A337938 Irregular triangle read by rows: T(n, k) gives the primitive period of the sequence {k (Modd n)}_{k >= 0}, for n >= 1.
0, 0, 1, 0, 1, 2, 0, 2, 1, 0, 1, 2, 3, 0, 3, 2, 1, 0, 1, 2, 3, 4, 0, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 0, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 0, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 0, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1
Offset: 1
Examples
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.
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.
Crossrefs
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.
Comments