---

Claim. The answer to Problem 449 is no. For every constant $K>0$ the integers $n$ with $r(n)>K\tau(n)$ contain a set of positive density, so already for $\epsilon=1$ the inequality $r(n)<\epsilon\tau(n)$ fails on a set of positive density and cannot hold for almost all $n$. The argument, which the site credits to Kevin Ford, compares $r(n)$ with the dyadic divisor count $\tau^+(n)$ of Problem 448, the number of $k$ with a divisor of $n$ in $[2^k,2^{k+1})$. Write $c_k$ for the number of divisors of $n$ in that block, so that $\sum_k c_k=\tau(n)$ with $\tau^+(n)$ nonzero terms. Two distinct divisors in one block satisfy $d_1<d_2<2d_1$, so

$$\sum_k c_k^2=\tau(n)+2\,\#\{d_1<d_2 \text{ in one block}\}\le\tau(n)+2r(n),$$

and the Cauchy-Schwarz inequality $\tau(n)^2\le\tau^+(n)\sum_k c_k^2$ gives

$$\frac{\tau(n)^2}{\tau^+(n)}\le\tau(n)+2r(n).$$

The site displays this bound without the factor $2$, a form that fails at $n=6$; the missing factor was pointed out in the problem's discussion thread on 2026-08-28 and is corrected in the Lean development linked above, and it does not affect the conclusion. The other input is the positive-density statement the site draws from the negative solution of Problem 448: for every $\alpha>0$ the integers with $\tau^+(n)\le\alpha\tau(n)$ contain a set of positive density. It is elementary: on the progression $n\equiv M\pmod{M^2}$ with $M=6^a$, the divisor count is $(a+1)^2$ times that of the cofactor while the dyadic count is at most $3a+2$ times it, so $\tau^+(n)\le\alpha\tau(n)$ on the whole progression once $a$ is large. The statement sits inside the study of the ratio $\tau^+(n)/\tau(n)$ by Erdős and Tenenbaum (1981) (card) and by Hall and Tenenbaum (1988), who prove that the ratio has a distribution function. On that set with $\alpha=1/(2K+2)$ the left side is at least $(2K+2)\tau(n)$, so $r(n)\ge(K+\tfrac12)\tau(n)>K\tau(n)$. The Lean development linked above works on the same progression with $M=6^a$ but in the other order: it proves $r(M)>K\tau(M)$ for $M$ itself from the Cauchy-Schwarz bound, using $\tau(6^a)=(a+1)^2$ and $\tau^+(6^a)\le3a+1$, and transfers the inequality to $n=Mq$ through $r(Mq)\ge r(M)\tau(q)$ for $q$ coprime to $M$, without bounding $\tau^+(n)$. The site adds that Hall and Tenenbaum, Divisors (Cambridge Tracts in Mathematics 90, 1988), Section 4.6, give the argument for an essentially identical problem; the book is not held here.

Depends on. No page of this wiki.

Claimant and date. The site's page credits the observation to Kevin Ford and thanks Ford for it, without a date. The remark is already present in an archived copy of the page from 2024-07-13, linked above, which is the earliest dated record of it known here and names this page.

Acceptance. Thomas Bloom, the site's curator, marks the problem disproved and credits Ford's observation on the problem page. No separate publication of the deduction is known here; the related result in Hall and Tenenbaum's book is cited above as the site cites it. Boris Alexeev's repository of formalized Erdős problems holds a Lean development, added on 2026-08-17, whose index page of 2026-08-22 is linked above at a pinned commit; its header names Ford as informal author and Codex and GPT-5.6 Sol as formal authors; it proves that for every $K>0$ the set of $n$ with $K\tau(n)<r(n)$ contains a set of positive density and hence the negation of the problem's statement. This corpus has not built or audited it, so no formalized evidence is listed. The claim is accepted on the curator's documented acceptance.