gpt pro 5.4 attempt on Math-729

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.