A333571 Square array T(n,k), n >= 1, k >= 2, read by antidiagonals, where T(n,k) is the number of Hamiltonian paths in the n X k grid graph which start at any of the n vertices on left side of the graph and terminate at any of the n vertices on the right side.
1, 1, 2, 1, 2, 4, 1, 2, 8, 6, 1, 2, 16, 14, 10, 1, 2, 32, 34, 38, 14, 1, 2, 64, 80, 162, 74, 20, 1, 2, 128, 190, 650, 426, 170, 26, 1, 2, 256, 450, 2728, 2166, 1594, 338, 34, 1, 2, 512, 1066, 11250, 12014, 12908, 4374, 724, 42, 1, 2, 1024, 2526, 46984, 62714, 119364, 47738, 14640, 1448, 52
Offset: 1
Examples
Square array T(n,k) begins: 1, 1, 1, 1, 1, 1, 1, ... 2, 2, 2, 2, 2, 2, 2, ... 4, 8, 16, 32, 64, 128, 256, ... 6, 14, 34, 80, 190, 450, 1066, ... 10, 38, 162, 650, 2728, 11250, 46984, ... 14, 74, 426, 2166, 12014, 62714, 340510, ...
Links
- Seiichi Manyama, Antidiagonals n = 1..14, flattened
Crossrefs
Programs
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Python
# Using graphillion from graphillion import GraphSet import graphillion.tutorial as tl def A(start, goal, n, k): universe = tl.grid(n - 1, k - 1) GraphSet.set_universe(universe) paths = GraphSet.paths(start, goal, is_hamilton=True) return paths.len() def A333571(n, k): if n == 1: return 1 s = 0 for i in range(1, n + 1): for j in range(k * n - n + 1, k * n + 1): s += A(i, j, k, n) return s print([A333571(j + 1, i - j + 2) for i in range(11) for j in range(i + 1)])