gpt pro 5.2 attempt on Math-729
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Erdős Problem #247
1) FORMAL RESTATEMENT
Let $a_1<a_2<\cdots$ be a strictly increasing sequence of positive integers such that \[ \limsup_{n\to\infty}\frac{a_n}{n}=\infty. \] Define the real number \[ \alpha \coloneqq \sum_{n=1}^{\infty} \frac{1}{2^{a_n}}. \] Question. Must $\alpha$ be transcendental (i.e., not algebraic over $\mathbb{Q}$)?2) QUICK LITERATURE/CONTEXT CHECK
- The Erdős Problems page for #247 states the problem as open and records that Erdős proved transcendence under a stronger growth hypothesis (citing [Erdos1975,Erdos1988]) [ErdosProblems247].
- The LaTeX view of the page lists references [Erdos1975,Erdos1988] [ErdosProblems247latex].
3) ATTACK PLAN
Possible routes to a proof (or disproof) include:- Diophantine approximation: Truncations give dyadic rationals $p_N/2^{a_N}$ approximating $\alpha$ with error $\ll 2^{-a_{N+1}}$. If one could force approximation exponents $>2$ infinitely often from the hypothesis $\limsup a_n/n=\infty$, then Roth's theorem would imply transcendence. However, $\limsup a_n/n=\infty$ does not directly control ratios $a_{n+1}/a_n$.
- Combinatorics/complexity of digit expansions: $\alpha$ has a binary expansion supported on the set $\{a_n\}$. If one could show that the subword complexity of this expansion is "too small" under the hypothesis, then deep theorems on the base-$b$ expansions of algebraic numbers (Adamczewski--Bugeaud type results) might yield transcendence. The hypothesis here is weak enough that the digit pattern can still be very complicated.
- Disproof attempt: Construct an algebraic number whose binary expansion has 1's at positions $a_n$ with $\limsup a_n/n=\infty$. I do not know any construction of algebraic numbers with such prescribed sparse digit structure.
4) WORK
4.1 The series is always irrational under the stated hypothesis
Although transcendence is open, the stated growth assumption immediately rules out rationality.Proposition (Irrationality under $\limsup a_n/n=\infty$): If $a_1<a_2<\cdots$ and $\limsup_{n\to\infty} a_n/n=\infty$, then \[
\alpha=\sum_{n\ge 1} 2^{-a_n}
\] is irrational.
Proof: Write the (unique) binary expansion of $\alpha$ that does not terminate in an infinite tail of 1's. Because the summands are distinct dyadic rationals, we have a well-defined binary digit sequence $(\varepsilon_m)_{m\ge 1}\in\{0,1\}^{\mathbb{N}}$ with \[
\alpha = \sum_{m=1}^{\infty} \varepsilon_m 2^{-m},
\qquad \varepsilon_m = 1 \iff m\in\{a_n:n\ge 1\}.
\] (There are no carries because each position $m$ receives at most one contribution $2^{-m}$.)
Assume for contradiction that $\alpha\in\mathbb{Q}$. Then its binary expansion is eventually periodic: there exist integers $M\ge 1$ and $\ell\ge 1$ and bits $b_1,\dots,b_\ell\in\{0,1\}$, not all zero, such that for all $m\ge M$ one has \[ \varepsilon_m = b_{((m-M)\bmod \ell)+1}. \] Let $s\coloneqq b_1+\cdots+b_\ell\ge 1$ be the number of ones in one period. Then among the digits $\varepsilon_M,\dots,\varepsilon_{M+N-1}$, the number of ones is at least $\lfloor N/\ell\rfloor\,s\ge (s/\ell)N-1$.
Equivalently, for all sufficiently large $L$, the number of indices $m\le L$ with $\varepsilon_m=1$ is at least $(s/\ell)(L-M)-O(1)$, i.e. grows linearly in $L$ with positive slope.
Now $a_n$ is the position of the $n$th 1 in this binary expansion. Linear growth of the count of ones implies a linear upper bound on $a_n$: there exists a constant $K>0$ such that for all large $n$, \[ a_n \le K n. \] (For instance, take $K=2\ell/s$ and compare $n$ to the count of ones up to $L=Kn$.) Thus $\sup_n a_n/n <\infty$, contradicting $\limsup a_n/n=\infty$.
Therefore $\alpha\notin\mathbb{Q}$. ∎
4.2 A sufficient condition for transcendence (stronger than the hypothesis)
The following criterion shows where "Liouville-type" arguments become available.Proposition (Unbounded ratio implies Liouville): If \[
\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}=\infty,
\] then $\alpha=\sum_{n\ge 1}2^{-a_n}$ is a Liouville number, hence transcendental.
Proof: Let $p_N/2^{a_N}$ be the $N$-term truncation: \[
\alpha_N\coloneqq\sum_{n=1}^{N}2^{-a_n}=\frac{p_N}{2^{a_N}}\in\mathbb{Q}.
\] The tail satisfies \[
0<\alpha-\alpha_N = \sum_{n>N} 2^{-a_n} \le 2^{-a_{N+1}}\sum_{j\ge 0}2^{-j}=2\cdot 2^{-a_{N+1}}.
\] Let $q_N\coloneqq 2^{a_N}$. Then \[
\left|\alpha-\tfrac{p_N}{q_N}\right| \le 2\,2^{-a_{N+1}} = 2\,q_N^{-a_{N+1}/a_N}.
\] By hypothesis, for every integer $m\ge 1$ there are infinitely many $N$ with $a_{N+1}/a_N>m+1$. For such $N$ (and $q_N\ge 2$) we have \[
\left|\alpha-\tfrac{p_N}{q_N}\right| \le 2\,q_N^{-(m+1)} < q_N^{-m}.
\] Thus for every $m$ there are infinitely many rational approximations $p_N/q_N$ satisfying $|\alpha-p_N/q_N|<q_N^{-m}$, which is exactly the definition of a Liouville number. Hence $\alpha$ is transcendental. ∎
5) VERIFICATION
- The irrationality proof only uses the standard fact that rationals have eventually periodic base-$2$ expansions, and that eventual periodicity forces a positive asymptotic density of 1's in the tail if the number is not dyadic-rational.
- The "no carries" claim is justified because the summands $2^{-a_n}$ have disjoint binary positions.
- The Liouville criterion is a standard truncation argument; the key inequality is the geometric-series bound on the tail.
6) FINAL
UNRESOLVED.- Strongest fully proved partial result obtained here: Under the stated hypothesis $\limsup a_n/n=\infty$, the number $\alpha=\sum 2^{-a_n}$ is irrational (Proposition 4.1). Also, if $\limsup a_{n+1}/a_n=\infty$ then $\alpha$ is Liouville and hence transcendental (Proposition 4.2).
- First gap preventing a full solution: The hypothesis $\limsup a_n/n=\infty$ does not force Liouville-type approximations (ratios $a_{n+1}/a_n$ may stay close to $1$) and does not obviously force low digit-complexity. I do not see a mechanism that excludes algebraicity from this weak sparsity condition.
- Most promising next move: Try to connect the growth condition to quantitative lower bounds on the block complexity (or other automaticity/regularity properties) of the binary expansion of $\alpha$, and then invoke known transcendence criteria for expansions with "too simple" structure. One must also handle the possibility that the expansion is very complex on most scales while still having occasional large gaps.
- Smallest plausible counterexample (if false): An algebraic irrational whose binary expansion contains arbitrarily long stretches with unusually few 1's (so that the 1-positions $a_n$ satisfy $\limsup a_n/n=\infty$) would refute the conjecture. No explicit candidate is known to me.