Erdős Problem #258

Erdős Problem #258

Statement and attribution

For positive integers $a_n\to\infty$, must \[ S=\sum_{n\ge1}\frac{\tau(n)}{a_1\cdots a_n} \] be irrational? Yes. The current problem page credits Chojecki, using GPT-5.4 Pro, with the deduction from Tao--Teräväinen below. This is an attributed reconstruction of that deduction, with its substantial external input stated explicitly; no novelty is claimed. The older local attempts prove only restricted cases and predate this resolution.

External input and plan

Tao and Teräväinen, Theorem 1.1 of arXiv:2512.01739, prove that there is an absolute $C>0$ and arbitrarily large integers $N$ such that \[ \Omega(N+k)\le Ck\qquad\hbox{for every integer }k\ge1. \tag{1} \] Here $\Omega$ counts prime factors with multiplicity. The simultaneous quantifier over the entire infinite tail is essential. Their sieve theorem is an external input, not reproved here. We prove completely that $1$ implies the answer, without monotonicity of the bases.

Work: a positive integer smaller than one

Write $Q_n=\prod_{j\le n}a_j$. Eventually $a_j\ge2$, and $\tau(n)\le n$, so comparison with $\sum n2^{-n}$ proves convergence. If $m=\prod p^{e_p}$, then \[ \tau(m)=\prod_p(e_p+1)\le\prod_p2^{e_p}=2^{\Omega(m)}. \] Consequently the shifts in $1$ satisfy $\tau(N+k)\le\Lambda^k$ for $\Lambda=2^C$.

Suppose $S=A/B$ with integers $A$ and $B\ge1$. Choose an integer $M>(B+1)\Lambda$. Since $a_n\to\infty$, all sufficiently late bases are at least $M$. Choose a shift $N$ from $1$ beyond that point and consider \[ I=BQ_N\left(S-\sum_{n\le N}\frac{\tau(n)}{Q_n}\right). \] This is an integer: $BQ_NS=AQ_N$, and $Q_N/Q_n$ is an integer for $n\le N$. It is strictly positive because the remaining summands are positive. But \[ 0<I=B\sum_{k\ge1}\frac{\tau(N+k)}{a_{N+1}\cdots a_{N+k}} \le B\sum_{k\ge1}(\Lambda/M)^k =\frac{B\Lambda}{M-\Lambda}<1, \] a contradiction. Hence $S$ is irrational.

Verification and outcome

The rational denominator $B$ need not divide any $Q_N$: multiplication by $B$ handles this. The argument requires small scaled tails only along the shifts provided by $1$, not at every index. No monotonicity or uniform bound on $\tau(n)/a_n$ was assumed.

SOLVED, relative to the explicitly stated Tao--Teräväinen theorem. The deduction is complete; the deep sieve input is not reproduced.

Sources

• T. Tao and J. Teräväinen, Quantitative correlations and some problems on prime factors of consecutive integers, Theorem 1.1, https://arxiv.org/abs/2512.01739.
• A short note on Erdős Problem #258, 14 April 2026, https://www.ulam.ai/research/erdos258.pdf.
• Current statement: https://www.erdosproblems.com/258.

COMPLETION ESTIMATE

COMPLETION: 100%

Erdős Problems #256--#258: Round 2 continuation

Erdős Problem #258

1) ROUND-2 OBJECTIVE

I am again using the fail-safe strict-progress option. The full problem remains open, but Round 2 will prove a broad new sufficient condition that is not covered by the known monotone case:

If $a_n\to\infty$ and \[ \frac{\tau(n)}{a_n}\to 0, \] then \[ \sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] is irrational.

This is the most promising direction because Round 1 showed that monotonicity is the hypothesis used by Erdős--Straus, but it did not identify a replacement mechanism. The new mechanism is a scaled-tail/fractional-part argument.

2) Round-1 FOUNDATION USED

I use the following Round-1 material.

1. The convergence of \[ T(a_1,a_2,\dots):=\sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] for every integer sequence with $a_n\to\infty$.
2. The known Erdős--Straus theorem: if $(a_n)$ is monotone nondecreasing and $a_n\to\infty$, then $T(a_1,a_2,\dots)$ is irrational.
3. The Round-1 gap: without monotonicity, there was no replacement for the mixed-radix control used in the classical proof.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is the following scaled-tail criterion.

Scaled-tail criterion. Let \[ Q_n:=a_1a_2\cdots a_n, \qquad T:=\sum_{n=1}^{\infty}\frac{\tau(n)}{Q_n}, \qquad R_n:=\sum_{m=n+1}^{\infty}\frac{\tau(m)}{a_{n+1}a_{n+2}\cdots a_m}. \] If $R_n\to 0$, then $T\notin\mathbb{Q}$.

This is genuinely new in the present investigation. It replaces monotonicity by a direct control of the tail after multiplying by $Q_n$.

4) ATTACK PLAN (ROUND-2)

The remaining Round-1 obstacle was: after multiplying by $Q_n$, how can one force a contradiction from rationality without assuming monotonicity? The plan is:

1. compute the fractional part of $Q_nT$;
2. show it equals the scaled tail $R_n$;
3. if $R_n\to 0$, compare this fractional part with the discrete set of possible fractional parts of a rational number.

Then I will prove $R_n\to 0$ under the concrete hypothesis $\tau(n)/a_n\to 0$.

5) WORK (ROUND-2)

Set \[ Q_n:=a_1a_2\cdots a_n, \qquad T:=\sum_{n=1}^{\infty}\frac{\tau(n)}{Q_n}. \] For each $n\ge 1$, define the scaled tail \[ R_n:=\sum_{m=n+1}^{\infty}\frac{\tau(m)}{a_{n+1}a_{n+2}\cdots a_m}. \] Because all summands are positive, we have $R_n>0$ for every $n$.

Lemma 1 (fractional-part identity).

For every $n\ge 1$, \[ Q_nT=\sum_{m=1}^{n}\tau(m)\frac{Q_n}{Q_m}+R_n, \] and hence, whenever $R_n<1$, \[ \{Q_nT\}=R_n, \] where $\{x\}$ denotes the fractional part of $x$.

Proof. Multiply the series defining $T$ by $Q_n$: \[ Q_nT =\sum_{m=1}^{n}\tau(m)\frac{Q_n}{Q_m}+\sum_{m=n+1}^{\infty}\tau(m)\frac{Q_n}{Q_m}. \] Since $Q_m=Q_n(a_{n+1}\cdots a_m)$ for $m>n$, the second sum is exactly $R_n$. For $m\le n$, the quotient $Q_n/Q_m=a_{m+1}\cdots a_n$ is an integer, so the first sum is an integer. Therefore, if $0<R_n<1$, the fractional part of $Q_nT$ is $R_n$. $\square$

Lemma 2 (scaled-tail criterion).

If $R_n\to 0$, then $T\notin\mathbb{Q}$.

Proof. Assume for contradiction that $T=p/r\in\mathbb{Q}$ in lowest terms with $r\ge 1$. For each $n$, the number $Q_nT$ is rational with denominator dividing $r$, so its fractional part belongs to the finite set \[ \left\{0,\frac1r,\frac2r,\dots,\frac{r-1}{r}\right\}. \] Since $R_n\to 0$ and $R_n>0$, there exists $N$ such that \[ 0<R_n<\frac1r\qquad (n\ge N). \] By Lemma 1, for all sufficiently large $n$ we have $\{Q_nT\}=R_n$. But no element of $ \{0,1/r,2/r,\dots,(r-1)/r\} $ lies strictly between $0$ and $1/r$. This contradiction proves $T\notin\mathbb{Q}$. $\square$

Lemma 3 (a concrete sufficient condition).

If \[ \frac{\tau(n)}{a_n}\to 0, \] then $R_n\to 0$.

Proof. Fix $\varepsilon>0$. Since $a_n\to\infty$, there exists $N_1$ such that $a_n\ge 2$ for all $n\ge N_1$. Since $\tau(n)/a_n\to 0$, there exists $N_2$ such that \[ \frac{\tau(n)}{a_n}\le \varepsilon\qquad (n\ge N_2). \] Let $N:=\max\{N_1,N_2\}$. For $n\ge N$ we get \[ R_n =\sum_{m=n+1}^{\infty}\frac{\tau(m)/a_m}{a_{n+1}a_{n+2}\cdots a_{m-1}} \le \varepsilon\sum_{m=n+1}^{\infty}\frac{1}{a_{n+1}a_{n+2}\cdots a_{m-1}}. \] For the term $m=n+1$ the denominator is the empty product, equal to $1$. For $m=n+1+j$ with $j\ge 1$, there are $j$ factors, each at least $2$, so \[ \frac{1}{a_{n+1}a_{n+2}\cdots a_{m-1}}\le 2^{-j}. \] Hence \[ R_n\le \varepsilon\sum_{j=0}^{\infty}2^{-j}=2\varepsilon. \] Since $\varepsilon>0$ was arbitrary, $R_n\to 0$. $\square$

Combining Lemmas 2 and 3 yields the main round-2 theorem.

Theorem. If $a_n\to\infty$ and $\tau(n)/a_n\to 0$, then \[ \sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] is irrational.

A concrete nonmonotone family.

Take \[ a_{2k-1}=k^3, \qquad a_{2k}=k^2 \qquad (k\ge 1). \] Then $a_n\to\infty$, but the sequence is not monotone since $a_{2k}=k^2<a_{2k-1}=k^3$ for every $k\ge 2$. Also $\tau(n)\le n$ for all $n$, so \[ \frac{\tau(2k-1)}{a_{2k-1}}\le \frac{2k-1}{k^3}\to 0, \qquad \frac{\tau(2k)}{a_{2k}}\le \frac{2k}{k^2}\to 0. \] Thus our theorem proves that \[ \sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] is irrational for this explicit nonmonotone sequence, a case not covered by Erdős--Straus.

Obstruction to any counterexample.

If a rational counterexample exists, then necessarily $R_n\not\to 0$. In particular, by contrapositive of Lemma 3, it cannot satisfy $\tau(n)/a_n\to 0$. Thus any rational counterexample must have $\tau(n)$ comparable to $a_n$ infinitely often, or at least enough mass in the scaled tails to prevent $R_n$ from tending to $0$.

6) ADVERSARIAL VERIFICATION

1. Did I use monotonicity anywhere? No. The proof only uses positivity, the integer nature of $Q_n/Q_m$ for $m\le n$, and the hypothesis $R_n\to 0$ (or the stronger concrete condition $\tau(n)/a_n\to 0$).
2. Could $R_n$ fail to be the fractional part because it is too large? Lemma 1 only asserts $\{Q_nT\}=R_n$ when $R_n<1$, and Lemma 2 uses exactly that for all sufficiently large $n$ once $R_n\to 0$.
3. Is positivity essential? Yes. The contradiction uses $0<R_n<1/r$. In the present problem all terms are positive, so this is valid.
4. Does the theorem really go beyond Round 1? Yes. The explicit family $a_{2k-1}=k^3$, $a_{2k}=k^2$ is nonmonotone, so it lies outside the known Erdős--Straus case, yet it is covered here.
5. Hidden dependence on stronger divisor estimates? None. The proof of the explicit family uses only the trivial bound $\tau(n)\le n$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The full problem remains open, but Round 2 proves a new theorem:

If $a_n\to\infty$ and $\tau(n)/a_n\to 0$, then \[ \sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n} \] is irrational.

Equivalently, any rational counterexample must violate the scaled-tail criterion; in particular it cannot have $\tau(n)=o(a_n)$. This gives a sharp obstruction and settles an explicit infinite family of nonmonotone sequences.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

1. P. Erdős and E. G. Straus, Some number theoretic results, Pacific Journal of Mathematics 36 $1971$, 635--646.
2. T. F. Bloom (ed.), Erdős Problem #258, erdosproblems.com/258, accessed 2026-03-07.

Erdős Problem #258

1) FORMAL RESTATEMENT

Let $(a_n)_{n\ge1}$ be a sequence of integers with $a_n\to +\infty$ as $n\to\infty$. (Interpreting the statement literally requires $a_n\neq0$ for all $n$; in practice one assumes $a_n\in\mathbb{N}$ and hence $a_n\ge 1$ with $a_n\ge2$ eventually.) Define \[ T(a_1,a_2,\dots):=\sum_{n=1}^{\infty}\frac{\tau(n)}{a_1a_2\cdots a_n}, \] where $\tau(n)$ is the divisor-counting function. The problem asks whether \[ (\ddagger)\qquad \forall (a_n)\subset\mathbb{Z}\text{ with }a_n\to\infty:\ T(a_1,a_2,\dots)\notin\mathbb{Q}. \]

2) QUICK LITERATURE/CONTEXT CHECK

• Erdős--Straus $1971$ proved $(\ddagger)$ holds if $(a_n)$ is monotone nondecreasing (i.e. $a_{n-1}\le a_n$ for all $n$) [ErSt71].
• Erdős $1948$ proved that for every integer $t\ge2$, the Lambert series $\sum_{n\ge1} d(n)/t^n$ is irrational [Er48]; this is related but does not directly cover $a_n\to\infty$.
• As of late 2025/early 2026 the unrestricted case (no monotonicity) remains listed as open on the Erdős Problems website [ErP258].

3) ATTACK PLAN

1. Recognise $T(a_1,a_2,\dots)$ as a Cantor-series type expansion with "bases" $a_n$ and "digits" $\tau(n)$.
2. Try two directions:
   • Extend the Erdős--Straus proof from monotone $a_n$ to general $a_n\to\infty$.
   • Attempt to construct a counterexample sequence $(a_n)$ forcing $T$ to be rational (e.g. via telescoping or mixed-radix periodicity).

I did not succeed in either direction.

4) WORK

Convergence sanity check.

Since $a_n\to\infty$, there exists $N$ such that $a_n\ge2$ for all $n\ge N$. Then for $n\ge N$, \[ \left\lvert a_1\cdots a_n\right\rvert\ge \left\lvert a_1\cdots a_N\right\rvert\,2^{n-N}. \] Also $\tau(n)\le n$ for all $n$. Hence the tail is bounded by a convergent series: \[ \sum_{n\ge N}\frac{\tau(n)}{\left\lvert a_1\cdots a_n\right\rvert} \le \frac{1}{\left\lvert a_1\cdots a_N\right\rvert}\sum_{n\ge N}\frac{n}{2^{n-N}}<\infty. \] So $T(a_1,a_2,\dots)$ is well-defined.

Tiny numerical experiment (illustrative only).

For the monotone sequence $a_n=n+1$ (which satisfies $a_n\to\infty$), the first $50$ terms give \[ \sum_{n=1}^{50}\frac{\tau(n)}{2\cdot 3\cdots (n+1)}\approx 0.94529965343349825190\dots \] This is consistent with convergence and with irrationality, but of course gives no proof.

Known conditional resolution under monotonicity.

Erdős--Straus [ErSt71] prove that if $a_1\le a_2\le\cdots$ and $a_n\to\infty$, then $T(a_1,a_2,\dots)$ is irrational. The present problem is to remove the monotonicity hypothesis.

A tempting (but invalid here) telescoping construction.

If one could take $a_n=\tau(n)+1$, then \[ \frac{\tau(n)}{\prod_{j\le n}(\tau(j)+1)} =\frac{1}{\prod_{j\le n-1}(\tau(j)+1)}-\frac{1}{\prod_{j\le n}(\tau(j)+1)}, \] so the series would telescope to $1$. However $\tau(n)$ does not tend to $\infty$ (it is merely unbounded), so $a_n=\tau(n)+1$ fails the hypothesis $a_n\to\infty$. This illustrates why some growth/regularity condition on $a_n$ may be necessary to rule out rationality.

5) VERIFICATION

• The convergence argument uses only $a_n\ge2$ eventually and the trivial bound $\tau(n)\le n$.
• The telescoping observation is algebraically correct, and it explains the remark (seen in Erdős--Straus) that for other arithmetic functions (e.g. $\phi$ or $\sigma$) one can force a rational sum by choosing $a_n=\phi(n)+1$ or $\sigma(n)+1$; for $\tau(n)$ this exact choice does not satisfy $a_n\to\infty$.

6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result obtained here. The series $T(a_1,a_2,\dots)$ converges for every integer sequence with $a_n\to\infty$ (proved above). The conjectured irrationality is known to hold when $(a_n)$ is monotone nondecreasing by Erdős--Straus [ErSt71].

(ii) First gap. Removing monotonicity breaks the existing proof strategy: one needs a way to control carries / mixed-radix expansions or to adapt divisor-function congruence arguments without assuming $a_{n-1}\le a_n$.

(iii) Top 3 next moves.

1. Re-express $T$ as a Cantor series and apply (or sharpen) known rationality criteria for Cantor-series expansions with varying bases $a_n$, attempting to exploit that $\tau(n)$ is "not too large" compared to $a_n\to\infty$.
2. Search for a counterexample sequence $(a_n)$ by imposing congruences on $a_n$ (and hence on partial products) that might force eventual periodicity of the expansion.
3. Attempt to weaken monotonicity to a growth condition (e.g. $a_{n+1}\ge a_n^{1+\delta}$ often enough) and see whether Erdős--Straus methods extend; this might provide intermediate theorems.

(iv) Minimal counterexample structure. A counterexample would be a sequence $a_n\to\infty$ with $\sum_n \tau(n)/(a_1\cdots a_n)\in\mathbb{Q}$. Because the terms are positive, such a sequence would have to enforce exact mixed-radix "eventual periodicity" or telescoping/cancellation at the level of partial products, without violating $a_n\to\infty$. Any such construction would likely require very non-monotone behaviour (since monotone sequences are known to give irrationality).

7) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 30%.

References

• [ErSz59] P. Erdős and G. Szekeres, On the product $\prod_{k=1}^n(1-z^{a_k})$, Publ. Math. Inst. Hung. Acad. Sci. 4 $1959$, 29--34.
• [At61] F. V. Atkinson, On a problem of Erdős and Szekeres, Canad. Math. Bull. 4 $1961$, 7--12.
• [Od82] A. Odlyzko, Minima of cosine sums and maxima of polynomials on the unit circle, J. London Math. Soc. 26 $1982$, 412--420.
• [BoCh18] J. Bourgain and M.-C. Chang, On a paper of Erdős and Szekeres, J. Anal. Math. 136 $2018$, 253--271.
• [BeKo96] A. S. Belov and S. V. Konyagin, An estimate of the free term of a non-negative trigonometric polynomial with integer coefficients, Izvestiya Math. 60 $1996$, 1123--1182.
• [Tang25] Q. Tang, An improved lower bound for Erdős--Szekeres products, arXiv:2509.14182 $2025$.
• [Er48] P. Erdős, On arithmetical properties of Lambert series, J. Indian Math. Soc. (N.S.) $1948$, 63--66.
• [Er68d] P. Erdős, Some problems and results on irrationality, Math. Student 36 $1968$, 222--226.
• [Er88c] P. Erdős, On the irrationality of certain series: problems and results, in New Advances in Transcendence Theory (Cambridge University Press, 1988).
• [KoTa24] V. Kovač and T. Tao, On several irrationality problems for Ahmes series, arXiv:2406.17593 (2024; version cited on arXiv as of 2025).
• [TaTe25] T. Tao and J. Teräväinen, Quantitative correlations and some problems on prime factors of consecutive integers, arXiv:2512.01739 $2025$.
• [ErSt71] P. Erdős and E. G. Straus, Some number theoretic results, Pacific J. Math. 36 $1971$, 635--646.
• [ErP258] T. F. Bloom (ed.), Erdős Problem #258 (website summary, accessed 2026-01-17), https://www.erdosproblems.com/258.