Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Statement
Setting (p. 216). A finite set of integers is admissible when, for every prime , at least one residue class modulo contains none of its elements, and is the size of the largest admissible set in .
Theorem 1 (p. 223, quoted). "If , then ."
The paper uses it in its exhaustive search (Section 3): when and , the value can be skipped, and when the search may require and to survive.
Proof pointer
P. 223. The proof uses the remark just before it (pp. 222--223) that endpoints which do not survive can be dropped, and the inequality stated on p. 222. Take an optimal sieve on . Both endpoints must survive, or else . The elements and must survive too, or else the interval (respectively ) would give , against the two strict inequalities. Then all survive; the classes and modulo are hit by and , so the removed class is , and must avoid it while does too, which leaves .
Read depth
Claims checked: the definition, Theorem 1 and its proof were read clause by clause on the page images of the copy named on the source card, and the proof was followed. Nothing here is independently reviewed.
Dependencies
None in the corpus. The proof uses only the search remarks and the inequality recorded on pp. 222--223.
Source. Daniel M. Gordon and Gene Rodemich, "Dense admissible sets," Algorithmic Number Theory, Lecture Notes in Computer Science 1423 (1998), 216--225, doi:10.1007/BFb0054864. Pages are the published pagination, p. 223 being p. 8 of the copy read, as the source card explains.
Bears on
- Problem 1204: is the largest with , by translation of admissible sets. Theorem 1 restricts where can rise at two consecutive steps of , so it constrains which endpoints can take in that pattern. The paper uses it only to prune its finite search; it gives no asymptotic information about or .