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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

All three statements concern a real additive function ff and are posed on p. 3 without proof.

Probable result (p. 3, quoted). "The following result probably holds, but I cannot prove it: Assume that f(n+1)−f(n)<c1f(n+1)-f(n)<c_1 for all nn. Then f(n)=clog⁡n+φ(n)f(n)=c\log n+\varphi(n), ∣φ(n)∣<c2|\varphi(n)|<c_2 for all nn." The paper adds that the converse is clearly true.

Conjecture 1 (p. 3, quoted). "if f(n+1)≥f(n)f(n+1)\ge f(n) for almost all nn (i.e., all nn except for a sequence of density 0), then f(n)=clog⁡nf(n)=c\log n"

Conjecture 2 (p. 3, quoted). "if f(n+1)−f(n)→0f(n+1)-f(n)\to0 when nn runs through a sequence of density 1 then f(n)=clog⁡nf(n)=c\log n."

The paper proves the cases without exceptional set: Theorem XI (p. 17) for Conjecture 1 and Theorem XIII (p. 18) for Conjecture 2.

Proof pointer

None: the paper poses these as open.

Read depth

Claims checked: the three statements read on the page image of p. 3. Nothing here is independently reviewed.

Dependencies

None.

Source. P. Erdős, On the distribution function of additive functions, Ann. of Math. (2) 47 (1946), 1--20, doi:10.2307/1969031; the edition read is named on the source card.

Bears on

  • Problem 491: the probable result has the one-sided hypothesis f(n+1)−f(n)<c1f(n+1)-f(n)<c_1, weaker than the problem's ∣f(n+1)−f(n)∣<c|f(n+1)-f(n)|<c, and the same conclusion; it is posed, not proved, here.
  • Problem 1122: Conjecture 1 is the problem's question; it is posed, not proved, here.