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Local Extrema and Infinite Derivatives of the Delange Function of Base 3

Published 15 Oct 2026 in math.CO and math.NT | (2610.21265v1)

Abstract: In the ternary Hamming cube Qn<sup>3Q_n<sup>3, the smallest number ξ<em>n(m)ξ<em>n(m) of edges leaving a set of mm vertices is 2(nm−S3(m))2(nm-S_3(m)), where $S_3(m)=\sum</em>{j&lt;m}s_3(j)$ and s3s_3 is the ternary digit sum. After rescaling, these numbers are the values at the points m/3<sup>nm/3<sup>n of a continuous function TT of Takagi type, the Delange function of base $3$. Both facts are known. In this paper, we study the local structure of TT. The local minima of TT are exactly the ternary rationals, that is, the numbers m/3<sup>nm/3<sup>n with integers mm and n≥0n\ge0. The local maxima are exactly the points (m+1/2)/3<sup>n+k(m+1/2)/3<sup>n+k, where kk, mm and nn are integers with n≥0n\ge0, $0\le m&lt;3<sup>n$ and s3(m)=ns_3(m)=n. We also characterize the points xx where TT has an infinite derivative from one or both sides, in terms of their ternary digits ξ<em>1,ξ2,…ξ<em>1,ξ_2,\ldots. At a ternary rational the derivative is +∞+\infty from the right and −∞-\infty from the left. At every other point, the derivative from the right is +∞+\infty exactly when the digit slopes 2∑</em>j≤n(1−ξj)2\sum</em>{j\le n}(1-ξ_j) tend to +∞+\infty. The derivative from the left is +∞+\infty exactly when a stronger condition holds, which also involves the lengths of the maximal strings of consecutive zeros among the digits ξjξ_j. For the value −∞-\infty the two sides are exchanged, and twos replace zeros. In particular, TT has an infinite derivative at every rational point that is not a ternary rational and whose period has different numbers of zeros and twos. For the finite cubes, we show that over every block of sizes a3<sup>k≤</sup>m≤(a+1)3<sup>ka3<sup>k\le</sup> m\le(a+1)3<sup>k, the maximum of ξn(m)ξ_n(m) exceeds the larger of its values at the two ends of the block by exactly max⁡3<sup>k−∣d∣−1,0\max{3<sup>{k-|d|}-1,0}, where d=n−k−s3(a)d=n-k-s_3(a). Finally, the variation VnV_n of TT over the partition of [0,1][0,1] into the $3n$ intervals [m/3<sup>n,(m+1)/3<sup>n][m/3<sup>n,(m+1)/3<sup>n] satisfies Vn/n→4/3πV_n/\sqrt n\to4/\sqrt{3π}, and we compute the Fourier coefficients of TT.

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