Erdős Problem #247

Erdős Problem #247 --- Round 2

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction.

Round 1 left the original statement open, but it also isolated the key gap: the only transcendence mechanism obtained there came from a very strong Liouville phenomenon, namely \[ \limsup_{n\to\infty}\frac{a_{n+1}}{a_n}=\infty. \] In this round I sharpen that gap in two different directions.

First, I prove a strictly stronger transcendence criterion: \[ \limsup_{n\to\infty}\frac{a_{n+1}}{a_n}>1 \quad\Longrightarrow\quad \alpha:=\sum_{n=1}^\infty 2^{-a_n}\ \text{is transcendental}. \] Second, I prove a degree obstruction: \[ \alpha\ \text{algebraic of degree }D \quad\Longrightarrow\quad a_n=O(n^D). \] These two results isolate the only possible location of any algebraic counterexample to the original Erdős problem: it would have to lie in the nonlacunary polynomial-growth regime \[ \frac{a_{n+1}}{a_n}\to 1, \qquad a_n=O(n^D)\ \text{for some finite }D\ge 2. \] That is the precise Round-2 obstruction/correction I will establish.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts without reproving them.

• Round-1 Proposition 4.1: under the original hypothesis \[ \limsup_{n\to\infty}\frac{a_n}{n}=\infty, \] we already know that \[ \alpha=\sum_{n\ge 1}2^{-a_n} \] is irrational.

• Round-1 truncation formula: for \[ \alpha_N:=\sum_{n=1}^N2^{-a_n}=\frac{p_N}{2^{a_N}}, \] we have \[ 0<\alpha-\alpha_N\le 2^{1-a_{N+1}}. \] (This was the tail estimate used in Round 1.)

• Round-1 gap diagnosis: the unresolved regime was exactly the one where the truncation denominators are dyadic, but the tail is not small enough to force a Liouville argument.

3) NEW INSIGHT / TOOL (ROUND-2)

There are two genuinely new inputs in this round.

• Dyadic Ridout bound. A consequence of Ridout's theorem, stated explicitly by Rivoal, is that if $\beta$ is an algebraic irrational number, then for every $\varepsilon>0$ one has \[ \left|\beta-\frac{p}{2^m}\right|\ge 2^{-m(1+\varepsilon)} \] for all integers $p$ and all sufficiently large integers $m$.

• Binary 1-bit lower bound for algebraic numbers. Bailey--Borwein--Crandall--Pomerance proved that if $\beta$ is a real algebraic number of degree $D>1$, then the number $B_\beta(N)$ of 1's among the first $N$ binary digits of $\beta$ satisfies \[ B_\beta(N)\ge c(\beta)N^{1/D} \] for all sufficiently large $N$, for some constant $c(\beta)>0$.

The first tool upgrades the Round-1 "Liouville if ratios are unbounded" argument to the much sharper threshold "transcendence if the ratio exceeds $1$ by any fixed amount infinitely often." The second tool converts algebraicity of fixed degree into a concrete upper bound on the growth of the 1-positions $a_n$.

4) ATTACK PLAN (ROUND-2)

After Round 1, the exact gap was this: the truncation estimate \[ 0<\alpha-\alpha_N\le 2^{1-a_{N+1}} \] only gives a Liouville contradiction when $a_{N+1}/a_N$ becomes arbitrarily large. That leaves open all sequences with bounded successive ratios.

I now attack that gap in three steps.

• Claim A. Strengthen Round 1 by proving \[ \limsup_{n\to\infty}\frac{a_{n+1}}{a_n}>1 \implies \alpha\ \text{transcendental}. \] This uses the dyadic Ridout estimate instead of a Liouville argument.

• Claim B. If $\alpha$ were algebraic of degree $D$, then its binary expansion would have at least $\gg N^{1/D}$ ones up to position $N$. Since for our number those 1-positions are exactly the $a_n$, this forces \[ a_n=O(n^D). \] This gives a new algebraic-degree barrier.

• Claim C. Show that the Round-2 dyadic-approximation method is sharp for truncations: if \[ \limsup_{n\to\infty}\frac{a_{n+1}}{a_n}=1, \] then the truncations $\alpha_N$ can never violate Ridout's dyadic lower bound for any fixed exponent $1+\varepsilon$. So the remaining unresolved regime is genuinely outside the reach of the simple truncation$+$Ridout strategy.

5) WORK (ROUND-2)

Let \[ A(N):=\#\{n\ge 1: a_n\le N\}. \] Because the binary expansion of $\alpha$ has a 1 exactly in positions $a_1,a_2,\dots$, the quantity $A(N)$ is exactly the number of 1's among the first $N$ binary digits of $\alpha$. In particular, \[ A(a_n)=n\qquad(n\ge 1). \]

Claim A: a Ridout upgrade of the Round-1 transcendence criterion

Proposition A. If \[ \Lambda:=\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}>1, \] then $\alpha$ is transcendental.

Proof.

I explicitly use two Round-1 facts here: first, $\alpha$ is irrational; second, \[ 0<\alpha-\alpha_N\le 2^{1-a_{N+1}} \qquad\text{for }\alpha_N=\frac{p_N}{2^{a_N}}. \]

Choose $\varepsilon>0$ such that \[ 1+2\varepsilon<\Lambda. \] Then for infinitely many $N$ we have \[ a_{N+1}\ge (1+2\varepsilon)a_N. \] For such $N$, \[ 0<\alpha-\frac{p_N}{2^{a_N}} \le 2^{1-a_{N+1}} \le 2^{1-(1+2\varepsilon)a_N} =2\cdot 2^{-(1+2\varepsilon)a_N}. \] For all sufficiently large such $N$, we also have $2<2^{\varepsilon a_N}$, hence \[ 2\cdot 2^{-(1+2\varepsilon)a_N}<2^{-(1+\varepsilon)a_N}. \] Therefore, for infinitely many $N$, \[ \left|\alpha-\frac{p_N}{2^{a_N}}\right|<2^{-(1+\varepsilon)a_N}. \]

Now assume for contradiction that $\alpha$ is algebraic. By the Round-1 irrationality result, it is then algebraic irrational. Applying Rivoal's dyadic form of Ridout's theorem to $\beta=\alpha$, we obtain that for this fixed $\varepsilon>0$, \[ \left|\alpha-\frac{p}{2^m}\right|\ge 2^{-m(1+\varepsilon)} \] for every integer $p$ and every sufficiently large integer $m$. Taking $m=a_N$ contradicts the infinitely many strict inequalities above.

Hence $\alpha$ is transcendental. $\square$

This strictly strengthens the Round-1 Liouville criterion \[ \limsup \frac{a_{n+1}}{a_n}=\infty, \] since the latter obviously implies the new hypothesis $\limsup a_{n+1}/a_n>1$.

Claim B: algebraic degree forces polynomial growth of the 1-positions

Proposition B. Assume that $\alpha$ is algebraic of degree $D>1$. Then there exists a constant $C=C(\alpha)>0$ such that \[ a_n\le Cn^D \qquad\text{for all sufficiently large }n. \] Equivalently, \[ A(N)\ge c(\alpha)N^{1/D} \qquad\text{for all sufficiently large }N, \] for some constant $c(\alpha)>0$.

Proof.

Bailey--Borwein--Crandall--Pomerance prove that for every real algebraic number $\beta$ of degree $D>1$, the number $B_\beta(N)$ of 1's among the first $N$ binary digits of $\beta$ satisfies \[ B_\beta(N)\ge c(\beta)N^{1/D} \] for all sufficiently large $N$.

Apply this with $\beta=\alpha$. Since the 1-positions of the binary expansion of $\alpha$ are exactly $a_1,a_2,\dots$, we have \[ B_\alpha(N)=A(N). \] Hence \[ A(N)\ge c(\alpha)N^{1/D} \] for all sufficiently large $N$.

Now substitute $N=a_n$. Since $A(a_n)=n$, we get \[ n\ge c(\alpha)a_n^{1/D} \] for all sufficiently large $n$, and therefore \[ a_n\le c(\alpha)^{-D}n^D. \] This is the required polynomial upper bound. $\square$

A useful reformulation is the following.

Corollary B1. Fix an integer $d\ge 2$. If \[ \limsup_{n\to\infty}\frac{a_n}{n^d}=\infty, \] then $\alpha$ is not algebraic of degree $\le d$.

Proof.

If $\alpha$ had algebraic degree $D\le d$, Proposition B would give \[ a_n=O(n^D)=O(n^d), \] contradicting $\limsup a_n/n^d=\infty$. $\square$

In particular:

• taking $d=2$, we obtain a rigorous nonquadraticity criterion: \[ \limsup_{n\to\infty}\frac{a_n}{n^2}=\infty \quad\Longrightarrow\quad \alpha\ \text{is not quadratic irrational}; \]
• if the condition $\limsup a_n/n^d=\infty$ holds for every $d\ge 1$, then $\alpha$ is transcendental.

The second bullet recovers the stronger sufficient condition already recorded on the Erdős problem page.

Claim C: the truncation$+$Ridout method is sharp at the threshold $a_{n+1}/a_n\to 1$

Proposition C. Assume \[ \limsup_{n\to\infty}\frac{a_{n+1}}{a_n}=1. \] Then for every $\varepsilon>0$, the truncation approximants $\alpha_N=p_N/2^{a_N}$ satisfy \[ \alpha-\frac{p_N}{2^{a_N}}\ge 2^{-(1+\varepsilon)a_N} \] for all sufficiently large $N$. In particular, the truncations never produce infinitely many dyadic approximations that violate Ridout's lower bound with exponent $1+\varepsilon$.

Proof.

Fix $\varepsilon>0$. Since $a_{n+1}/a_n\ge 1$ for every $n$ and the limsup is $1$, in fact \[ \frac{a_{n+1}}{a_n}\to 1. \] Hence, for all sufficiently large $N$, \[ a_{N+1}\le (1+\varepsilon)a_N. \] Now the truncation error is the positive tail \[ \alpha-\frac{p_N}{2^{a_N}}=\sum_{n>N}2^{-a_n}, \] so it is bounded below by its first omitted term: \[ \alpha-\frac{p_N}{2^{a_N}}\ge 2^{-a_{N+1}}\ge 2^{-(1+\varepsilon)a_N}. \] Thus for every fixed $\varepsilon>0$, the truncations fail to satisfy \[ \left|\alpha-\frac{p_N}{2^{a_N}}\right|<2^{-(1+\varepsilon)a_N} \] for all sufficiently large $N$. $\square$

This shows that Proposition A is sharp for the simple truncation strategy. The unresolved regime really begins when successive ratios tend to $1$.

A sharp reduction of the original Erdős problem

Combining Propositions A and B yields the most useful Round-2 reduction.

Corollary D. Suppose the original Erdős problem had an algebraic counterexample, i.e. suppose there exists an increasing sequence $(a_n)$ with \[ \limsup_{n\to\infty}\frac{a_n}{n}=\infty \] such that \[ \alpha=\sum_{n=1}^\infty 2^{-a_n} \] is algebraic. Then necessarily:

1. $\alpha$ has algebraic degree $D\ge 2$;
2. $a_{n+1}/a_n\to 1$;
3. $a_n=O(n^D)$.

Proof.

Item $1$ is exactly the Round-1 irrationality result. Item $2$ is the contrapositive of Proposition A, together with the trivial inequality $a_{n+1}/a_n\ge 1$. Item $3$ is Proposition B with $D=\deg(\alpha)$. $\square$

Thus the original problem reduces to the following strictly narrower subclass:

Any method that only exploits truncations as dyadic rational approximants, or only uses the power-law lower bound on the density of 1's coming from algebraicity of bounded degree, cannot by itself settle the full problem outside this reduced class.

6) ADVERSARIAL VERIFICATION

I now check the new work against the main failure modes.

• Did Proposition A accidentally assume more than Round 1 proved? No. It uses only two Round-1 inputs: irrationality of $\alpha$ and the dyadic truncation tail bound. Everything else is new.

• Does the Rivoal--Ridout bound apply in exactly the needed form? Yes. The cited statement is for every integer $p$ and all sufficiently large $m$, so there is no coprimality issue with the truncation numerators $p_N$.

• Could the constant factor $2$ in the tail estimate spoil Proposition A? No. After choosing $\varepsilon>0$ with $1+2\varepsilon<\Lambda$, one has \[ 2\cdot 2^{-(1+2\varepsilon)a_N}<2^{-(1+\varepsilon)a_N} \] whenever $a_N>1/\varepsilon$, which holds for all sufficiently large $N$.

• Does Proposition B really identify $A(N)$ with the binary 1-count? Yes. By construction, \[ \alpha=\sum_{n\ge 1}2^{-a_n} \] has a 1 in binary position $a_n$ and a 0 elsewhere; there are no carries because each power $2^{-m}$ appears at most once.

• Did I overstate Proposition C? I have stated it carefully: it rules out only the truncation$+$Ridout strategy, not all conceivable dyadic approximants or all possible transcendence methods.

• What happens in the boundary case $\limsup a_{n+1}/a_n=1$? Since the sequence is increasing, all ratios are at least $1$, so limsup $=1$ forces \[ \frac{a_{n+1}}{a_n}\to 1. \] Thus the boundary case is handled correctly.

• Is Corollary D logically stronger than what was proved? No. It is just the direct combination of Round-1 irrationality, Proposition A, and Proposition B.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Erdős problem remains open, but Round 2 gives three substantive advances beyond Round 1.

1. New transcendence criterion. I proved \[ \limsup_{n\to\infty}\frac{a_{n+1}}{a_n}>1 \quad\Longrightarrow\quad \sum_{n=1}^\infty 2^{-a_n}\ \text{is transcendental}. \] This strictly strengthens the Round-1 Liouville criterion.

2. New degree obstruction. If $\alpha$ is algebraic of degree $D$, then necessarily \[ a_n=O(n^D). \] Equivalently, if for some $d\ge 2$ one has \[ \limsup_{n\to\infty}\frac{a_n}{n^d}=\infty, \] then $\alpha$ cannot have algebraic degree $\le d$.

3. Sharp reduction of the open core. Any hypothetical algebraic counterexample to the original problem must satisfy \[ \frac{a_{n+1}}{a_n}\to 1 \qquad\text{and}\qquad a_n=O(n^D) \] for some finite $D\ge 2$. So the unresolved part of the problem has now been localized to the nonlacunary polynomial-growth regime.

The first genuine remaining gap is therefore no longer vague: one must handle sequences whose growth is superlinear in the Erdős sense, but whose successive ratios tend to $1$ and whose growth remains bounded by some fixed power of $n$.

8) COMPLETION ESTIMATE

COMPLETION: 45%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #247, Erdős Problems website, accessed 2026-03-07. The page records the problem statement, its open status, and Erdős' stronger sufficient condition.

2. T. Rivoal, On the binary expansion of irrational algebraic numbers, Actes des Rencontres du C.I.R.M. 1 $2009$, no. 1, 55--60. Used for the dyadic consequence of Ridout's theorem, especially Eq. $2.5$.

3. D. H. Bailey, J. M. Borwein, R. E. Crandall, and C. Pomerance, On the binary expansions of algebraic numbers, Journal de Théorie des Nombres de Bordeaux 16 $2004$, no. 3, 487--518. Used for Theorem 7.1 on lower bounds for the number of 1's in the binary expansion of an algebraic number.

Problem 247: periodic approximants inside the nonlacunary regime

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. The target is transcendence of $\alpha=\sum_n2^{-a_n}$ for increasing positive integers with $\limsup a_n/n=\infty$. I read gpt_pro_5.4/247.tex in full. Its Ridout and digit-count deductions leave nonlacunary polynomial growth. I give a concrete transcendental family within that surviving regime, using periodic approximants rather than truncations. This is an example, not a proof for every such sequence.

2) WORK

Put $L_j=2^{2^j}$ for $j\ge1$. After any fixed initial binary prefix of length $L_1$, fill positions $L_{j-1}+1,\ldots,L_j$ for each $j\ge2$ by repetitions of the length-$j$ word $0^{j-1}1$, truncating the last repetition if necessary. Let $a_n$ be the positions of the ones.

The ones have density zero. Indeed, for any fixed $J$, the stages after $J$ contribute at most $X/J+O(\log\log X)$ ones up to position $X$, while the first $J$ stages contribute a constant. Let $X\to\infty$ and then $J\to\infty$. Thus $a_n/n\to\infty$. There are infinitely many ones, so the expansion is not eventually periodic: any eventually periodic binary sequence with infinitely many ones has positive one-density. Hence $\alpha$ is irrational.

Inside stage $j$ gaps between ones are $j$; across its boundary they are at most $2j+2$. Their positions are at least $L_{j-1}-j$, so $a_{n+1}/a_n\to1$. Also $j=O(\log\log a_n)$ and the number of ones up to $a_n$ is at least $a_n/j-O(j+L_1)$: in each preceding stage its period is at most $j$, with at most one lost incomplete period per stage. Thus $a_n=O(n\log\log a_n)$, in particular $a_n=O(n^2)$.

Let $R_j$ retain the first $L_{j-1}$ digits and then repeat the same length-$j$ word forever. It is rational with reduced denominator $q_j\le2^{L_{j-1}}(2^j-1)$ and \[ 0<|\alpha-R_j|\le2^{-L_j}. \] For every fixed $D$, $L_j-D(L_{j-1}+j)\to\infty$. These approximations contradict algebraicity of degree $D$: if $P\in\mathbb Z[x]$ is a minimal polynomial of such an $\alpha$, then $P(R_j)\ne0$ and $|P(R_j)|\ge q_j^{-D}$; the mean-value bound on a fixed neighborhood gives $|P(R_j)|\le C|R_j-\alpha|$. Therefore $|R_j-\alpha|\ge C^{-1}q_j^{-D}$, incompatible with the displayed estimates. The constructed number is transcendental.

3) ADVERSARIAL VERIFICATION

The upper denominator estimate is sufficient in the Liouville comparison and remains valid after reduction. Irrationality excludes equality with any periodic approximant. Polynomial growth and consecutive-position ratios tending to one show that neither of the supplied sufficient conditions alone explains this family. The imposed long repetitions are an additional structural assumption absent from the original question.

4) FINAL

UNRESOLVED. This handles an explicit subclass within the previously isolated regime. A general sparse binary sequence need not contain these long periodic blocks, so the universal transcendence assertion remains unproved.

5) SOURCES CHECKED

Complete supplied version, read 2026-09-23. The additional construction and elementary polynomial approximation obstruction are proved directly; the supplied deep external theorems are not invoked as premises here.

6) COMPLETION ESTIMATE

Subjective progress toward the universal sparse-digit assertion. COMPLETION: 15%

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:

1. 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 $a_{n+1}/a_n$.
2. 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.
3. 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.

4.2 A sufficient condition for transcendence (stronger than the hypothesis)

The following criterion shows where "Liouville-type" arguments become available.

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.

1. 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).
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.
3. 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.
4. 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.

7) COMPLETION ESTIMATE

Estimated likelihood of completion with additional work: Moderate-to-low ($\approx 20\%$). The problem seems to require either a new transcendence criterion tailored to sparse digit sets with weak hypotheses, or new information about possible sparsity patterns in algebraic base-$2$ expansions.