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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

Item 6 of the new-problems half (printed p. 150, in Hungarian): let a1<⋯<ak≤na_1<\dots<a_k\le n be an arbitrary sequence and b1<…b_1<\dots the sequence of those numbers that are multiples of at least one aa ("azon számok sorozata, melyek legalább egy aa-nak többszörösei"). Is it true that for every m>nm>n

B(m)m<2B(n)n,B(x)=∑bi≤x1 ?(1)\frac{B(m)}{m}<\frac{2B(n)}{n},\qquad B(x)=\sum_{b_i\le x}1\,? \tag{1}

In translation: "(1), if true, clearly cannot be improved: let the sequence a1<…a_1<\dots consist only of a1a_1, n=2a1−1n=2a_1-1, m=2a1m=2a_1. We may also remark that there is no ε>0\varepsilon>0 for which B(m)/m>εB(n)/nB(m)/m>\varepsilon B(n)/n holds for every sequence a1<⋯<ak≤na_1<\dots<a_k\le n and m>nm>n": take the aa's to be the numbers between n/2n/2 and nn and m=m(n)m=m(n) large; if n>n0(ε)n>n_0(\varepsilon) then B(m)/m<ε/2B(m)/m<\varepsilon/2. The item cites Erdős, Note on sequences of integers no one of which is divisible by any other, J. London Math. Soc. 10 (1935), 126--128.

So the 1966 paper defines BB as the count of multiples (a∣ba\mid b), the site's reading of Problem 488; the 1961 problem paper prints the count of non-multiples at its item (I.27.1) (printed p. 236; filed as erdos_1961_unsolved_problems).

Source. P. Erdős, Számelméleti megjegyzések, V. Extremális problémák a számelméletben, II, Mat. Lapok 17 (1966), 135--155; item 6 on printed p. 150 (PDF p. 16 of the Rényi archive's 21-page scan), read on the page image.

Read depth. Claims checked: the definition, display (1), the single-element example and the reverse-inequality remark were read clause by clause on the page image. A question with two elementary remarks; no proof.

Proof pointer

None; a question. The two remarks are elementary: for the single set {a1}\{a_1\} with n=2a1−1n=2a_1-1 and m=2a1m=2a_1, B(n)/n=1/(2a1−1)B(n)/n=1/(2a_1-1) and B(m)/m=2/(2a1)=1/a1B(m)/m=2/(2a_1)=1/a_1, so (B(m)/m)/(B(n)/n)=(2a1−1)/a1→2(B(m)/m)/(B(n)/n)=(2a_1-1)/a_1\to2, and no constant below 22 works in (1); the reverse example follows from the multiples of the numbers in (n/2,n](n/2,n] thinning out.

Dependencies

None (a question).

Bears on

  • Problem 488: the problem's statement in the 1966 wording with BB the count of multiples, and the sharpness example the site's page records; the 1961 paper's item prints the opposite divisibility condition.