A300699 Irregular triangle read by rows: T(n, k) = number of vertices with rank k in concertina n-cube.
1, 1, 1, 1, 2, 2, 1, 1, 3, 6, 6, 6, 3, 1, 1, 4, 12, 18, 28, 24, 28, 18, 12, 4, 1, 1, 5, 20, 40, 80, 95, 150, 150, 150, 150, 95, 80, 40, 20, 5, 1, 1, 6, 30, 75, 180, 270, 506, 660, 840, 1080, 1035, 1035, 1080, 840, 660, 506, 270, 180, 75, 30, 6, 1, 1, 7, 42, 126, 350, 630, 1337, 2107, 3192, 4760
Offset: 0
Examples
First rows of the triangle: k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 n 0 1 1 1 1 2 1 2 2 1 3 1 3 6 6 6 3 1 4 1 4 12 18 28 24 28 18 12 4 1 5 1 5 20 40 80 95 150 150 150 150 95 80 40 20 5 1 6 1 6 30 75 180 270 506 660 840 1080 1035 ... The ranks of vertices of a concertina cube (n=3) can be seen in the linked Hasse diagrams. T(3, 4) = 6, so there are 6 vertices with rank 4. Ey Ez Ax P(x, y, z) implies Ey Ax Ez P(x, y, z), and their ranks are 3 and 4. As the difference in rank is 1, this implication is an edge in the Hasse diagram.
Links
- Tilman Piesk, Rows 0..9, flattened
- Tilman Piesk, Rows 0..9
- Tilman Piesk, Formulas in predicate logic (Wikiversity)
- Tilman Piesk, Concertina cube Hasse diagram with labels and with highlighted ranks
- Tilman Piesk, Lists of vertices ordered by rank for n=2..6
- Tilman Piesk, Python code used to generate the sequence
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