

A305185


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


1



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
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OFFSET

1,3


COMMENTS

From Jianing Song, May 05 2019:
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)


LINKS

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


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*k1) = 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)


EXAMPLE

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, w1, w+2, w2}, so a(3) = max_{s in S(3)} N(s) = 3.


PROG

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


CROSSREFS

Cf. A004016, A033687.
Sequence in context: A286728 A300332 A244819 * A083561 A018195 A086417
Adjacent sequences: A305182 A305183 A305184 * A305186 A305187 A305188


KEYWORD

nonn,easy


AUTHOR

Jianing Song, May 27 2018


EXTENSIONS

Entry rewritten by Jianing Song, May 05 2019


STATUS

approved



