char_counterexample |
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casas_alvero.prime_power
The Casas-Alvero conjecture holds for polynomials of prime power degree. This was proved by Graf von Bothmer, Labs, Schicho, and van de Woestijne. Reference: [The Casas-Alvero conjecture for infinitely many degrees](https://arxiv.org/pdf/math/0605090)
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Solved | Paper__CasasAlvero__casas_alvero.prime_power |
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catalans_conjecture
The only natural number solution to the equation $x^a - y^b = 1$ such that $a, b > 1$ and $x, y > 0$ is given by $a = 2$, $b = …
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Solved | Wikipedia__Catalan__catalans_conjecture |
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charmichaelTotient
*Carmichael's totient function conjecture*: For every positive natural number $n$, there exists a natural number $m$ with $m ≠ n$, such that $φ(n) = φ(m)$.
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Open | Wikipedia__CarmichaelTotient__charmichaelTotient |
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CH.existsWeaklyFirstCountableCompactNotFirstCountable
Under CH, such a space exists as constructed in [Ya1976] by Yakovlev.
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Solved | Paper__WeaklyFirstCountable__CH.existsWeaklyFirstCountableCompactNotFirstCountable |
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chromaticNumber_johnson_2k_k_lower_bound
It is known that for $3 \leq k \leq 8$, the chromatic number of $J(2k, k)$ is greater than $k+1$, see [Johnson graphs](https://aeb.win.tue.nl/graphs/Johnson.html).
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Solved | ErdosProblems__835__chromaticNumber_johnson_2k_k_lower_bound |
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chromaticNumber_johnson_2k_k_lower_bound_odd
It is also known that for $3 \leq k \leq 203$ odd, the chromatic number of $J(2k, k)$ is greater than $k+1$, see [Johnson graphs](https://aeb.win.tue.nl/graphs/Johnson.html).
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Solved | ErdosProblems__835__chromaticNumber_johnson_2k_k_lower_bound_odd |
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c_inf_lower
Best-known lower bound for $c(\infty)$ due to Cloninger and Steinerberger ([CS17]).
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Solved | GreensOpenProblems__35__c_inf_lower |
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c_inf_upper
Best-known upper bound for $c(\infty)$ due to Matolcsi and Vinuesa ([MV10]).
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Solved | GreensOpenProblems__35__c_inf_upper |
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class_number_problem
There are infinitely many real quadratic fields `ℚ(√d)` with class number one, where `d > 1` is a squarefree integer.
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Open | Wikipedia__ClassNumberProblem__class_number_problem |
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class_number_problem.variants.imaginary
**Stark–Heegner theorem** : For any squarefree integer `d < 0`, the class number of the imaginary quadratic field Q(√d) is one if and only if `d ∈ {-1, -2, -3, -7, -11, …
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Solved | Wikipedia__ClassNumberProblem__class_number_problem.variants.imaginary |
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CollatzLike
For $n > 8$, $2^n$ is not the the sum of distinct powers of $3$. Expressed here in terms of the base $3$ digits of $n$. This conjecture is equivalent to the …
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Open | Arxiv__CollatzLike__CollatzLike |
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comesFromPrimeQuadruple_of_a
All members of the sequence A56777 come from prime quadruples.
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Open | OEIS__56777__comesFromPrimeQuadruple_of_a |
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complete_for_alpha_in_Ioo_one_to_goldenRatio
It seems likely that the sequence is complete for all for all $t>0$ and all $1 < \alpha < \frac{1+\sqrt{5}}{2}$.
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Open | ErdosProblems__349__complete_for_alpha_in_Ioo_one_to_goldenRatio |
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complexity_five_pow
Is `5n` the complexity of `5^n` for `0 < n`? Answer: No.
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Solved | Mathoverflow__75792__complexity_five_pow |
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complexity_three_pow
Is `3n` the complexity of `3^n` for `0 < n`? Answer: Yes, by John Selfridge. Reference: https://arxiv.org/abs/1207.4841
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Solved | Mathoverflow__75792__complexity_three_pow |
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complexity_two_pow
Is `2n` the complexity of `2^n` for `0 < n`?
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Open | Mathoverflow__75792__complexity_two_pow |
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conditional_artin_primitive_roots.parts.i
**Artin's Conjecture on Primitive Roots**, first half, conditional on GRH.
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Solved | Wikipedia__ArtinPrimitiveRootsConjecture__conditional_artin_primitive_roots.parts.i |