login
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.
3
1, 16, 95, 426, 1745, 6838, 25897, 95292, 342505, 1208392, 4201765, 14445130, 49221691, 166563454, 560595853, 1878809676, 6275993883, 20910561068
OFFSET
1,2
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
KEYWORD
nonn
AUTHOR
Seiichi Manyama, Mar 25 2020
STATUS
approved