A001184
Number of simple Hamiltonian paths connecting opposite corners of a 2n+1 X 2n+1 grid.
Original entry on oeis.org
1, 2, 104, 111712, 2688307514, 1445778936756068, 17337631013706758184626, 4628650743368437273677525554148, 27478778338807945303765092195103685118924
Offset: 0
A333520
Triangle read by rows: T(n,k) is the number of self-avoiding paths of length 2*(n-1+k) connecting opposite corners in the n X n grid graph (0 <= k <= floor((n-1)^2/2), n >= 1).
Original entry on oeis.org
1, 2, 6, 4, 2, 20, 36, 48, 48, 32, 70, 224, 510, 956, 1586, 2224, 2106, 732, 104, 252, 1200, 3904, 10560, 25828, 58712, 121868, 217436, 300380, 280776, 170384, 61336, 10180, 924, 5940, 25186, 88084, 277706, 821480, 2309402, 6140040, 15130410, 33339900, 62692432, 96096244, 116826664, 110195700, 78154858, 39287872, 12396758, 1879252, 111712
Offset: 1
T(3,1) = 4;
S--* S--*--* S *--* S
| | | | | |
*--* *--* *--* * * *--*
| | | | | |
*--*--E *--E E *--* E
Triangle starts:
=======================================================
n\k| 0 1 2 3 4 ... 8 ... 12
---|---------------------------------------------------
1 | 1;
2 | 2;
3 | 6, 4, 2;
4 | 20, 36, 48, 48, 32;
5 | 70, 224, 510, 956, 1586, ... , 104;
6 | 252, 1200, 3904, 10560, ................. , 10180;
T(n,floor((n-1)^2/2)) gives
A121788(n-1).
T(2*n-1,2*(n-1)^2) gives
A001184(n-1).
-
# Using graphillion
from graphillion import GraphSet
import graphillion.tutorial as tl
def A333520(n):
if n == 1: return [1]
universe = tl.grid(n - 1, n - 1)
GraphSet.set_universe(universe)
start, goal = 1, n * n
paths = GraphSet.paths(start, goal)
return [paths.len(2 * (n - 1 + k)).len() for k in range((n - 1) ** 2 // 2 + 1)]
print([i for n in range(1, 8) for i in A333520(n)])
Showing 1-2 of 2 results.