Erdős Problem #264

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.

Problem 264: at most countably many exceptional factorial perturbations

1) OBJECTIVE AND SOURCE REVIEW

2) WORK

Thus two tails whose first differing digit occurs sufficiently late cannot have the same sum: the first difference dominates everything after it. Once the initial finite perturbation prefix is fixed, the map from infinite tails to sums is injective. There are only countably many rational numbers, so at most countably many such tails have rational sums. There are finitely many prefixes for each fixed $B$; taking the countable union over $B$ proves that at most countably many bounded nonzero integer perturbation sequences can yield a rational sum. In particular this applies to $a_n=n!$.

3) ADVERSARIAL VERIFICATION

4) FINAL

5) SOURCES CHECKED

6) COMPLETION ESTIMATE

Erdős Problem #264

1) FORMAL RESTATEMENT

The question asks whether the specific choices $a_n=2^n$ and $a_n=n!$ are Type 3 irrationality sequences.

2) QUICK LITERATURE/CONTEXT CHECK (only if browsing available)

• Kovač--Tao $2024/2025$ introduced a quantitative criterion implying a sequence is not Type 3; in particular they show $a_n=2^n$ fails to be Type 3 (indeed, there exists bounded $b_n$ for which the sum is rational). They also show any strictly increasing $a_n$ with $\limsup a_{n+1}/a_n<\infty$ is not Type 3, so "pure exponential" growth cannot work.
• In the opposite direction, Kovač--Tao prove that for any function $F$ with $F(n+1)/F(n)\to\infty$ there exists a Type 3 irrationality sequence with $a_n\sim F(n)$; in particular there exist Type 3 sequences asymptotic to $n!$, but this does not decide whether the exact sequence $a_n=n!$ itself is Type 3.

3) ATTACK PLAN

1. Disprove the Type 3 property for $a_n=2^n$ by constructing (or at least proving existence of) a bounded integer sequence $b_n$ such that \[\sum_{n\ge 1}\frac{1}{2^n+b_n}\in\mathbb{Q}. \] It suffices to produce one rational value.
2. Show the existence of such a $b_n$ by an "overlapping-interval"/digit-expansion argument: allow $b_n\in\{1,2,3,4,5\}$, prove the set of achievable sums is an interval, then pick a rational point inside that interval.
3. For $a_n=n!$, summarize what is known and why the same interval method fails for fixed bounded $b_n$ (the gaps between $1/(n!+b)$ values dwarf the tail width). Record the open part.

4) WORK (with full details)

4.1. A concrete counterexample for $a_n=2^n$ (bounded perturbations can give a rational sum)

Fix $B=5$. For $n\ge 1$ define the "tail bounds" \[ L_n := \sum_{k=n}^{\infty}\frac{1}{2^k+B},\qquad U_n := \sum_{k=n}^{\infty}\frac{1}{2^k+1}, \] so $0<L_n\le U_n<\infty$ and $L_n,U_n\to 0$ as $n\to\infty$.

For each $n\ge 1$ and each $b\in\{1,2,3,4,5\}$ set \[ I_{n}(b):=\left[\frac{1}{2^n+b}+L_{n+1},\ \frac{1}{2^n+b}+U_{n+1}\right]. \] Intuitively, if we choose the $n$th "digit" $b_n=b$, then the remaining tail sum lies in $[L_{n+1},U_{n+1}]$, so the total sum lies in $I_n(b)$.

We now exhibit a rational $x\in[L_1,U_1]$.

4.2. The case $a_n=n!$

The above interval-overlap method is tailored to exponential denominators: the "digit gaps" \[ \frac{1}{2^n+b}-\frac{1}{2^n+(b+1)}\asymp 4^{-n} \] are of the same order as the tail flexibility $U_{n+1}-L_{n+1}\asymp 4^{-n}$.

For $a_n=n!$, if we again restricted to a fixed bounded range $b\in\{1,\dots,B\}$, then the digit gaps are on the order of $1/(n!)^2$, but the tail flexibility is on the order of \[ \sum_{k\ge n+1}\frac{1}{(k!)^2}\asymp \frac{1}{((n+1)!)^2}=\frac{1}{(n+1)^2}\cdot\frac{1}{(n!)^2}, \] which is smaller by a factor $(n+1)^{-2}$. Thus no fixed $B$ can yield the kind of uniform overlap we used above.

At present, the exact Type 3 status of $a_n=n!$ is not settled by the argument above. What is known (from the general existence theorem mentioned in the literature check) is that one can construct Type 3 sequences $(a_n)$ with $a_n\sim n!$ (even with $a_n=n!+o(n!)$), but this does not imply that the unperturbed factorial sequence itself has the Type 3 property.

5) VERIFICATION / SANITY CHECK

• All series considered with denominators $2^n+b_n$ and bounded $b_n$ converge absolutely, since $2^n+b_n\asymp 2^n$.
• The constructed $b_n$ satisfy the constraints $b_n\neq 0$ and $2^n+b_n\neq 0$ automatically (here $b_n\in\{1,2,3,4,5\}$).
• The overlap inequality was reduced to an explicit polynomial inequality $1+6x+17x^2\ge 0$, so no hidden analytic assumptions remain.

6) FINAL

UNRESOLVED.

• (i) What is proved here: $a_n=2^n$ is not Type 3, by exhibiting a bounded choice of $b_n\in\{1,\dots,5\}$ for which $\sum_{n\ge 1}\frac{1}{2^n+b_n}=\tfrac{3}{4}\in\mathbb{Q}$.
• (ii) What remains open: whether $a_n=n!$ is Type 3 (i.e., whether $\sum\frac{1}{n!+b_n}$ is irrational for every bounded integer perturbation $b_n\neq 0$ with $n!+b_n\neq 0$).
• (iii) Most plausible next step: develop methods that exploit the arithmetic structure of $n!$ (e.g., strong divisibility and factorial-base expansions) to force irrationality for all bounded perturbations; the interval-overlap method above cannot work uniformly for fixed bounded $b_n$.
• (iv) Completion estimate: resolving the factorial case likely requires a new invariant or approximation argument; at minimum, one would need either (a) a universal irrationality criterion that applies to all bounded perturbations of $n!$, or (b) an explicit bounded perturbation producing a rational sum.

7) COMPLETION ESTIMATE

For the remaining $a_n=n!$ case, one would need to either:

1. produce an explicit bounded $b_n$ with $\sum 1/(n!+b_n)\in\mathbb{Q}$ (a counterexample), or
2. prove a uniform irrationality criterion that works against every bounded perturbation of $n!$.