Open-problem statements, with their sources attached.

Is there an absolute constant $K$ such that, for every $C > 0$, if $n$ is sufficiently large then $n$ has at most $K$ divisors in $(n^{\frac{1}{2}}, n^{\frac{1}{2}} + C n^{\frac{1}{4}})$.

Erdős and Rosenfeld, ask whether $4$ is the best possible $K$ for the infinitude of $n$ with (at least) $K$ divisors in $(n^{\frac{1}{2}}, n^{\frac{1}{2}} + n^{\frac{1}{4}})$.

Let $v(n,k)$ count the prime factors of $n+k$ which do not divide $n+i$ for $0\leq i < k$. Is it true that $v_0(n)=\max_{k\geq 0}v(n,k)\to \infty$ as $n\to \infty$?

Let $v_l(n) = \max_{k\geq l} v(n,k)$. For every fixed $l$, $v_l(n) \to \infty$ as $n \to \infty$

[ErSe67] Erdős, P. and Selfridge, J. L., Some problems on the prime factors of consecutive integers. Illinois J. Math. (1967), 428--430.

Does $v_1(n) = 1$ have finite solutions?

[ErSe67] Erdős, P. and Selfridge, J. L., Some problems on the prime factors of consecutive integers. Illinois J. Math. (1967), 428--430.

Does $V_1(n) = 1$ have finite solutions?

This is a modification of erdos_889.variants.v1_eq_1_finite, which might make it more amenable to attack according to [ErSe67].

[ErSe67] Erdős, P. and Selfridge, J. L., Some problems on the prime factors of consecutive integers. Illinois J. Math. (1967), 428--430.

If $\omega_k(n)$ counts the number of distinct prime factors of $n$ which are $>k$, then is it true that, for every $k\geq 1$, $\liminf_{n\to \infty}\sum_{0\leq i < k}\omega_k(n+i)\leq k?$

Is it true that $\limsup_{n\to \infty}\left(\sum_{0\leq i < k}\omega(n+i)\right) \frac{\log\log n}{\log n}=1,$ where $\omega$ counts the number of distinct prime factors without restriction?

Let $2=p_1 < p_2 < \cdots$ be the primes and $k\geq 2$. Is it true that, for all sufficiently large $n$, there must exist an integer in $[n,n+p_1\cdots p_k)$ with $>k$ many prime factors?

This is unknown even for $k=2$ - that is, is it true that in every interval of $6$ (sufficiently large) consecutive integers there must exist one with at least $3$ prime factors?

Weisenberg has observed that Dickson's conjecture implies the answer is no if we replace $p_1\cdots p_k$ with $p_1\cdots p_k-1$. Indeed, let $L_k$ be the lowest common multiple of all integers at most $p_1\cdots p_k$. By Dickson's conjecture [Wikipedia], there are infinitely many $n'$ such that $\frac{L_k}{m}n'+1$ is prime for all $1\leq m < p_1\cdots p_k$. It follows that, if $n=L_kn'+1$, then all integers in $[n,n+p_1\cdots p_k-1)$ have at most $k$ prime factors.

Let $f(n)$ be an additive function (so that $f(ab)=f(a)+f(b)$ if $(a,b)=1$) such that $\limsup_{p,k} f(p^k) / \log(p^k) = ∞$ and $f(p^k) = f(p)$ or $f(p^k) = kf(p)$. Is it true that $\limsup_n (f(n+1)−f(n))/ \log n = ∞$?

The known counterexample does not satisfy either of these extra hypotheses, so this variant remains open.

Let $f(n)$ be an additive function (so that $f(ab)=f(a)+f(b)$ if $(a,b)=1$) such that $\limsup_{p,k} f(p^k) / \log(p^k) = ∞$ and $f(p^k) = f(p)$ or $f(p^k) = kf(p)$. Is it true that $\limsup_n f(n+1)/f(n) = ∞$?

The known counterexample does not satisfy either of these extra hypotheses, so this variant remains open.

Prove that there exists some $c>0$ such that $h(n) \sim c \left(\frac{n}{\log n}\right)^{1/2}$ as $n\to \infty$.

A heuristic of Tao using the Cramér model for the primes suggests this is true with $c=\sqrt{2\pi}$.

Are there infinitely many $n$ such that if

is the factorisation into distinct primes then all exponents $k_i$ are distinct?

It is likely that there are infinitely many primes $p$ such that $8p^2 - 1$ is also prime.

Is it true that, for every $r$, there is a $k$ such that if $I_1,\ldots,I_r$ are disjoint intervals of consecutive integers, all of length at least $k$, then

is not a perfect power?

Let $k_1 \geq k_2 \geq 3$. Are there only finitely many $n_2\geq n_1 + k_1$ such that

have the same prime factors?

Erdős thought perhaps if the two products have the same factors then $n_2 > 2(n_1 + k_1)$. It is an open question whether this is true when allowing a finite number of counterexamples.