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Math-730-A2 · revision r1 · on Math-730

gpt pro 5.2 attempt on Math-730

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #249
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK
  4. 3) ATTACK PLAN
  5. 4) WORK
  6. 4.1 Convergence and a high-precision value
  7. 4.2 Two exact alternative series representations
  8. Representation A (Möbius-weighted rational functions).
  9. Representation B (Lambert series with A007431).
  10. 5) VERIFICATION
  11. 6) FINAL
  12. 7) COMPLETION ESTIMATE
  13. References

Erdős Problem #249

1) FORMAL RESTATEMENT

Let $\varphi(n)$ denote Euler's totient function. Consider the convergent series \[ S \coloneqq \sum_{n=1}^{\infty}\frac{\varphi(n)}{2^n}. \] Question. Is $S$ irrational?

2) QUICK LITERATURE/CONTEXT CHECK

  • The Erdős Problems database lists this question as open (last edited Sept 28, 2025) [ErdosProblems249].
  • OEIS entry A256936 records the decimal expansion of $S$ and gives alternative series representations [OEISA256936].
  • Pratt $2024$ notes that proving irrationality for the analogous Lambert-series-type sums involving $\varphi(n)$ and $\sigma(n)$ "seems to present difficulties" and explicitly states these questions are open [Pratt2024].

3) ATTACK PLAN

A natural attempt is to adapt Erdős' classical "blocks of zeros" method used for some Lambert series (e.g. divisor-function series) to the base-$2$ expansion of $S$. The obstacle is that $\varphi(n)$ is typically of size comparable to $n$, so carries from far-out terms are harder to control than in cases where the coefficients grow very slowly.

Two other angles are:

  1. Find a transformation expressing $S$ as a linear combination of more standard constants (e.g. sums of $1/(2^k-1)$ and its derivatives), then apply known irrationality/transcendence results.
  2. Use exact identities for $S$ to reduce irrationality to nontrivial statements about cancellations in series with Möbius weights.

4) WORK

4.1 Convergence and a high-precision value

Since $0\le \varphi(n)\le n$, we have \[ 0\le \sum_{n>N}\frac{\varphi(n)}{2^n} \le \sum_{n>N}\frac{n}{2^n} = \frac{N+2}{2^N}, \] so the series converges absolutely and very rapidly.

A direct computation (summing to $n=400$ and using the tail bound above) yields \[ S = 1.3676308019850223507905081462130881390748919989627948529565984637621567103976692\ldots \] which agrees with the digits listed in OEIS A256936 [OEISA256936].

4.2 Two exact alternative series representations

Representation A (Möbius-weighted rational functions).
Proposition: For $|x|<1$, \[ \sum_{n=1}^{\infty} \varphi(n) x^n = \sum_{k=1}^{\infty} \mu(k)\,\frac{x^k}{(1-x^k)^2}, \] where $\mu$ is the Möbius function. In particular, \[ S = \sum_{k=1}^{\infty} \mu(k)\,\frac{2^{k}}{(2^{k}-1)^2}. \]
Proof: Using the Dirichlet-convolution identity $\varphi = \mu * \mathrm{id}$ (equivalently $\varphi(n)=\sum_{d\mid n}\mu(d)\,n/d$), write \[ \sum_{n\ge 1}\varphi(n)x^n =\sum_{n\ge 1}\sum_{d\mid n}\mu(d)\,\frac{n}{d}\,x^n. \] Set $n=dm$ and interchange sums (justified by absolute convergence for $|x|<1$): \[ \sum_{n\ge 1}\varphi(n)x^n =\sum_{d\ge 1}\mu(d)\sum_{m\ge 1} m\,x^{dm}. \] For $|y|<1$ we have $\sum_{m\ge 1} m y^m = y/(1-y)^2$. Taking $y=x^d$ gives \[ \sum_{m\ge 1} m\,x^{dm} = \frac{x^d}{(1-x^d)^2}. \] Thus \[ \sum_{n\ge 1}\varphi(n)x^n = \sum_{d\ge 1}\mu(d)\frac{x^d}{(1-x^d)^2}. \] Setting $x=1/2$ and simplifying $\frac{2^{-k}}{(1-2^{-k})^2}=\frac{2^{k}}{(2^{k}-1)^2}$ yields the claimed formula for $S$. ∎
Representation B (Lambert series with A007431).
Define \[ a(n)\coloneqq \sum_{d\mid n} \varphi(d)\,\mu\left(\frac{n}{d}\right)=(\varphi*\mu)(n). \] This sequence is OEIS A007431 [OEISA256936,OEISA007431]. Then:
Proposition: For $|x|<1$, \[ \sum_{n=1}^{\infty} \varphi(n) x^n = \sum_{m=1}^{\infty} a(m)\,\frac{x^m}{1-x^m}. \] In particular, \[ S = \sum_{m=1}^{\infty} \frac{a(m)}{2^m-1}. \]
Proof: Using the geometric series, for $|x|<1$, \[ \sum_{m\ge 1} a(m)\frac{x^m}{1-x^m} =\sum_{m\ge 1} a(m)\sum_{j\ge 1} x^{mj} =\sum_{n\ge 1}\left(\sum_{m\mid n} a(m)\right)x^n. \] But $\sum_{m\mid n} a(m) = (a*1)(n) = ((\varphi*\mu)*1)(n)=\varphi(n)$ because $\mu*1=\varepsilon$ (the identity for Dirichlet convolution). Hence the coefficient of $x^n$ equals $\varphi(n)$, proving the identity.

Setting $x=1/2$ gives \[ S=\sum_{m\ge 1} a(m)\frac{2^{-m}}{1-2^{-m}}=\sum_{m\ge 1}\frac{a(m)}{2^m-1}. \] ∎

These exact identities are useful structural information, but none by itself resolves irrationality.

5) VERIFICATION

  • The tail bound $\sum_{n>N} n/2^n=(N+2)/2^N$ is standard and can be verified by differentiating $\sum_{n\ge 0} x^n=1/(1-x)$ and substituting $x=1/2$.
  • The two exact representations were numerically cross-checked: truncating both at $200$ terms gives agreement to about $\sim 10^{-58}$.

6) FINAL

UNRESOLVED.
  1. Strongest fully proved partial result obtained here: I proved two exact identities for the constant \[S=\sum_{n\ge 1}\varphi(n)2^{-n},\] namely $S=\sum_{k\ge 1}\mu(k)\,\frac{2^k}{(2^k-1)^2}$ and $S=\sum_{m\ge 1}\frac{a(m)}{2^m-1}$ with $a=\varphi*\mu$ (OEIS A007431).
  2. First gap preventing a full solution: None of these identities gives a clear route to proving $S\notin\mathbb{Q}$. Standard "block of zeros" arguments for dyadic expansions become delicate because coefficients $\varphi(n)$ are typically of order $n$.
  3. Most promising next move: Try to combine a sieve-based construction that forces long stretches of strong $2$-adic divisibility in $\varphi(n)$ with a method to control carries contributed by the remaining tail. Alternatively, seek a reduction from irrationality of $S$ to known irrationality results for related constants (e.g. Erdős--Borwein type sums).
  4. Smallest plausible counterexample (if false): A rational value $S=P/Q$ would impose strong periodicity constraints on the binary expansion of the partial sums modulo $1$. Given the apparent randomness of $\varphi(n)$, this seems unlikely, but I have no rigorous obstruction.

7) COMPLETION ESTIMATE

Estimated likelihood of completion with additional work: Low-to-moderate ($\approx 15\%$). There is substantial evidence that this question is genuinely hard and resistant to the classical methods used for "slower" arithmetic functions.

References

  • [Romanoff1934] N. P. Romanoff, Über einige Sätze der additiven Zahlentheorie, Mathematische Annalen 109 $1934$, 668--678.
  • [WikipediaRomanov] Wikipedia contributors, Romanov's theorem, Wikipedia (accessed 2026-01-17).
  • [Ding2025] Y. Ding, On a Romanoff type problem of Erdős and Kalmár, arXiv:2503.22700 $2025$.
  • [ErdosProblems249] T. F. Bloom (ed.), Erdős Problem #249, Erdős Problems website (last edited 28 Sep 2025; accessed 2026-01-17).
  • [ErdosGraham1980] P. Erdős and R. L. Graham, Old and new problems and results in combinatorial number theory, Monographies de L'Enseignement Mathématique, Vol. 28, Université de Genève, 1980.
  • [Erdos1975] P. Erdős, Some problems and results on the irrationality of the sum of infinite series, J. Math. Sci. 10 $1975$, 1--7 (often cited as appearing in 1976).
  • [Erdos1988] P. Erdős, On the irrationality of certain series: problems and results, in New advances in transcendence theory (Durham, 1986), Cambridge Univ. Press, 1988, 102--109.
  • [Pratt2024] K. Pratt, The irrationality of a prime factor series under a prime tuples conjecture, arXiv:2409.15185 $2024$.
  • [OEISA256936] N. J. A. Sloane et al., OEIS A256936: Decimal expansion of $\sum_{k\ge 1}\varphi(k)/2^k$, https://oeis.org/A256936 (accessed 2026-01-17).
  • [OEISA007431] N. J. A. Sloane et al., OEIS A007431: $a(n)=\sum_{d\mid n}\varphi(d)\mu(n/d)$, https://oeis.org/A007431 (accessed 2026-01-17).