For a finite alphabet A, determine the spectrum operatornameSpec_rD( A)=operatornamerD(w):w∈ A^ N of the refined Diophantine exponent operatornamerD on infinite words over A, including which values in [1,∞] occur. In the ternary case, A=-1,0,1, the spectrum is known to be [1,∞]; the corresponding question for broader alphabets remains open.

Spectrum problem for the refined Diophantine exponent

For a finite alphabet $\mathcal A$ , determine the spectrum $\operatorname{Spec}_{\mathrm{rD}}(\mathcal A)=\{\operatorname{rD}(w):w\in\mathcal A^{\mathbb N}\}$ of the refined Diophantine exponent $\operatorname{rD}$ on infinite words over $\mathcal A$ , including which values in $[1,\infty]$ occur. In the ternary case, $\mathcal A=\{-1,0,1\}$ , the spectrum is known to be $[1,\infty]$ ; the corresponding question for broader alphabets remains open.

References

Primary source

arXiv

Additional references

Spectrum of the refined Diophantine exponent — arXiv — Nguyen, Quang-Khai

combinatorics-on-words diophantine-approximation

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A new preprint proves the complete range of the invariant in the ternary setting, while broader alphabet cases remain open.

The problem asks for the possible values of the refined Diophantine exponent. Nguyen and Quang-Khai’s preprint, dated August 20, 2026, gives the complete ternary-alphabet answer.

August 2026 spectrum theorem

Nguyen and Quang-Khai prove that the ternary spectrum is $[1,\infty]$ . Their construction produces, for every $C>1$ , an infinite word over $\{-1,0,1\}$ with exponent $(C+1)/2$ , including the endpoint case. They also give bounds for Thue–Morse and Rudin–Shapiro words, each at most $25$ , and identify further open questions for automatic and overlap-free words.

Current status (as of August 2026): The ternary-alphabet spectrum is settled as $[1,\infty]$ by a proof-bearing preprint; spectra for other alphabet settings remain open.

Sources

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