OFFSET
1,6
COMMENTS
The length of row n is 1 for n = 1, 2 for n = 2, and 2*n for n >= 3.
The modified modular equivalence relation Modd n is defined, for integer k and positive integer n, by k (Modd n) = k (mod n) if floor(k/n) is even, and -k (mod n) if floor(k/n) is odd. The smallest nonnegative complete residue system modulo n, namely RS(n) = {0, 1, ..., n-1}, is used. See the W. Lang link, Definition 4, eq. (69), p. 25 - 26.
In order to have row length 2*n for all n >= 1 one could use for n = 1 and 2 the imprimitive periods 0, 0 and 0, 1, 0, 1, respectively.
The name Modd n derives from the fact that the multiplicative (but not additive ) group Modd n has the smallest positive reduced residue system with only odd numbers, named RRSodd(n), as elements (for n = 0 RRS(n) = {0}, but here it is taken as {1}). This group is isomorphic to the Galois group G(rho(n)) = Gal(Q(rho(n))/Q), with rho(n) = 2*cos(pi/n). See the W. Lang link.
LINKS
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012, 2017.
FORMULA
T(n,k) = k (Modd n), for n >= 1, and k = 0 for n = 1, k = 0, 1 for n = 2, and k = 0, 1, ..., 2*n - 1, for n >= 3. For k (Modd n) see the comment above.
EXAMPLE
The irregular triangle begins:
n \ k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ..
1: 0
2: 0 1
3: 0 1 2 0 2 1
4: 0 1 2 3 0 3 2 1
5: 0 1 2 3 4 0 4 3 2 1
6: 0 1 2 3 4 5 0 5 4 3 2 1
7: 0 1 2 3 4 5 6 0 6 5 4 3 2 1
8: 0 1 2 3 4 5 6 7 0 7 6 5 4 3 2 1
9: 0 1 2 3 4 5 6 7 8 0 8 7 6 5 4 3 2 1
10 :0 1 2 3 4 5 6 7 8 9 0 9 8 7 6 5 4 3 2 1
...
T(1, 0) = 0 because {k (Modd 1)}_{k >= 0} is the 0 sequence A000007: 0 (Modd 1) = 0 (mod 1) = 0, 1 (Modd 1) = -1 (mod 1) = 0, 2 (Modd 1) = 2 (mod 1) = 0, ... .
T(7, 6) = 6 because floor(6/7) = 0, which is even, hence 6 (Modd 7) = 6 (mod 7) = 6.
T(7, 8) = 6 because floor(8/7) = 1, which is odd, hence 8 (Modd 7) = -8 (mod 7) = 6.
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Wolfdieter Lang, Oct 25 2020
STATUS
approved