A333513 Square array T(n,k), n >= 2, k >= 2, read by antidiagonals, where T(n,k) is the number of self-avoiding closed paths on an n X k grid which pass through four corners ((0,0), (0,k-1), (n-1,k-1), (n-1,0)).

1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 7, 11, 7, 1, 1, 17, 49, 49, 17, 1, 1, 41, 229, 373, 229, 41, 1, 1, 99, 1081, 3105, 3105, 1081, 99, 1, 1, 239, 5123, 26515, 44930, 26515, 5123, 239, 1, 1, 577, 24323, 227441, 674292, 674292, 227441, 24323, 577, 1

Offset: 2

Author

Seiichi Manyama, Mar 25 2020

Keywords

• nonn
• tabl

Examples

			Square array T(n,k) begins:
  1,  1,    1,     1,      1,        1, ...
  1,  1,    3,     7,     17,       41, ...
  1,  3,   11,    49,    229,     1081, ...
  1,  7,   49,   373,   3105,    26515, ...
  1, 17,  229,  3105,  44930,   674292, ...
  1, 41, 1081, 26515, 674292, 17720400, ...
		

Links

• Seiichi Manyama, Antidiagonals n = 2..15, flattened

Crossrefs

Column k=2-7 give: A000012, A001333(n-2), A333514, A333515, A358712, A358713.

Main diagonal gives A333466.

Cf. A333758.

Programs

Python

# Using graphillion
from graphillion import GraphSet
import graphillion.tutorial as tl
def A333513(n, k):
    universe = tl.grid(n - 1, k - 1)
    GraphSet.set_universe(universe)
    cycles = GraphSet.cycles()
    for i in [1, k, k * (n - 1) + 1, k * n]:
        cycles = cycles.including(i)
    return cycles.len()
print([A333513(j + 2, i - j + 2) for i in range(11 - 1) for j in range(i + 1)])

Formula

T(n,k) = T(k,n).