Wiki
Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Problem 1200
Statement. There exists a constant such that for all large there is a collection of primes with together with a system of congruences such that every integer satisfies at least one of these congruences.
Status. Open.
Source. erdosproblems.com/1200, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #1200, https://www.erdosproblems.com/1200.
References.
- [Er80] Erdős, Paul, A survey of problems in combinatorial number theory. Ann. Discrete Math. (1980), 89-115.
- [ErRu80] Erdős, P. and Ruzsa, I. Z., On the small sieve. I. Sifting by primes. J. Number Theory (1980), 385-394.
Formalization. None recorded.
Progress
Not yet compiled.
Known Results
Not yet compiled.
Linked library material
These entries are derived from explicit links on library pages. They are navigation only and do not by themselves record mathematical progress.
- erdos_1980_survey_problems_combinatorial_number_theory
- erdos_1980_small_sieve
- erdos_1980_small_sieve / problem_2
- ruzsa_1982_small_sieve_ii_sifting_composite_numbers
- ruzsa_1982_small_sieve_ii_sifting_composite_numbers / theorem_iii
- warlimont_1991_problem_posed_i_z_ruzsa
- warlimont_1991_problem_posed_i_z_ruzsa / equation_1
- warlimont_1991_problem_posed_i_z_ruzsa / inequality_2
Linked from (10)
Primesnumber_theory/erdos_1980_survey_problems_combinatorial_number_theoryPrimesprimes/erdos_1980_small_sieveProblem 2 (p. 386): does sifting by arbitrary residue classes of primes of reciprocal sum at most K leave at least c(K) x integers?primes/ruzsa_1982_small_sieve_ii_sifting_composite_numbersTheorem III (p. 262): 1/2 < μ(x) < log(5/2) + o(1/x) for covering systems with distinct moduli in (1, x]primes/warlimont_1991_problem_posed_i_z_ruzsaEquation (1) (p. 54): the linear-programming relaxation nu*(n) of Ruzsa's covering cost equals log(2^5 3^6/23^3) + O(1/n)Inequality (2) (p. 54): the 0-1 relaxation nu(n) exceeds its linear-programming relaxation nu*(n) by O(1/n)
Graph