Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Statement
Notation: is the number of distinct prime factors of (p. 1), and (p. 5). Erdős had called a "barrier" if for all (p. 6); thus is a barrier if and only if .
Conjecture (p. 6, unnumbered, quoted). "We conjecture that there are infinitely many barriers, that is, the minimal order of is 1."
Minimal order. The authors state (p. 6) that they can prove that the minimal order of is , and indicate the method only: a sieve arrangement making the integers just below free of primes in the interval . No proof is given.
Maximal order. The paper states (p. 6), without proof beyond naming the Chinese remainder theorem and the prime number theorem, that
for infinitely many , probably close to best possible; that the trivial bound holds for all ; and that the latter is easily improved to .
Related open question (p. 6). For , the number of solutions of , clearly ; the authors expect to be unbounded, and the best they record is infinitely often, the main result of the first paper of the series.
Source. Paul Erdős, Carl Pomerance and András Sárközy, On locally repeated values of certain arithmetic functions. III, Proc. Amer. Math. Soc. 101 (1987), no. 1, 1--7; p. 6. The edition is identified in the source digest.
Read depth. Claims checked: the conjecture and the surrounding statements were read on p. 6. None of the minimal- or maximal-order statements is proved in the paper; nothing here is independently reviewed.
Proof pointer
None: a conjecture, and statements given without proof.
Dependencies
The function of Theorem 3.2.
Bears on
- Problem 413: the conjecture is the problem's first question, with for the problem's . The minimal-order bound , which the paper states without proof, says that for infinitely many at most integers have ; it settles neither of the problem's questions.