Erdős Problem #262
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Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolvedStatement
An increasing sequence of positive integers $(a_n)$ is called an irrationality sequence here if $\sum_{n\ge1}1/(t_na_n)$ is irrational for every sequence of positive integers $(t_n)$. How slowly can $a_n$ grow?
Hančl's resolution gives the necessary threshold \[ \limsup_{n\to\infty}\frac{\log_2\log_2 a_n}{n}\ge1. \tag{1} \] We prove a constructive obstruction implying this threshold and the stronger condition recorded on the problem page. For the matching example we explicitly use Erdős's known theorem that $a_n=2^{2^n}$ is an irrationality sequence. Thus the answer concerns the double-exponential growth scale, not a classification of every sequence at the boundary. No novelty is claimed.
Restatement from Math-5537-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
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GPT 6 Astra Ultra attempt on Math-5537
Erdős Problem #262
Statement and scope
An increasing sequence of positive integers $(a_n)$ is called an irrationality sequence here if $\sum_{n\ge1}1/(t_na_n)$ is irrational for every sequence of positive integers $(t_n)$. How slowly can $a_n$ grow?Hančl's resolution gives the necessary threshold \[ \limsup_{n\to\infty}\frac{\log_2\log_2 a_n}{n}\ge1. \tag{1} \] We prove a constructive obstruction implying this threshold and the stronger condition recorded on the problem page. For the matching example we explicitly use Erdős's known theorem that $a_n=2^{2^n}$ is an irrationality sequence. Thus the answer concerns the double-exponential growth scale, not a classification of every sequence at the boundary. No novelty is claimed.
Work: a greedy construction of a rational sum
We prove the following useful sufficient condition for failure: \[ L:=\sum_{n\ge1}2^{-n}\log a_n<\infty \quad\Longrightarrow\quad (a_n)\hbox{ is not an irrationality sequence}. \tag{2} \] Choose a positive rational $r_0$ with $\log r_0+L<-1$. Recursively define \[ t_n=\left\lfloor\frac1{a_nr_{n-1}}\right\rfloor+1, \qquad r_n=r_{n-1}-\frac1{a_nt_n}. \] The strict choice of the next integer ensures $t_n>1/(a_nr_{n-1})$, so $r_n>0$, even when the quantity inside the floor is an integer. Also \[ t_n\le\frac1{a_nr_{n-1}}+1, \qquad r_n\le r_{n-1}-\frac{r_{n-1}}{1+a_nr_{n-1}} \le a_nr_{n-1}^2. \] Taking logarithms and iterating gives \[ \log r_n\le 2^n\left(\log r_0+ \sum_{j=1}^n2^{-j}\log a_j\right)<-2^n. \] Thus $r_n\to0$, and telescoping proves \[ \sum_{n\ge1}\frac1{t_na_n}=r_0\in\mathbb Q. \] All $t_n$ are positive integers and every summand is nonzero. This proves $2$.Consequences for the precise growth question
If the limsup in $1$ were less than one, then for some $\varepsilon>0$ and all sufficiently large $n$, $\log a_n\le (\log2)2^{(1-\varepsilon)n}$. The series in $2$ would converge geometrically, contradicting irrationality. This proves $1$.More generally, suppose $a_n\le K2^{2^{n-F(n)}}$ eventually and $\sum_n2^{-F(n)}<\infty$. Then \[ 2^{-n}\log a_n\le 2^{-n}\log K+(\log2)2^{-F(n)}, \] so $2$ applies. This gives the stronger obstruction in the current editorial, with no regularity required of $F$.
The known Erdős example $a_n=2^{2^n}$ has $\log_2\log_2 a_n=n$. Its irrationality for arbitrary integer multipliers is the external positive input; it does not follow merely from its rate of growth. Indeed double-exponential sequences can also have rational reciprocal sums. For the simpler non-example $a_n=n!$, choosing $t_n=n+2$ gives the exact telescoping identity \[ \frac1{(n+2)n!}=\frac1{(n+1)!}-\frac1{(n+2)!}, \qquad \sum_{n\ge1}\frac1{(n+2)n!}=\frac12. \]
Verification and outcome
The greedy proof does not rely on numerical approximation, divisibility between different $a_n$, or monotonicity. It constructs a rational sum and bounds the residual by $e^{-2^n}$. Together with Erdős's example it determines the threshold $1$ sharply on the stated logarithmic scale.SOLVED at the growth-scale resolution in the question, with the positive example explicitly cited. No claim is made that $1$ is sufficient for an arbitrary sequence.