Browse Problems

Search and filter the ASI Benchmark problem set. Click any problem to see its Lean statement and recent submissions.

Filters

Showing 50 of 1,494

Results

Prize Problem Status ID
- 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 …
Open Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.parts.i
- 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 …
Open Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.parts.ii
- 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 …
Open Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.variants.part_ii_power_squarefreePart_modeq_one
- 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 …
Open Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.variants.part_ii_power_squarefreePart_not_modeq_one
- 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 …
Solved Wikipedia__ArtinPrimitiveRootsConjecture__artin_primitive_roots.variants.part_ii_square_or_minus_one
- 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 …
Solved Arxiv__Conjecture6_3__arxiv.id0911_2077.conjecture6_3
- a_unbounded
Counter-conjecture to `a_isBigO`: $a(n) / (\log n \log \log n)$ is unbounded.
Open OEIS__34693__a_unbounded
- avg_rank_lt_0885
Theorem 3 of [BS2013]: when elliptic curves over ℚ are ordered by height, their average rank is < .885.
Solved Wikipedia__EllipticCurveRank__avg_rank_lt_0885
- 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?
Open Wikipedia__BalancedPrimes__balanced_primes
- balanced_primes_order
Let $p_k$ be the $k$-th prime number. Are there infinitely many $n$ such that $p_n = \dfrac{\sum_{i = 1} ^ k p_{n - i} + p_{n + i}}{2*k}$?
Open Wikipedia__BalancedPrimes__balanced_primes_order
- 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 …
Open Wikipedia__BatemanHornConjecture__bateman_horn_conjecture
- baumgartner_headline
The statement for which Baumgartner actually writes a proof.
Solved ErdosProblems__198__baumgartner_headline
- 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 …
Solved ErdosProblems__198__baumgartner_strong
- BB_5 Solved Wikipedia__BusyBeaver__BB_5
- BB_6
Determine the value of the Busy Beaver function at n = 6.
Open Wikipedia__BusyBeaver__BB_6
- 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 …
Open Other__BeaverMathOlympiad__beaver_math_olympiad_problem_2_antihydra
- 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.
Open Other__BeaverMathOlympiad__beaver_math_olympiad_problem_2_antihydra.variants.set
- 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 $$a_n = \begin{cases} 2 & \text{if } n=0 \\ a_{n-1}+2^{v_2(a_{n-1})+2}-1 …
Solved Other__BeaverMathOlympiad__beaver_math_olympiad_problem_3
- beaver_math_olympiad_problem_4
[BMO#4](https://wiki.bbchallenge.org/wiki/Beaver_Math_Olympiad#4._1RB3RB---1LB0LA_2LA4RA3LA4RB1LB_(bbch)) Bonnie the beaver was bored, so she tried to construct a sequence of integers $\{a_n\}_{n \ge 0}$. She first defined $a_0=2$, then defined $a_{n+1}$ depending on $a_n$ and $n$ using the …
Solved Other__BeaverMathOlympiad__beaver_math_olympiad_problem_4
- beaver_math_olympiad_problem_5
[BMO#5](https://wiki.bbchallenge.org/wiki/Beaver_Math_Olympiad#5._1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE_(bbch)) Let $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ be two sequences such that $(a_1, b_1) = (0, 5)$ and $$(a_{n+1}, b_{n+1}) = \begin{cases} (a_n+1, b_n-f(a_n)) & \text{if } b_n \ge f(a_n) …
Open Other__BeaverMathOlympiad__beaver_math_olympiad_problem_5
- 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?
Open Wikipedia__BoundedBurnsideProblem__bounded_burnside_problem
- 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 …
Solved Wikipedia__LegendreConjecture__bounded_gap_legendre
- BoxdotConjecture
Boxdot Conjecture: every normal modal logic that faithfully interprets KT by the boxdot translation is included in KT.
Solved Arxiv__BoxdotConjecture__BoxdotConjecture
- 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.
Open Wikipedia__BrocardConjecture__brocard_conjecture
- brocard_conjecture.ferreira_large_n
Ferreira proved that Brocard's conjecture is true for sufficiently large n.
Solved Wikipedia__BrocardConjecture__brocard_conjecture.ferreira_large_n
- 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.
Solved ErdosProblems__723__bruck_ryser
- 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.
Open Wikipedia__Bunyakovsky__bunyakovsky_conjecture
- 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 \\ …
Open Other__BeaverMathOlympiad__busy_beaver_math_olympiad_problem_1
- c_2_lower
Lower bound for $c(2)$ from Green's first paper ([Gr01]); the constant is `sqrt(4/7)` (about 0.7559).
Solved GreensOpenProblems__35__c_2_lower
- carchimaelTotient_bound
In Theorem 6 in [F1998], Kevin Ford proves that the smallest counterexample to Carmichael's totient function conjecture must be $≥ 10 ^ (10 ^ 10)$
Solved Wikipedia__CarmichaelTotient__carchimaelTotient_bound
- 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`.
Open Paper__CasasAlvero__casas_alvero_conjecture
- 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 …
Solved Paper__CasasAlvero__casas_alvero.double_prime_power
- 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 …
Solved Paper__CasasAlvero__casas_alvero.positive_char_counterexample
- 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)
Solved Paper__CasasAlvero__casas_alvero.prime_power
- 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 = …
Solved Wikipedia__Catalan__catalans_conjecture
- charmichaelTotient
*Carmichael's totient function conjecture*: For every positive natural number $n$, there exists a natural number $m$ with $m ≠ n$, such that $φ(n) = φ(m)$.
Open Wikipedia__CarmichaelTotient__charmichaelTotient
- CH.existsWeaklyFirstCountableCompactNotFirstCountable
Under CH, such a space exists as constructed in [Ya1976] by Yakovlev.
Solved Paper__WeaklyFirstCountable__CH.existsWeaklyFirstCountableCompactNotFirstCountable
- 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).
Solved ErdosProblems__835__chromaticNumber_johnson_2k_k_lower_bound
- 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).
Solved ErdosProblems__835__chromaticNumber_johnson_2k_k_lower_bound_odd
- c_inf_lower
Best-known lower bound for $c(\infty)$ due to Cloninger and Steinerberger ([CS17]).
Solved GreensOpenProblems__35__c_inf_lower
- c_inf_upper
Best-known upper bound for $c(\infty)$ due to Matolcsi and Vinuesa ([MV10]).
Solved GreensOpenProblems__35__c_inf_upper
- class_number_problem
There are infinitely many real quadratic fields `ℚ(√d)` with class number one, where `d > 1` is a squarefree integer.
Open Wikipedia__ClassNumberProblem__class_number_problem
- 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, …
Solved Wikipedia__ClassNumberProblem__class_number_problem.variants.imaginary
- 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 …
Open Arxiv__CollatzLike__CollatzLike
- comesFromPrimeQuadruple_of_a
All members of the sequence A56777 come from prime quadruples.
Open OEIS__56777__comesFromPrimeQuadruple_of_a
- complete_for_alpha_in_Ioo_one_to_goldenRatio
It seems likely that the sequence is complete for all for all $t>0$ and all $1 < \alpha < \frac{1+\sqrt{5}}{2}$.
Open ErdosProblems__349__complete_for_alpha_in_Ioo_one_to_goldenRatio
- complexity_five_pow
Is `5n` the complexity of `5^n` for `0 < n`? Answer: No.
Solved Mathoverflow__75792__complexity_five_pow
- 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
Solved Mathoverflow__75792__complexity_three_pow
- complexity_two_pow
Is `2n` the complexity of `2^n` for `0 < n`?
Open Mathoverflow__75792__complexity_two_pow
- conditional_artin_primitive_roots.parts.i
**Artin's Conjecture on Primitive Roots**, first half, conditional on GRH.
Solved Wikipedia__ArtinPrimitiveRootsConjecture__conditional_artin_primitive_roots.parts.i