

A322041


Triangle read by rows: let E denote the standard triangular 6valent grid in the plane, regarded as a graph with the Eisenstein integers as vertices; row n gives the coordination sequence of the quotient graph E/nE.


2



1, 1, 3, 1, 6, 2, 1, 6, 9, 0, 1, 6, 12, 6, 0, 1, 6, 12, 15, 2, 0, 1, 6, 12, 18, 12, 0, 0, 1, 6, 12, 18, 21, 6, 0, 0, 1, 6, 12, 18, 24, 18, 2, 0, 0, 1, 6, 12, 18, 24, 27, 12, 0, 0, 0, 1, 6, 12, 18, 24, 30, 24, 6, 0, 0, 0, 1, 6, 12, 18, 24, 30, 33, 18, 2, 0, 0, 0, 1, 6, 12, 18, 24, 30, 36, 30, 12, 0, 0, 0, 0
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OFFSET

1,3


COMMENTS

The Eisenstein integers E are the complex numbers r+s*omega, where r, s in Z and omega = exp(2*Pi*i/3) is a complex cube root of unity.
Denote the entries in the triangle by T(n,k), for n >= 1, 0 <= k <= n1. Then T(n,k) <= 6*k for k >= 1, and Sum_{k=0..n1} T(n,k) = n^2.
When E is regarded as a lattice in R^2, E/nE has packing radius roughly n/2, but covering radius roughly n/sqrt(3) > n/2 (see ConwaySloane, Chapter 4). This means that as n increases, the number of terms in the nth row of the triangle will increase linearly with n. The largest k such that T(n,k) is nonzero is A322042(n), which is conjecturally n  ceiling(n/3).


REFERENCES

J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", SpringerVerlag, 3rd. ed., 1993. Fig. 7.1, p. 199. Illustrates row 2 (note that E/2E is isomorphic to GF(4)).


LINKS

Table of n, a(n) for n=1..91.
N. J. A. Sloane, Rows 1 through 36.
N. J. A. Sloane, Illustration for n=2
N. J. A. Sloane, Illustration for n=3
N. J. A. Sloane, Illustration for n=4
N. J. A. Sloane, Illustration for n=6
N. J. A. Sloane, Illustration for n=8


FORMULA

Examination of the first 36 rows (see link) shows an obvious quasiperiodic structure. Call an entry T(n,k) "full" if k=0 or T(n,k)=6*k. Then it appears that column k>0 is full starting at n=2k+1. It also appears that the number of trailing 0's is floor((n1)/3) (see A322042). Combining these two observations suggests that the rows of the triangle are quasiperiodic with period 6.
One can now formulate a specific conjecture for what row n is, for each of the six residue classes of n mod 6.
For example, suppose n=6t. Then it appears that row n is [1, 6, 18, 24, ..., 18t6, 18t3, 18(t1), 18(t2), 18(t3), ..., 36, 18, 2, 0 (2t1 times)].
For t=3, for example, we get:
[1, 6, 12, 18, 24, 30, 36, 42, 48, 51, 36, 18, 2, 0, 0, 0, 0, 0]
There are similar conjectures for n = 6t+1, ..., 6t+5.


EXAMPLE

The first 18 rows are
1 [1]
2 [1, 3]
3 [1, 6, 2]
4 [1, 6, 9, 0]
5 [1, 6, 12, 6, 0]
6 [1, 6, 12, 15, 2, 0]
7 [1, 6, 12, 18, 12, 0, 0]
8 [1, 6, 12, 18, 21, 6, 0, 0]
9 [1, 6, 12, 18, 24, 18, 2, 0, 0]
10 [1, 6, 12, 18, 24, 27, 12, 0, 0, 0]
11 [1, 6, 12, 18, 24, 30, 24, 6, 0, 0, 0]
12 [1, 6, 12, 18, 24, 30, 33, 18, 2, 0, 0, 0]
13 [1, 6, 12, 18, 24, 30, 36, 30, 12, 0, 0, 0, 0]
14 [1, 6, 12, 18, 24, 30, 36, 39, 24, 6, 0, 0, 0, 0]
15 [1, 6, 12, 18, 24, 30, 36, 42, 36, 18, 2, 0, 0, 0, 0]
16 [1, 6, 12, 18, 24, 30, 36, 42, 45, 30, 12, 0, 0, 0, 0, 0]
17 [1, 6, 12, 18, 24, 30, 36, 42, 48, 42, 24, 6, 0, 0, 0, 0, 0]
18 [1, 6, 12, 18, 24, 30, 36, 42, 48, 51, 36, 18, 2, 0, 0, 0, 0, 0]
...


MAPLE

# We work in a fundamental region for E/nE and calculate the edgedistance of each point to the nearest point of nE.
hist:=proc(n) local A, i, j, m, d1, d2, d3, d4;
A:=Array(0..n, 0);
for i from 0 to n1 do
for j from 0 to n1 do
d1:=i+j; d2:=ni; d3:=2*nij; d4:=nj;
if i+j<n then m:=min(d1, d2, d3, d4);
elif i+j=n then m:=min(i, j);
else m:=min(i, j, d1, d3);
fi;
A[m]:=A[m]+1;
od: od:
[seq(A[i], i=0..n1)];
end;
for n from 1 to 14 do lprint(hist(n)); od:


CROSSREFS

The rows converge to A008458.
Cf. A322038 (an analog for the square grid), A322042.
Sequence in context: A067231 A218971 A145791 * A130302 A270103 A134546
Adjacent sequences: A322038 A322039 A322040 * A322042 A322043 A322044


KEYWORD

nonn,tabl


AUTHOR

N. J. A. Sloane, Dec 05 2018; corrected and extended Dec 06 2018


STATUS

approved



