A333795 Number of self-avoiding closed paths on an n X n grid which pass through all points on the two diagonals of the grid.

1, 0, 6, 68, 6102, 1404416, 1094802826, 2524252113468

Offset: 2

Author

Seiichi Manyama, Apr 05 2020

Keywords

• nonn
• more

Comments

a(11) = 407977071391342237828.

Examples

			a(2) = 1;
   +--+
   |  |
   +--+
a(4) = 6;
   +--*--*--+   +--*--*--+   +--*--*--+
   |        |   |        |   |        |
   *--+--+  *   *--+  +--*   *  +--+--*
         |  |      |  |      |  |
   *--+--+  *   *--+  +--*   *  +--+--*
   |        |   |        |   |        |
   +--*--*--+   +--*--*--+   +--*--*--+
   +--*--*--+   +--*  *--+   +--*  *--+
   |        |   |  |  |  |   |  |  |  |
   *  +--+  *   *  +--+  *   *  +  +  *
   |  |  |  |   |        |   |  |  |  |
   *  +  +  *   *  +--+  *   *  +--+  *
   |  |  |  |   |  |  |  |   |        |
   +--*  *--+   +--*  *--+   +--*--*--+
		

Crossrefs

Cf. A333455, A333464, A333466, A333796.

Programs

Python

# Using graphillion
from graphillion import GraphSet
import graphillion.tutorial as tl
def A333795(n):
    universe = tl.grid(n - 1, n - 1)
    GraphSet.set_universe(universe)
    cycles = GraphSet.cycles()
    points = [i + 1 for i in range(n * n) if i % n - i // n == 0 or i % n + i // n == n - 1]
    for i in points:
        cycles = cycles.including(i)
    return cycles.len()
print([A333795(n) for n in range(2, 10)])