Is it true that \[\sum_{n\in A}\frac{1}{n}<\infty?\]
Is it true that \[\sum_{n\in A}\frac{1}{n}<\infty?\]
An example of an $A$ with this property where \[\liminf \frac{\lvert A\cap\{1,\ldots,N\}\rvert}{N^{1/2}}\log N>0\] is given by the set of $p^2$, where $p\equiv 3\pmod{4}$ is prime.
For the finite version see [13].
For the infinite version see [12].
Solved by Tao [Ta23b], who proved that \[ \lvert A\rvert \leq \left(1+O\left(\frac{(\log\log x)^5}{\log x}\right)\right)\pi(x).\]
In [Er95c] Erdős further asks about the situation when $\phi(a_1)\leq \cdots \leq \phi(a_t)$.
Can one show that there exists an $\epsilon>0$ such that there are infinitely many $n$ where $m+\epsilon \omega(m)\leq n$ for all $m<n$?
Erdős also believed that $\Omega$, the count of the number of prime factors with multiplicity), should have infinitely many barriers. Selfridge found the largest barrier for $\Omega$ which is $<10^5$ is $99840$.
In [ErGr80] this problem is suggested as a way of showing that the iterated behaviour of $n\mapsto n+\omega(n)$ eventually settles into a single sequence, regardless of the starting value of $n$ (see also [412] and [414]).
Erdős and Graham report it could be attacked by sieve methods, but 'at present these methods are not strong enough'.
Ryan Alweiss has provided the following simple argument showing that the answer is yes: suppose we have some red/blue colouring without this property. Without loss of generality, suppose $1$ is coloured red, and then either $3$ or $5$ must be blue.
Suppose first that $3$ is blue. If $n\geq 6$ is red then (considering $1,n,2n-1$) we deduce $2n-1$ is blue, and then (considering $3,n+1,2n-1$) we deduce that $n+1$ is red. In particular the colouring must be eventually constant, and we are done.
Now suppose that $5$ is blue. Arguing similarly (considering $1,n,2n-1$ and $5,n+2,2n-1$) we deduce that if $n\geq 8$ is red then $n+2$ is also red, and we are similarly done, since the colouring must be eventually constant on some congruence class modulo $2$.
In [Er79] Erdős says 'it is extremely doubtful' that there are infinitely many such $n$, and in fact suggets that \[\lim_{n\to \infty}\max_{m<n}(\tau(m)+m-n)=\infty.\]
In [Er79d] Erdős says it 'seems certain' that for every $k$ there are infinitely many $n$ for which \[\max_{n-k<m<n}(m+\tau(m))\leq n+2,\] but 'this is hopeless with our present methods', although it follows from Schinzel's Hypothesis H.
See also [413].
Sampaio has observed that the answer to this question is no if one of the primes can be $2$ - for example this is false with $p=19$ and $q=2$, since if $n+1=2^k$ and $19\mid n$ then (since $2$ is a primitive root modulo $19$) we must have $18\mid k$, and hence $73\mid 2^{18}-1\mid n$. A similar argument works with $19$ replaced by any prime $p>13$ for which $2$ is a primitive root, using a result of Rotkiewicz [Ro64b] that for every prime $p>13$ there is a prime $q>p$ which divides $2^{p-1}-1$.
It is unclear whether Erdős was aware of this obstacle; certainly in [Er95c] he asks for any two primes, but may have intended to rule out such small prime obstacles.
More generally, one can ask about whether for any primes $p_1,\ldots,p_k$ there exists some $n$ such that the largest prime factor of $n+i$ is $p_i$. Erdős writes this is 'clearly impossible' if the $p_i$ are the first $k$ primes and $k$ is sufficiently large, but does not know what happens if all of the primes are sufficiently large compared to $k$.
Estimate $f(m)$. In particular is it true that $f(m)\ll m^{1/2}$?