A333511 Number of self-avoiding walks in the n X 3 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, 16, 95, 426, 1745, 6838, 25897, 95292, 342505, 1208392, 4201765, 14445130, 49221691, 166563454, 560595853, 1878809676, 6275993883, 20910561068

Offset

1,2

Links

Table of n, a(n) for n=1..18.

Example

a(1) = 1;
   +--*--+
a(2) = 16;
   +  *--+   +  *  +   +--*  +   +--*--+
   |  |      |     |      |  |
   *--*  *   *--*--*   *  *--*   *  *  *
   -------------------------------------
   +  *--*   +  *  *   +--*  *   +--*--*
   |  |  |   |            |            |
   *--*  +   *--*--+   *  *--+   *  *  +
   -------------------------------------
   *--*  +   *--*--+   *  *--+   *  *  +
   |  |  |   |            |            |
   +  *--*   +  *  *   +--*  *   +--*--*
   -------------------------------------
   *--*  *   *--*--*   *  *--*   *  *  *
   |  |      |     |      |  |
   +  *--+   +  *  +   +--*  +   +--*--+

Prog

(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)
    return paths.len()
def A333509(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
def A333511(n):
    return A333509(n, 3)
print([A333511(n) for n in range(1, 15)])

Crossrefs

Column k=3 of A333509.

Sequence in context: A159245 A066487 A318021 * A320406 A241936 A321338

Adjacent sequences: A333508 A333509 A333510 * A333512 A333513 A333514

Keyword

nonn

Author

Seiichi Manyama, Mar 25 2020

Status

approved