Erdős Problem #1049

Erdős Problem 1049 --- collection record

PROVENANCE AND STATUS

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1049.tex.

REUSED SOURCE 1

1) FORMAL RESTATEMENT

Let $t>1$ be a rational number. Define the Lambert series \[ S(t):=\sum_{n=1}^\infty \frac{1}{t^n-1}. \] The problem states the identity \[ S(t)=\sum_{n=1}^\infty \frac{\tau(n)}{t^n}, \] where $\tau(n)$ is the number of positive divisors of $n$. Question: Is $S(t)$ irrational for every rational $t>1$?

2) QUICK LITERATURE/CONTEXT CHECK

The provided text states this is a conjecture of Chowla, and that Erdős proved irrationality when $t\ge 2$ is an integer. I will not assume Erdős' proof; instead I give partial structural lemmas and identify the precise blocker to completing an irrationality proof.

3) ATTACK PLAN

Proof track: - Prove the divisor-sum identity rigorously from geometric series (Lemma 4.1). - Try to adapt a base-$t$ (or base-$b$) digit-block argument: produce long stretches where the coefficients $\tau(n)$ have large divisibility properties to force zeros in an expansion.

Disproof track: - Attempt to find a rational $t>1$ with algebraic relations that might cause telescoping or known rationality; attempt numerical recognition for small rationals (fast check).

4) WORK

FAST REALITY CHECK (numerics)

Using partial sums (computed in Python with high precision): - $t=2$: $S(2)\approx 1.6066951524152917637833015231909\dots$ - $t=3/2$: $S(3/2)\approx 3.8971550754986773894172694282364\dots$ No obvious rationality pattern is visible from truncations.

Lemma 4.1 (Divisor-sum identity)

For every real $t>1$, \[ \sum_{n=1}^\infty \frac{1}{t^n-1}=\sum_{m=1}^\infty \frac{\tau(m)}{t^m}. \]

Proof. For $t>1$ and each $n\ge 1$ we have the absolutely convergent geometric series \[ \frac{1}{t^n-1} = \frac{t^{-n}}{1-t^{-n}} = \sum_{k=1}^\infty t^{-nk}. \] Summing over $n$ and using absolute convergence to justify rearrangement, \[ \sum_{n=1}^\infty \frac{1}{t^n-1} = \sum_{n=1}^\infty\sum_{k=1}^\infty t^{-nk}. \] Let $m=nk$. For each fixed $m\ge 1$, the number of pairs $(n,k)$ with $nk=m$ is exactly $\tau(m)$. Therefore the double sum equals \[ \sum_{m=1}^\infty \tau(m) t^{-m} = \sum_{m=1}^\infty \frac{\tau(m)}{t^m}. \] $\square$

Lemma 4.2 (CRT construction: long blocks where $\tau$ is divisible by a given integer)

Fix an integer $b\ge 2$ and an integer $k\ge 1$. There exists an integer $N$ such that \[ b\mid \tau(N+1),\ \dots,\ b\mid\tau(N+k). \] More strongly, there exists $N$ such that \[ b^k \mid \tau(N+1),\ \dots,\ b^k\mid\tau(N+k). \]

Proof. We prove the stronger statement. Choose distinct primes $p_1,\dots,p_k$ with $p_i>k$ and $p_i> b$. For each $i$, impose the congruence \[ N+i \equiv p_i^{\,b^k-1} \pmod{p_i^{\,b^k}}. \] This implies $p_i^{b^k-1}\mid (N+i)$ but $p_i^{b^k}\nmid (N+i)$, i.e. the exact $p_i$-adic valuation of $N+i$ is $b^k-1$. Write the prime factorization of $N+i$ as $N+i=p_i^{b^k-1}m_i$ with $p_i\nmid m_i$. Then \[ \tau(N+i)=\tau(p_i^{b^k-1})\,\tau(m_i)=b^k\,\tau(m_i), \] so $b^k\mid\tau(N+i)$.

The moduli $p_i^{b^k}$ are pairwise coprime, so by the Chinese remainder theorem there exists $N$ solving all $k$ congruences simultaneously. $\square$

What this lemma does NOT yet give: To turn Lemma 4.2 into an irrationality proof for $S(b)=\sum \tau(n)/b^n$, one typically needs to control carries in a base-$b$ expansion (or to build exceptionally good rational approximations with controlled denominators). Lemma 4.2 alone does not control the contribution of $\tau(N+j)$ for $j>k$ to the first $k$ digits of $\{b^N S(b)\}$.

5) VERIFICATION

- Lemma 4.1 uses only absolute convergence for $t>1$ and a divisor-counting bijection. - Lemma 4.2 is fully constructive and uses only CRT and the formula $\tau(p^a)=a+1$. - No step assumes unproved distribution of primes in arithmetic progressions.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. - The divisor-sum identity $S(t)=\sum \tau(n)/t^n$ for all $t>1$ (Lemma 4.1). - For each integer base $b\ge 2$ and each $k$, there exist arbitrarily long blocks of consecutive integers on which $\tau$ is divisible by $b^k$ (Lemma 4.2).

(ii) First gap (crisp statement). Give a complete, carry-controlled digit-block argument (or another rational approximation argument) showing that for every integer $b\ge 2$, \[ S(b)=\sum_{n\ge 1}\frac{\tau(n)}{b^n}\ \text{is irrational}. \] In particular: starting from Lemma 4.2, show that one can force a block of base-$b$ digits of $S(b)$ (equivalently of $\{b^N S(b)\}$) to be all $0$ while ensuring the expansion is not eventually periodic.

(iii) Top 3 next moves. 1. Strengthen Lemma 4.2 to impose a structured pattern on $\tau(N+j)$ for $j>k$ that provably prevents carry into the first $k$ digits of $\{b^N S(b)\}$. 2. Replace base-$b$ expansion with a Diophantine approximation approach: find infinitely many $N$ for which $\|b^N S(b)\|$ (distance to nearest integer) is strictly between $0$ and $b^{-k}$, contradicting rationality. 3. For rational $t=a/b$, attempt to clear denominators to transform $S(t)$ into a similar Lambert series in an integer base (e.g. express in terms of $a^{-n}$) and see whether an Erdős-style argument can be adapted.

(iv) Minimal counterexample structure. If Chowla's conjecture fails, a counterexample would be a rational $t>1$ for which the digit/carry structure of the coefficients $\tau(n)$ in base $t$ conspires to become eventually periodic. Any such $t$ would likely have special algebraic structure (e.g. a low-height fraction) and might be detectable by high-precision computation plus rational reconstruction; no such candidate has emerged from the quick numerical checks above.

COMPLETION ESTIMATE

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.

1) FORMAL RESTATEMENT

Let $t>1$ be a rational number. Define the Lambert series \[ S(t):=\sum_{n=1}^\infty \frac{1}{t^n-1}. \] The problem states the identity \[ S(t)=\sum_{n=1}^\infty \frac{\tau(n)}{t^n}, \] where $\tau(n)$ is the number of positive divisors of $n$. Question: Is $S(t)$ irrational for every rational $t>1$?

2) QUICK LITERATURE/CONTEXT CHECK

The provided text states this is a conjecture of Chowla, and that Erdős proved irrationality when $t\ge 2$ is an integer. I will not assume Erdős' proof; instead I give partial structural lemmas and identify the precise blocker to completing an irrationality proof.

3) ATTACK PLAN

Proof track: - Prove the divisor-sum identity rigorously from geometric series (Lemma 4.1). - Try to adapt a base-$t$ (or base-$b$) digit-block argument: produce long stretches where the coefficients $\tau(n)$ have large divisibility properties to force zeros in an expansion.

Disproof track: - Attempt to find a rational $t>1$ with algebraic relations that might cause telescoping or known rationality; attempt numerical recognition for small rationals (fast check).

4) WORK

FAST REALITY CHECK (numerics)

Using partial sums (computed in Python with high precision): - $t=2$: $S(2)\approx 1.6066951524152917637833015231909\dots$ - $t=3/2$: $S(3/2)\approx 3.8971550754986773894172694282364\dots$ No obvious rationality pattern is visible from truncations.

Lemma 4.1 (Divisor-sum identity)

For every real $t>1$, \[ \sum_{n=1}^\infty \frac{1}{t^n-1}=\sum_{m=1}^\infty \frac{\tau(m)}{t^m}. \]

Proof. For $t>1$ and each $n\ge 1$ we have the absolutely convergent geometric series \[ \frac{1}{t^n-1} = \frac{t^{-n}}{1-t^{-n}} = \sum_{k=1}^\infty t^{-nk}. \] Summing over $n$ and using absolute convergence to justify rearrangement, \[ \sum_{n=1}^\infty \frac{1}{t^n-1} = \sum_{n=1}^\infty\sum_{k=1}^\infty t^{-nk}. \] Let $m=nk$. For each fixed $m\ge 1$, the number of pairs $(n,k)$ with $nk=m$ is exactly $\tau(m)$. Therefore the double sum equals \[ \sum_{m=1}^\infty \tau(m) t^{-m} = \sum_{m=1}^\infty \frac{\tau(m)}{t^m}. \] $\square$

Lemma 4.2 (CRT construction: long blocks where $\tau$ is divisible by a given integer)

Fix an integer $b\ge 2$ and an integer $k\ge 1$. There exists an integer $N$ such that \[ b\mid \tau(N+1),\ \dots,\ b\mid\tau(N+k). \] More strongly, there exists $N$ such that \[ b^k \mid \tau(N+1),\ \dots,\ b^k\mid\tau(N+k). \]

Proof. We prove the stronger statement. Choose distinct primes $p_1,\dots,p_k$ with $p_i>k$ and $p_i> b$. For each $i$, impose the congruence \[ N+i \equiv p_i^{\,b^k-1} \pmod{p_i^{\,b^k}}. \] This implies $p_i^{b^k-1}\mid (N+i)$ but $p_i^{b^k}\nmid (N+i)$, i.e. the exact $p_i$-adic valuation of $N+i$ is $b^k-1$. Write the prime factorization of $N+i$ as $N+i=p_i^{b^k-1}m_i$ with $p_i\nmid m_i$. Then \[ \tau(N+i)=\tau(p_i^{b^k-1})\,\tau(m_i)=b^k\,\tau(m_i), \] so $b^k\mid\tau(N+i)$.

The moduli $p_i^{b^k}$ are pairwise coprime, so by the Chinese remainder theorem there exists $N$ solving all $k$ congruences simultaneously. $\square$

What this lemma does NOT yet give: To turn Lemma 4.2 into an irrationality proof for $S(b)=\sum \tau(n)/b^n$, one typically needs to control carries in a base-$b$ expansion (or to build exceptionally good rational approximations with controlled denominators). Lemma 4.2 alone does not control the contribution of $\tau(N+j)$ for $j>k$ to the first $k$ digits of $\{b^N S(b)\}$.

5) VERIFICATION

- Lemma 4.1 uses only absolute convergence for $t>1$ and a divisor-counting bijection. - Lemma 4.2 is fully constructive and uses only CRT and the formula $\tau(p^a)=a+1$. - No step assumes unproved distribution of primes in arithmetic progressions.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. - The divisor-sum identity $S(t)=\sum \tau(n)/t^n$ for all $t>1$ (Lemma 4.1). - For each integer base $b\ge 2$ and each $k$, there exist arbitrarily long blocks of consecutive integers on which $\tau$ is divisible by $b^k$ (Lemma 4.2).

(ii) First gap (crisp statement). Give a complete, carry-controlled digit-block argument (or another rational approximation argument) showing that for every integer $b\ge 2$, \[ S(b)=\sum_{n\ge 1}\frac{\tau(n)}{b^n}\ \text{is irrational}. \] In particular: starting from Lemma 4.2, show that one can force a block of base-$b$ digits of $S(b)$ (equivalently of $\{b^N S(b)\}$) to be all $0$ while ensuring the expansion is not eventually periodic.

(iii) Top 3 next moves. 1. Strengthen Lemma 4.2 to impose a structured pattern on $\tau(N+j)$ for $j>k$ that provably prevents carry into the first $k$ digits of $\{b^N S(b)\}$. 2. Replace base-$b$ expansion with a Diophantine approximation approach: find infinitely many $N$ for which $\|b^N S(b)\|$ (distance to nearest integer) is strictly between $0$ and $b^{-k}$, contradicting rationality. 3. For rational $t=a/b$, attempt to clear denominators to transform $S(t)$ into a similar Lambert series in an integer base (e.g. express in terms of $a^{-n}$) and see whether an Erdős-style argument can be adapted.

(iv) Minimal counterexample structure. If Chowla's conjecture fails, a counterexample would be a rational $t>1$ for which the digit/carry structure of the coefficients $\tau(n)$ in base $t$ conspires to become eventually periodic. Any such $t$ would likely have special algebraic structure (e.g. a low-height fraction) and might be detectable by high-precision computation plus rational reconstruction; no such candidate has emerged from the quick numerical checks above.