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ERDőS #265 · PARTIAL

Erdős problem #265 — wave w039

Accessed and worked on 2026-07-29 (UTC).

Claim labels

Result of this run

[A] Theorem (verified sharpening of the available block construction). Put

\[ \gamma=\sqrt{\frac32}=1.224744871391\ldots. \]

There is a strictly increasing sequence of integers \(a_1<a_2<\cdots\) for which

\[ \sum_{n\geq1}\frac1{a_n}\in\mathbb Q,\qquad \sum_{n\geq1}\frac1{a_n-1}\in\mathbb Q, \]

and

\[ \liminf_{n\to\infty}a_n^{\,1/\gamma^n}>1. \]

Consequently, for every \(1<\beta<\sqrt{3/2}\),

\[ \lim_{n\to\infty}a_n^{\,1/\beta^n}=\infty. \]

[B] Comparison with the cited result. Equation (7.9) in Kovač--Tao gives, when \(d=2\), every

\[ 1<\beta<\sqrt{\frac65}=1.095445115010\ldots. \]

Thus the argument below raises the proved per-term exponent from the cited proof's \(\sqrt{6/5}\) to the endpoint \(\sqrt{3/2}\). I found no source in the search described below that states this sharpening, but that search is not a proof of novelty.

This does not solve the live remaining question, whose exponent is \(2\).

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Step 0: mandatory live-page check

Exact live statement

I fetched the rendered page through the Bright Data browser route, then separately fetched its discussion thread. The exact statement displayed on the page was:

Let \(1 \leq a_1<a_2<\cdots\) be an increasing sequence of integers. How fast can \(a_n\to\infty\) grow if \[ > \sum\frac1{a_n}\quad\text{and}\quad\sum\frac1{a_n-1} > \] are both rational?

Source: Erdős Problems #265.

Status and collision markers

Therefore the requested skip condition was not triggered.

All two comments

The rendered discussion thread contained exactly two comments.

  1. [B] Vjeko_Kovac, 19 January 2026: the problem needs strict monotonicity. He explains that this is implicit in Erdős--Graham, appears as at least non-decreasing in Erdős's 1985 memorial paper, and was assumed in Kovač--Tao. Without monotonicity, rearranging any valid sequence to place an extremely fast subsequence at selected indices would trivialize the limsup question. The comment records that the site was updated.
  2. [B] DanielChin, 5 March 2026: the earlier link to the 1985 paper had broken; the working link is 1985-10.pdf.

Neither comment claims a proof or says that anyone is currently working on the problem.

Known results shown on the live page

\[ \sum_{n=3}^{\infty}\frac1{\binom n2} =\sum_{n=3}^{\infty}\frac2{n(n-1)}=1, \] and \[ \sum_{n=3}^{\infty}\frac1{\binom n2-1} =\sum_{n=3}^{\infty}\frac2{(n-2)(n+1)}=\frac{11}{9}. \]

\(p(n)=n^3+6n^2+5n\): both \(\sum_{n\ge2}1/p(n)\) and \(\sum_{n\ge2}1/(p(n)-12)\) are rational.

\[ \limsup_{n\to\infty}a_n^{1/2^n}>1 \] can occur.

\(\sum1/a_n\) is irrational if \(a_n^{1/2^n}\to\infty\).

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Primary-source audit

Original formulations

Current cited construction

\[ 1<\beta< \left(\frac{2d+2}{2d+1}\right)^{1/d}. \] For the two sums relevant here, \(d=2\), so this is \(\beta<\sqrt{6/5}\).

Post-2024 search and honest misses

I searched the exact shifted-sum phrase, the Kovač--Tao title and citations, the \(1/2^n\) limsup, “simultaneous rationality,” and Erdős problem #265.

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Proof of the sharpening

1. A rational change of coordinates

Let \(b_n=a_n-1\). Define

\[ r=\sum_n\frac1{b_n},\qquad t=\sum_n\frac1{b_n(b_n+1)}. \]

[A] Term by term,

\[ \frac1{b}-\frac1{b(b+1)}=\frac1{b+1}. \]

Thus it is enough to make \(r,t\in\mathbb Q\), because then

\[ \sum_n\frac1{a_n-1}=r,\qquad \sum_n\frac1{a_n}=r-t. \]

For real \(x,y>0\), put

\[ \Phi(x,y)= \left( \frac1x+\frac1y,\, \frac1{x(x+1)}+\frac1{y(y+1)} \right). \]

One block will consist of two integer denominators, one near \(N\) and one near \(2N\).

2. Quantitative nonlinear block lemma

Fix

\[ \varepsilon=\frac1{20000}. \]

For an integer \(N\ge1000\), define

\[ s_N=\Phi(N,2N), \]
\[ S_N=\left\{\Phi(p,q): \begin{array}{l} p,q\in\mathbb Z,\\ |p-N|\le N/100,\\ |q-2N|\le N/100 \end{array}\right\}, \]
\[ R_N= \left[-\frac{\varepsilon}{N},\frac{\varepsilon}{N}\right] \times \left[-\frac{\varepsilon}{N^2},\frac{\varepsilon}{N^2}\right], \]

and

\[ E_N= \left[-\frac4{N^2},\frac4{N^2}\right] \times \left[-\frac4{N^3},\frac4{N^3}\right]. \]

[A] Block lemma.

\[ s_N+R_N\subseteq S_N+E_N. \tag{1} \]
Proof

Scale the variables and the two output coordinates:

\[ H_N(X,Y)= \left( N\Phi_1(NX,NY),\, N^2\Phi_2(NX,NY) \right). \]

Writing \(\eta=1/N\),

\[ H_N(X,Y)= \left( \frac1X+\frac1Y,\, \frac1{X(X+\eta)}+\frac1{Y(Y+\eta)} \right). \]

As \(N\to\infty\), this converges in \(C^1\), locally, to

\[ H_\infty(X,Y)= \left(\frac1X+\frac1Y,\frac1{X^2}+\frac1{Y^2}\right). \]

At \(z_0=(1,2)\),

\[ A:=DH_\infty(z_0)= \begin{pmatrix} -1&-1/4\\ -2&-1/4 \end{pmatrix}, \qquad A^{-1}= \begin{pmatrix} 1&-1\\ -8&4 \end{pmatrix}. \tag{2} \]

The determinant is \(-1/4\), and the row-sum norm of \(A^{-1}\) is \(12\).

Work on the closed box

\[ B=z_0+[-10^{-3},10^{-3}]^2. \]

For \(N\ge1000\), direct one-variable estimates give the following uniform bounds for

\[ E(z)=DH_N(z)-A: \]
\[ |E_{11}|<\frac1{400},\quad |E_{12}|<\frac1{3000},\quad |E_{21}|<\frac1{100},\quad |E_{22}|<\frac1{1000}. \tag{3} \]

For completeness, the estimates used are:

\[ |X^{-2}-1| \le \frac{\rho(2+\rho)}{(1-\rho)^2}, \quad |Y^{-2}-1/4| \le \frac{2\rho}{(2-\rho)^3}, \]

where \(\rho=1/1000\), and, for \(c=1,2\),

\[ \left| \frac{2u+\eta}{u^2(u+\eta)^2}-\frac2{c^3} \right| \le \frac{6\rho}{(c-\rho)^4} + \frac{\eta(3(c+\rho)+2\eta)}{(c-\rho)^5}. \]

These are just the mean-value theorem plus

\[ \frac2{u^3} - \frac{2u+\eta}{u^2(u+\eta)^2} = \frac{\eta(3u+2\eta)}{u^3(u+\eta)^2}. \]

Multiplying the coarse bounds (3) by (2) gives, row by row,

\[ \|A^{-1}E\|_\infty < \max\left\{ \frac1{400}+\frac1{3000}+\frac1{100}+\frac1{1000},\, 8\left(\frac1{400}+\frac1{3000}\right) +4\left(\frac1{100}+\frac1{1000}\right) \right\} =\frac1{15}<\frac1{10}. \]

Using the underlying exact fractional bounds rather than their coarse roundings, the verifier obtains the slightly tighter value

\[ \sup_{z\in B}\|A^{-1}(DH_N(z)-A)\|_\infty \le 0.056475096428\ldots<\frac1{10}. \tag{4} \]

For any \(\delta\in[-\varepsilon,\varepsilon]^2\), consider

\[ T_\delta(z) =z-A^{-1}\bigl(H_N(z)-H_N(z_0)-\delta\bigr). \]

By (4), \(T_\delta\) contracts distances on \(B\) by a factor less than \(1/10\). Also,

\[ \|T_\delta(z_0)-z_0\|_\infty \le12\varepsilon=\frac{12}{20000}, \]

and hence, for \(z\in B\),

\[ \|T_\delta(z)-z_0\|_\infty < \frac1{10}\frac1{1000}+\frac{12}{20000} =\frac7{10000}<\frac1{1000}. \]

Thus \(T_\delta\) maps \(B\) into itself. Iterating it gives a Cauchy sequence (successive differences shrink geometrically), whose limit \(z\in B\) is a fixed point. Therefore

\[ H_N(z)=H_N(z_0)+\delta. \tag{5} \]

This paragraph is also a proof of the only fixed-point fact used here.

Write \(z=(X,Y)\), and round \(NX,NY\) independently to nearest integers \(p,q\). Every point along either rounding segment is at least

\[ \frac{1997}{2000}N. \]

The mean-value theorem gives

\[ \left|\Delta\Phi_1\right| \le \frac{1.003007}{N^2}<\frac4{N^2} \]

and, because

\[ \left|\frac{d}{du}\frac1{u(u+1)}\right| =\frac{2u+1}{u^2(u+1)^2}\le\frac3{u^3}, \]
\[ \left|\Delta\Phi_2\right| \le \frac{3.013541}{N^3}<\frac4{N^3}. \tag{6} \]

The rounded \(p,q\) lie in the wider \(1\%\) windows defining \(S_N\). Rescaling (5) and using (6) proves (1). \(\square\)

3. Nesting the blocks

[A] Suppose a scale sequence satisfies, from some point onward,

\[ \frac4{N_k^2}\le\frac{\varepsilon}{N_{k+1}}, \qquad \frac4{N_k^3}\le\frac{\varepsilon}{N_{k+1}^2}. \tag{7} \]

Then \(E_{N_k}\subseteq R_{N_{k+1}}\), so the block lemma gives

\[ s_{N_k}+R_{N_k}\subseteq S_{N_k}+R_{N_{k+1}}. \tag{8} \]

For instance, if

\[ \log N_k\asymp\alpha^k \]

with \(1<\alpha<3/2\), then (7) eventually holds because

\[ \frac{N_{k+1}}{N_k^2}\to0, \qquad \frac{N_{k+1}^2}{N_k^3}\to0. \]

Since a block contributes two sequence terms, any \(\beta\) with \(\beta^2<\alpha\) then satisfies \(a_n^{1/\beta^n}\to\infty\).

The endpoint \(\alpha=3/2\) is also attainable by inserting a small fixed coefficient. The following exact choice will be used:

\[ N_0=80000^4=40960000000000000000, \qquad N_{k+1}=\left\lfloor\frac{N_k^{3/2}}{400}\right\rfloor. \tag{9} \]

For every \(k\),

\[ 80000N_{k+1}\le N_k^2,\qquad 80000N_{k+1}^2\le N_k^3. \tag{10} \]

Indeed, the second follows from \(80000/400^2=1/2\), while the first follows from \(\sqrt{N_k}\ge200\). Since \(4/\varepsilon=80000\), (10) is exactly (7).

Also, once the argument of the floor is at least \(2\),

\[ N_{k+1}\ge\frac{N_k^{3/2}}{800}. \]

At \(N_0\) this gives

\[ \frac{N_{k+1}}{N_k}\ge\frac{\sqrt{N_k}}{800}>45000, \tag{11} \]

and the lower bound only improves. Hence the integer windows are strictly ordered: every integer near \(N_k\) is below every integer near \(2N_k\), which is below every integer near \(N_{k+1}\).

4. An exact rational target

Let

\[ C=\sum_{k=0}^{\infty}s_{N_k}. \]

The ratios (11) imply the coordinate-wise tail estimates

\[ \sum_{k=1}^{\infty}(s_{N_k})_1 \le\frac9{4N_1} \le\frac{\varepsilon}{N_0}, \]
\[ \sum_{k=1}^{\infty}(s_{N_k})_2 \le\frac{45}{32N_1^2} \le\frac{\varepsilon}{N_0^2}. \tag{12} \]

Here the first geometric ratio was weakened to \(1/3\), and the second to \(1/9\); the actual ratio is far smaller.

Therefore the explicit rational point

\[ x=s_{N_0} \]

belongs to \(C+R_{N_0}\).

[A] Nested selection. Start with

\[ x=C+r_0,\qquad r_0\in R_{N_0}. \]

If blocks \(y_0,\ldots,y_{k-1}\) have been chosen so that

\[ x=\sum_{j<k}y_j+\sum_{j\ge k}s_{N_j}+r_k, \qquad r_k\in R_{N_k}, \]

then (8) selects \(y_k\in S_{N_k}\) and \(r_{k+1}\in R_{N_{k+1}}\) with

\[ s_{N_k}+r_k=y_k+r_{k+1}. \]

The invariant continues. The rectangles \(R_{N_k}\) shrink to a point, so passing to the limit yields

\[ x=\sum_{k=0}^{\infty}y_k. \tag{13} \]

Write \(y_k=\Phi(p_k,q_k)\), and enumerate

\[ b_{2k+1}=p_k,\qquad b_{2k+2}=q_k,\qquad a_n=b_n+1. \]

The ordering just proved makes \((a_n)\) strictly increasing.

Equation (13) gives the fully explicit rational values

\[ \sum_n\frac1{a_n-1} =\sum_n\frac1{b_n} =\frac{3}{2N_0}, \tag{14} \]
\[ \sum_n\frac1{a_n} =\frac{3}{2N_0} - \left( \frac1{N_0(N_0+1)} + \frac1{2N_0(2N_0+1)} \right). \tag{15} \]

Thus the target values are exact even though the individual nested block choices are not a closed-form list.

5. Endpoint growth

Let \(L_k=\log N_k\). From (9),

\[ L_{k+1}\ge\frac32L_k-\log800. \]

Induction gives

\[ L_k\ge \left(\frac32\right)^k\bigl(L_0-2\log800\bigr)+2\log800. \tag{16} \]

The coefficient \(L_0-2\log800\) is positive. Each selected \(b_{2k+1}\) is at least \(0.99N_k\), and each \(b_{2k+2}\) is at least \(1.99N_k\). Since

\[ \gamma^2=\frac32, \]

(16) implies

\[ \liminf_{n\to\infty}\frac{\log a_n}{\gamma^n}>0. \]

Exponentiating proves

\[ \liminf_{n\to\infty}a_n^{1/\gamma^n}>1. \]

This completes the theorem.

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Exact obstruction for this block architecture

The live target \(2\) is not close to what fixed-window finite blocks can deliver.

[A] Counting/area obstruction. Consider any version of this architecture with \(q\) denominators per scale, each having \(O(N)\) integer choices, and with the same two natural residual radii \(N^{-1}\) and \(N^{-2}\). There are only \(O(N^q)\) block values. The current residual rectangle has area \(\asymp N^{-3}\). If the next scale is \(N'=N^\alpha\), each translate of the next residual rectangle has area \(\asymp N'^{-3}=N^{-3\alpha}\). Even with no overlaps, coverage requires

\[ N^qN^{-3\alpha}\gtrsim N^{-3}, \]

so necessarily

\[ \alpha\le\frac{q+3}{3}. \tag{17} \]

With \(q\) sequence terms per scale, the associated per-term base is at most

\[ \left(\frac{q+3}{3}\right)^{1/q}. \]

This decreases for \(q\ge2\): differentiating \(\log(1+q/3)/q\) and using \(\log(1+x)>x/(1+x)\) proves the claim. Its maximum is therefore

\[ \sqrt{\frac53}=1.290994448736\ldots \quad(q=2). \tag{18} \]

So even a hypothetically optimal discrete covering inside this whole fixed-window block family cannot reach exponent \(2\). The precise missing ingredient for the live problem is not another constant optimization of Kovač--Tao's blocks. It must evade (17), for example through an exact algebraic cancellation, a cross-scale correction that reuses old degrees of freedom, or a substantially different nonlocal construction.

[C] Within the present two-term architecture, there may still be room between the proved \(\sqrt{3/2}\) and the counting ceiling \(\sqrt{5/3}\), through an anisotropic lattice-covering lemma. No such lemma is proved here, and even its optimal form would remain far below \(2\).

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Independent verification

The standalone checker is:

runs/erdos265_wavew039_verify.py

It uses only the Python standard library. It independently:

  1. recomputes Cantor's two sums by exact Fraction arithmetic;
  2. checks the coordinate identity \(1/b-1/(b(b+1))=1/(b+1)\);
  3. multiplies the Jacobian and its proposed inverse exactly;
  4. derives all four uniform derivative-error bounds as exact fractions;
  5. verifies the contraction and self-map inequalities;
  6. verifies both nearest-integer rounding constants;
  7. checks the endpoint scale inequalities and the explicit tail target;
  8. checks the published and improved exponent arithmetic; and
  9. checks the fixed-\(q\) counting ceiling exactly for \(2\le q\le100\) (the all-\(q\) proof is the analytic argument above).

Command run:

python runs/erdos265_wavew039_verify.py

Output:

PASS: all exact assertions succeeded
Cantor sums: 1 and 11/9
Jacobian error bounds: e11=0.00200500801101, e12=0.000250375375313, e21=0.0090441303006, e22=0.000563877033596, q=0.0564750964274
Rounding constants: first=1.00300676353 second=3.01354060148
Exponent comparison: published beta<1.095445115010 new beta<1.224744871392
Sub-endpoint exponent check: beta=1.200000 alpha=1.450000
Endpoint scale checked: gamma=sqrt(3/2)=1.224744871392 N_(k+1)=floor(N_k^(3/2)/400)
Fixed-window block counting cap: gamma<=sqrt(5/3)=1.290994448736 (exactly checked for q=2..100; analytic monotonicity is in the report)
N_0: 40960000000000000000
Explicit transformed target r: 3/81920000000000000000
Explicit transformed target t: formula checked exactly
Scale digit counts: [20, 27, 38, 54, 79]

The script also passed python -m py_compile.

[D] Independent numerical sanity check. Separately from the exact checker, I solved the nonlinear fixed-point equation at all \(25\) points of a \(5\times5\) target grid at \(N=10^9\), using 100-digit arithmetic, rounded both denominators, and checked (6). The worst observed fractions of the proved error budgets were \(0.125422\) and \(0.249947\) in the two coordinates. This finite test is not used in the proof.

Honest stopping point

[A] This run establishes a strict, uniform improvement of the known quantitative exponent and gives exact rational values for the two sums. [A] It does not establish \(\limsup a_n^{1/2^n}>1\). [A] The elementary area count proves that optimizing any fixed-size, fixed-relative-window version of this block method cannot reach that threshold. [C] A solution of the live problem therefore needs a genuinely different mechanism; no verified candidate for that mechanism emerged in this run.

PARTIAL: proved simultaneous rationality with the sharper endpoint growth \(\liminf a_n^{1/(\sqrt{3/2})^n}>1\), exact target sums, a standalone checker, and a \(\sqrt{5/3}\) counting ceiling for the fixed-window two-coordinate block architecture; the live exponent \(2\) remains open.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger