数論セミナーのお知らせ
今回は三人の方にお話いただきます。
日時: 2024 年 2 月 19 日 (月)場所: D814

14:00- William Mance (Adam Mickiewicz University)
Title: Borel complexity of sets of normal numbers via generic points insubshifts with speciﬁcation
Abstract: We study the Borel complexity of sets of normal numbers in several numeration systems. Taking a dynamical point of view, we oﬀer a uniﬁed
treatment for continued fraction expansions and base r expansions, and theirvarious generalisations: generalised Lüroth series expansions and β-expansions.
In fact, we consider subshifts over a countable alphabet generated by all possible
expansions of numbers in [0, 1). Then normal numbers correspond to genericpoints of shift-invariant measures. It turns out that for these subshifts the setof generic points for a shift-invariant probability measure is precisely at thethird level of the Borel hierarchy (it is a Π03 -complete set, meaning that it is acountable intersection of Fσ -sets, but it is not possible to write it as a countable union of Gδ -sets). We also solve a problem of Sharkovsky–Sivak on theBorel complexity of the basin of statistical attraction. The crucial dynamical
feature we need is a feeble form of speciﬁcation. All expansions named above
generate subshifts with this property. Hence the sets of normal numbers underconsideration are Π03 -complete.
15:15- Xiang Gao (Hubei University)
Title: On an Erdős similarity problem in the large
Abstract: In a recent paper, Kolountzakis and Papageorgiou ask if for everyρ ∈ (0, 1), there exists a set S ⊂ R such that |S ∩ I| ≥ 1 − ρ for every interval
I ⊂ R with unit length, but that does not contain any aﬃne copy of a given
increasing sequence of exponential growth or faster. This question is an analogueof the well-known Erdős similarity problem. In this talk, we show that for eachsequence of real numbers whose integer parts form a set of positive upper Banachdensity, one can explicitly construct such a set S that contains no aﬃne copyof that sequence. Since there exist sequences of arbitrarily rapid growth thatsatisfy this condition, our result answers Kolountzakis and Papageorgiou squestion in the aﬃrmative. A key ingredient of our proof is a generalization ofresults by Amice, Kahane, and Haight from metric number theory. This is joint
work with Yuveshen Mooroogen and Chi Hoi Yip.

16:30- Tetsuro Kamae (Osaka Metropolitan Univ)
Title: Some word problems
Abstract: Prof. Shigeki Akiyama (Tsukuba) asked me the following problem:(*) Given a nondecreasing sequence of integers {an ; n ∈ N} (N = {0, 1, 2, · · · })having bounded gaps an+1 − an (n ∈ N). Does there exists {i < j < k} ⊂ Nsuch that 2j = i + k and 2aj = ai + ak ?
It was originally posed by Pirillo and Varricchio in 1994 and seems to be still
open. I cannot answer to this problem, but I introduce here some related word
problems and experimental result, one suggesting “no” and the other suggestion“yes” to this problem.
Let A be a nonempty ﬁnite set of symbols. For a ﬁnite word ξ = ξ1 ξ2 · · · ξlover A and a ∈ A, denote |ξ|a = #{i; ξi = a}. Then, the abelinization of ξ isdeﬁned to be the vector (|ξ|a ; a ∈ A). Take any A with #A = #{an+1 −an ; n ∈N} and a bijection τ : {an+1 − an ; n} → A. We deﬁne α ∈ AN so thatα(n) = τ (an+1 − an ) (∀n ∈ N).
We say that α is repetition free if ξξ is not a factor of α for any nonemptyﬁnite word ξ. We also say that α is abelian repetition free if ξη is not a factor of α
for any nonempty ﬁnite words ξ, η having the same abelinization. It holds that ifα has an abelian repetition, then the answer to (*) is “yes”. In the case #A = 3,
it is well known that there exists a repetition free α ∈ AN but there does notexist an abelian repetition free α ∈ AN . Hence, if #{an+1 −an ; n ∈ N} ≤ 3, thenthe answer to (*) is “yes”. In the case #A = 4, it is known that there exists anabelian repetition free α ∈ AN (V. Keranen). This suggests that the answer to(*) in general is “no”. In fact, there is no 2-dimensional arithmetic progressionof length 4 among {(n, an ); n ∈ N} if {an+1 − an ; n ∈ N} = {0, 1, 3, 4} (J.
Cassaigne et al.), whereas (*) is asking about length 3.

On the other hand, starting with a0 = a1 = 0, deﬁne an (n = 2, 3, · · · )inductively. Assume that ai (i = 0, 1, · · · , n − 1) are already deﬁned. Deﬁnean to be the minimum integer a ≥ an−1 such that there is no {i < j < n} ⊂{0, 1, · · · , n} with 2j = i + n and 2aj = ai + a. The sequence (an+1 − an ; n ∈ N)
thus obtained is:01020103101410201023010612010401071312104326103142010230103421063515250512101720 · · ·The above graph is the value #{ai − ai−1 ; i = 1, 2, · · · , n} deﬁned like thisas a function of n. It looks like this function increases unboundedly so that aslong as an+1 − an (n ∈ N) stays bounded, the relation that 2j = i + k and2aj = ai + ak for some i < j < k is inevitable and the answer to (*) is “yes”.
Acknowledgment: I appreciate Prof. Luca Zamboni (Lyon) for giving meimportant informations about repetition of words.
References:
Veikko Keranen, Abelian squares are avoidable on 4 letters, Lecture Note in
Computer Science, July 1992
Julien Cassaigne, James D. Currie, Luke Schaeﬀer and Jeﬀrey Shallit, Avoiding
Three consecutive blocks of the same size and same sum, arXiv:1106.5204v3[cs.DM],
Aug. 2011
世話人: 秋山茂樹 (内: 4395)