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A322041
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Triangle read by rows: let E denote the standard triangular 6-valent 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.
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2
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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
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1,3
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COMMENTS
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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 <= n-1. Then T(n,k) <= 6*k for k >= 1, and Sum_{k=0..n-1} 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 Conway-Sloane, Chapter 4). This means that as n increases, the number of terms in the n-th 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).
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REFERENCES
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J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, 3rd. ed., 1993. Fig. 7.1, p. 199. Illustrates row 2 (note that E/2E is isomorphic to GF(4)).
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LINKS
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FORMULA
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Examination of the first 36 rows (see link) shows an obvious quasi-periodic 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((n-1)/3) (see A322042). Combining these two observations suggests that the rows of the triangle are quasi-periodic 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, ..., 18t-6, 18t-3, 18(t-1), 18(t-2), 18(t-3), ..., 36, 18, 2, 0 (2t-1 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.
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EXAMPLE
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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]
...
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MAPLE
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# We work in a fundamental region for E/nE and calculate the edge-distance 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 n-1 do
for j from 0 to n-1 do
d1:=i+j; d2:=n-i; d3:=2*n-i-j; d4:=n-j;
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..n-1)];
end;
for n from 1 to 14 do lprint(hist(n)); od:
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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