A305185 a(n) minimizes the maximum norm of elements in a complete residue system of Eisenstein integers modulo n.

0, 1, 3, 4, 7, 12, 13, 19, 27, 28, 37, 48, 49, 61, 75, 76, 91, 108, 109, 127, 147, 148, 169, 192, 193, 217, 243, 244, 271, 300, 301, 331, 363, 364, 397, 432, 433, 469, 507, 508, 547, 588, 589, 631, 675, 676, 721, 768, 769, 817, 867, 868, 919, 972, 973, 1027, 1083, 1084, 1141, 1200

Offset: 1

Author

Jianing Song, May 27 2018

Keywords

• nonn
• easy

Comments

From Jianing Song, May 05 2019: (Start)

For any Eisenstein integer w != 0, let R(w) be any set of N(w) Eisenstein integers such that no two numbers are congruent modulo w, then we intend to find the smallest possible value of max_{s in R(w)} N(s). Here N(w) is the norm of w.

If we can find a set of complex numbers A such that: (i) for any Eisenstein integer x, r in A, |r| <= |r - x|; (ii) every complex number z can be uniquely represented as z = x + r, where x is an Eisenstein integer, r is in A, then S(w) = {r*w : r is in A} is a complete residue system modulo w formed by choosing one element with the minimal norm in each residue class modulo w (there may be more than one element whose norms are minimal in one residue class). As a result, the smallest possible value of max_{s in R(w)} N(s) is max_{s in S(w)} N(s). For more details, see my further notes in the Link section.

Now, for positive integers n, we find the value of max_{s in S(n)} N(s) over the ring of Eisenstein integers. Let A be the set shown in Page 5, Figure 2 in my further notes on this sequence (see Links section below), and S(w) = {r*w : r in A}. For n >= 2, note that for any s in S(n), s != 0, there exists some s' in S(n) such that s/s' is an Eisenstein unit and arg(s') is in the range [-Pi/6, Pi/6]. Let s' = (x + y*sqrt(3)*i)/2 where x and y have the same parity, 0 < x <= n and -x/3 <= y <= x/3, then N(s) = N(s') = (x^2 + 3*y^2)/4. For fixed x >= 2, we have max |y| = x - 2*ceiling(x/3) so max N(s') = max_{x=2..n} (x^2 + 3*(x - 2*ceiling(x/3))^2)/4 = (n^2 + 3*(n - 2*ceiling(n/3))^2)/4. (End)

Examples

			In the following examples let w = (-1 + sqrt(-3))/2. Let A be the set shown in Page 5, Figure 2 in my further notes on this sequence, and S(w) = {r*w : r is in A}.
n = 1: S(1) = {0}, so a(1) = max_{s in S(1)} N(s) = 0.
n = 2: S(2) = {0, 1, w, w+1}, so a(2) = max_{s in S(2)} N(s) = 1.
n = 3: S(3) = {0, 1, -1, w, w+1, -w, -w-1, w+2, -w-2}, so a(3) = max_{s in S(3)} N(s) = 3.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..10000
• Jianing Song, Further notes on A305185

Crossrefs

Cf. A004016, A033687.

Programs

PARI

a(n) = if(n>1, n^2 - 3*n*ceil(n/3) + 3*ceil(n/3)^2, 0) \\ Jianing Song, May 12 2019

Formula

From Jianing Song, May 05 2019: (Start)

a(1) = 0; for n >= 2, a(n) = (n^2 + 3*(n - 2*ceiling(n/3))^2)/4 = n^2 - 3*n*ceiling(n/3) + 3*ceiling(n/3)^2.

For k >= 1, a(3*k-1) = 3*k^2 - 3*k + 1, a(3*k) = 3*k^2, a(3*k+1) = 3*k^2 + 1.

G.f.: (x^2*(1 + x^2)*(1 + 2*x - x^3 + x^4))/((1 - x)^3*(1 + x + x^2)^2). (End)

Extensions

Entry rewritten by Jianing Song, May 05 2019