gpt pro 5.4 attempt on Math-736

Erdős Problem #264 (Round 2)

1. ROUND-2 OBJECTIVE

I pursue (C) obstruction/correction. Round 1 already gave a negative answer for $a_n=2^n$, but the factorial case remained open. The most promising genuine advance is to (i) sharpen the $2^n$ construction, and (ii) prove a structural theorem showing that for every superexponential skeleton, and in particular for $a_n=n!$, the set of all bounded-perturbation sums is extremely thin: it has Hausdorff dimension $0$. This gives a rigorous barrier against any "interval-filling" strategy in the factorial case.

2. Round-1 FOUNDATION USED

I use the following Round 1 facts.

1. Round 1 already established that $a_n=2^n$ is not Type 3 by an interval-overlap argument with perturbations $b_n\in\{1,2,3,4,5\}$.
2. Round 1 observed heuristically that the same interval method cannot work uniformly for $a_n=n!$, because one-step gaps are of order $(n!)^{-2}$ while the remaining tail flexibility is only of order $((n+1)!)^{-2}$.
3. Round 1 left the exact factorial question unresolved.

I will reuse the Round 1 coding idea for $2^n$, but with a sharper overlap estimate, and I will replace the factorial heuristic by a theorem.

3. NEW INSIGHT / TOOL (ROUND-2)

There are two genuinely new ingredients in this round.

1. The Round 1 bound $b_n\in\{1,2,3,4,5\}$ for the $2^n$ counterexample is not optimal for that argument: one can already work with \[ b_n\in\{1,2,3,4\}. \]
2. More importantly, for every strictly increasing positive integer sequence $(a_n)$ with \[ \frac{a_{n+1}}{a_n}\to\infty, \] and every fixed perturbation bound $B$, the set of values \[ \sum \frac{1}{a_n+b_n},\qquad |b_n|\le B, \] has Hausdorff dimension $0$. In particular, the full set of bounded-perturbation sums for $a_n=n!$ is a countable union of Hausdorff-dimension-zero sets.

This is a strict advance beyond Round 1: it turns the earlier "the overlap method should fail" heuristic into a precise geometric obstruction.

4. ATTACK PLAN (ROUND-2)

The exact gaps left after Round 1 were:

1. the Round 1 $2^n$ counterexample was existence-only with perturbation bound $5$;
2. the factorial case had no rigorous structural theorem, only a size comparison heuristic.

I now prove the following two claims.

Claim A. For $a_n=2^n$, perturbations in $\{1,2,3,4\}$ already suffice to realize the rational value $3/4$.

Claim B. If $a_{n+1}/a_n\to\infty$, then every fixed bounded perturbation family gives a value set of Hausdorff dimension $0$; hence for $a_n=n!$ the bounded-perturbation value set has empty interior.

These two claims together sharpen the negative half of #264 and isolate a real obstruction in the positive half.

5. WORK (ROUND-2)

5.1. Sharpening the $2^n$ counterexample from bound $5$ to bound $4$.

Define \[ L_n:=\sum_{k=n}^{\infty}\frac{1}{2^k+4}, \qquad U_n:=\sum_{k=n}^{\infty}\frac{1}{2^k+1}, \] and, for $b\in\{1,2,3,4\}$, \[ I_n(b):=\left[\frac{1}{2^n+b}+L_{n+1},\ \frac{1}{2^n+b}+U_{n+1}\right]. \] As in Round 1, $I_n(b)$ is the set of all possible total sums once the $n$th perturbation is fixed to be $b$.

Lemma 5.1. For every $n\ge 1$ one has \[ \bigcup_{b=1}^4 I_n(b)=[L_n,U_n]. \]

Proof. The endpoints agree: \[ \min I_n(4)=\frac{1}{2^n+4}+L_{n+1}=L_n, \qquad \max I_n(1)=\frac{1}{2^n+1}+U_{n+1}=U_n. \] So it suffices to prove overlap of consecutive intervals. Because $b\mapsto 1/(2^n+b)$ is decreasing, it is enough to show that \[ U_{n+1}-L_{n+1}\ge \frac{1}{2^n+b}-\frac{1}{2^n+b+1} \] for $b=1,2,3$; the right-hand side is largest at $b=1$, so it is enough to prove \[ U_{n+1}-L_{n+1}\ge \frac{1}{(2^n+1)(2^n+2)}. \] Now \[ U_{n+1}-L_{n+1} =\sum_{k=n+1}^{\infty}\left(\frac{1}{2^k+1}-\frac{1}{2^k+4}\right) =\sum_{k=n+1}^{\infty}\frac{3}{(2^k+1)(2^k+4)}. \] Write $x:=2^{-(n+1)}$. For every $k\ge n+1$ we have $2^{-k}\le x$, hence \[ (2^k+1)(2^k+4) =2^{2k}(1+2^{-k})(1+4\cdot 2^{-k}) \le 2^{2k}(1+x)(1+4x), \] so \[ \frac{3}{(2^k+1)(2^k+4)} \ge \frac{3\cdot 4^{-k}}{(1+x)(1+4x)}. \] Summing over $k\ge n+1$ gives \[ U_{n+1}-L_{n+1} \ge \frac{3}{(1+x)(1+4x)}\sum_{k=n+1}^{\infty}4^{-k} =\frac{4^{-n}}{(1+x)(1+4x)}. \] On the other hand, \[ \frac{1}{(2^n+1)(2^n+2)} =\frac{4^{-n}}{(1+2x)(1+4x)}. \] Since $1+x\le 1+2x$, the lower bound above is at least this target quantity. Thus all consecutive intervals overlap, and the union is exactly $[L_n,U_n]$. $\square$

Lemma 5.2. Every $x\in [L_1,U_1]$ can be written in the form \[ x=\sum_{n=1}^{\infty}\frac{1}{2^n+b_n} \qquad\mbox{with } b_n\in\{1,2,3,4\}. \]

Proof. Exactly as in Round 1: recursively choose $b_n$ so that the current remainder lies in one of the intervals $I_n(b_n)$, and use Lemma 5.1 at each step. The tail remainders tend to $0$ because $U_n\to 0$. $\square$

Lemma 5.3. The rational number $3/4$ belongs to $[L_1,U_1]$.

Proof. For the lower endpoint, \[ L_1=\sum_{n=1}^{\infty}\frac{1}{2^n+4} \le \frac16+\sum_{n=2}^{\infty}\frac{1}{2^n} =\frac16+\frac12=\frac23<\frac34. \] For the upper endpoint, \[ U_1\ge \sum_{n=1}^{7}\frac{1}{2^n+1} =\frac{3559541}{4703985}>\frac34. \] Hence $3/4\in[L_1,U_1]$. $\square$

Corollary 5.4. There exists a bounded sequence $b_n\in\{1,2,3,4\}$ such that \[ \sum_{n=1}^{\infty}\frac{1}{2^n+b_n}=\frac34. \] Therefore $a_n=2^n$ is not a Type 3 irrationality sequence.

This improves the explicit perturbation bound from Round 1.

5.2. A general thin-set theorem for bounded perturbations of superexponential sequences.

Let $(a_n)$ be a strictly increasing sequence of positive integers with \[ \frac{a_{n+1}}{a_n}\to\infty. \] Fix $B\ge 1$, and let \[ D_B:=\{-B,-B+1,\dots,-1,1,\dots,B-1,B\}. \] Choose $n_0=n_0(B)$ so large that \[ a_n>2B \qquad\mbox{and}\qquad a_{n+1}\ge 2a_n \quad\mbox{for all } n\ge n_0. \] Define the tail value set \[ E_{B}(a;n_0):= \left\{ \sum_{n=n_0}^{\infty}\frac{1}{a_n+b_n} :\ b_n\in D_B\ \mbox{for all }n\ge n_0 \right\}. \]

Lemma 5.5. If the digits $b_{n_0},\dots,b_M$ are fixed, then all sums in $E_B(a;n_0)$ with this prefix lie in an interval of length at most \[ W_M:=\sum_{k=M+1}^{\infty} \left(\frac{1}{a_k-B}-\frac{1}{a_k+B}\right) \le \frac{32B}{9a_{M+1}^2}. \]

Proof. For $k\ge n_0$, \[ \frac{1}{a_k-B}-\frac{1}{a_k+B} =\frac{2B}{a_k^2-B^2} \le \frac{2B}{a_k^2-a_k^2/4} =\frac{8B}{3a_k^2}, \] because $a_k>2B$. Hence \[ W_M\le \frac{8B}{3}\sum_{k=M+1}^{\infty}\frac{1}{a_k^2}. \] Since $a_{k+1}\ge 2a_k$ for $k\ge n_0$, the tail of $(a_k^{-2})$ is dominated by a geometric series with ratio at most $1/4$: \[ \sum_{k=M+1}^{\infty}\frac{1}{a_k^2} \le \frac{1}{a_{M+1}^2}\sum_{j=0}^{\infty}4^{-j} =\frac{4}{3a_{M+1}^2}. \] Combining these estimates gives the claim. $\square$

Theorem 5.6. For every fixed $B\ge 1$, \[ \dim_{\rm H} E_B(a;n_0)=0. \] Consequently the full bounded-perturbation value set \[ E(a):= \left\{ \sum_{n=1}^{\infty}\frac{1}{a_n+b_n} :\ (b_n)\ \mbox{is a bounded integer sequence},\ b_n\ne 0,\ a_n+b_n\ne 0 \right\} \] is a countable union of sets of Hausdorff dimension $0$, and in particular has Hausdorff dimension $0$ and empty interior.

Proof. At level $M\ge n_0$, the set $E_B(a;n_0)$ is covered by at most \[ (2B)^{M-n_0+1} \] intervals, one for each choice of the prefix $(b_{n_0},\dots,b_M)$, and Lemma 5.5 shows that every such interval has length at most $W_M\le 32B/(9a_{M+1}^2)$.

Fix $s>0$. Choose a real number $\lambda>(2B)^{1/(2s)}$. Since $a_{n+1}/a_n\to\infty$, we have $a_{n+1}\ge \lambda a_n$ for all sufficiently large $n$, hence for some constant $C=C(s,B)>0$, \[ a_{M+1}\ge C\lambda^M \qquad\mbox{for all large }M. \] Therefore the total $s$-content of the above cover is bounded by \[ (2B)^{M-n_0+1}\left(\frac{32B}{9a_{M+1}^2}\right)^s \le C'\left(\frac{2B}{\lambda^{2s}}\right)^M, \] which tends to $0$ because $\lambda^{2s}>2B$. Since this holds for every $s>0$, the Hausdorff dimension is $0$.

Now fix $B$ and consider all bounded perturbation sequences with $|b_n|\le B$. The terms with $n<n_0(B)$ contribute only finitely many possibilities, so this larger set is a finite union of translates of $E_B(a;n_0(B))$ and still has Hausdorff dimension $0$. Finally, \[ E(a)=\bigcup_{B=1}^{\infty} E_B(a) \] is a countable union of Hausdorff-dimension-zero sets, so it also has Hausdorff dimension $0$. $\square$

Corollary 5.7. For $a_n=n!$, the full set of all sums \[ \sum_{n=2}^{\infty}\frac{1}{n!+b_n} \qquad ((b_n)\ \mbox{bounded},\ b_n\ne 0,\ n!+b_n\ne 0) \] has Hausdorff dimension $0$.

Proof. Here $a_{n+1}/a_n=n+1\to\infty$, so Theorem 5.6 applies. $\square$

Thus the factorial case has a rigorous geometric obstruction: unlike the $2^n$ case, bounded perturbations of $n!$ do not generate intervals at all.

6. ADVERSARIAL VERIFICATION

1. Small-index issues. In Theorem 5.6 I started at $n_0(B)$ precisely to avoid accidental zeros or sign changes in $a_n+b_n$ and to guarantee $a_n>2B$ and $a_{n+1}\ge 2a_n$.
2. Countable union issue. The full bounded-perturbation set is not itself a single fixed-$B$ set, so I explicitly decomposed it as a countable union over $B$ and finite initial prefixes.
3. No contradiction with the $2^n$ half. The superexponential theorem requires $a_{n+1}/a_n\to\infty$, which fails for $a_n=2^n$. So there is no inconsistency with the interval phenomenon proved above for $2^n$.
4. Is Hausdorff dimension really enough to rule out the old strategy? Yes. Any set containing a non-degenerate interval has Hausdorff dimension $1$, so dimension $0$ completely forbids the interval-overlap mechanism.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 2 makes two substantive advances:

1. the negative $2^n$ result is sharpened to an explicit perturbation range $b_n\in\{1,2,3,4\}$;
2. the factorial side is no longer supported only by a heuristic obstruction: for every superexponential skeleton, and hence for $n!$, the bounded-perturbation value set has Hausdorff dimension $0$.

The exact Type 3 status of $a_n=n!$ remains open, but one now knows that any eventual resolution cannot come from an interval-filling argument of the sort that works for $2^n$.

8. COMPLETION ESTIMATE

COMPLETION: 62%

9. REFERENCES

1. V. Kovač and T. Tao, On several irrationality problems for Ahmes series, arXiv:2406.17593.