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Source. G. Tenenbaum, Some of Erdős' unconventional problems in number theory, thirty-four years later, in L. Lovász, I. Z. Ruzsa and V. T. Sós (eds), Erdős Centennial, Bolyai Society Mathematical Studies 25 (2013), 651--681. Labels and pages here are those of the author's version identified on the source card, paginated 1--22; the published chapter was not read. The definition and conjecture (13) are on p. 6, equation (15) on p. 7.

Read depth. Claims checked: the statement was read clause by clause on the printed page. The survey reports (15) from another work and does not prove it.

Statement

Setting (p. 6, from the passage of Erdős that the survey quotes). $\tau^+(n)$ is the number of integers $k$ for which $n$ has a divisor $d$ with $2^k<d\le2^{k+1}$, and $\tau(n)$ the number of divisors of $n$. Erdős conjectured (13) that $\tau^+(n)/\tau(n)\to0$ for almost all $n$.

Equation (15) (p. 7). The survey states (p. 7) that (13) is wrong, and reports, from Hall and Tenenbaum's book Divisors (Cambridge Tracts in Mathematics 90, 1988), Chapter 4, improving on an estimate of Erdős and Tenenbaum (Ann. Inst. Fourier 31 (1981), 17--37; card) that was already enough to invalidate (13): the function $\tau^+(n)/\tau(n)$ has a limiting distribution $\nu$ with

$$\frac{z}{\sqrt{\log(2/z)}}\ll\nu(z)\ll z\log(2/z)\qquad(0<z<1).\qquad(15)$$

The survey draws from (15) that $\nu$ is continuous at the origin, and names two open problems: to improve (15), and to find the discontinuity points of $\nu$, if any. Its Theorem 1 (page) answers the second at $z=1$.

Proof pointer

None in the survey; the source is Chapter 4 of Divisors.

Dependencies

Hall and Tenenbaum, Divisors (1988), Chapter 4.

Bears on

• Problem 448: the problem asks whether, for every $\epsilon>0$, $\tau^+(n)<\epsilon\tau(n)$ for almost all $n$, which is (13). The survey states that (13) is wrong, and the upper bound in (15) makes $\nu(z)$ tend to $0$ with $z$; at a continuity point $\epsilon$ of $\nu$ the integers with $\tau^+(n)\le\epsilon\tau(n)$ have density $\nu(\epsilon)$, which is below $1$ once $\epsilon$ is small, so the answer is no. The survey counts dyadic intervals $(2^k,2^{k+1}]$, where the problem's statement uses $[2^k,2^{k+1})$.