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).

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

Author

Seiichi Manyama, Mar 29 2020

Keywords

• nonn
• tabf

Examples

			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;
		

Links

• Seiichi Manyama, Rows n = 1..9, flattened

Crossrefs

Row sums give A007764.

T(n,0) gives A000984(n-1).

T(n,1) gives A257888(n).

T(n,floor((n-1)^2/2)) gives A121788(n-1).

T(2*n-1,2*(n-1)^2) gives A001184(n-1).

Cf. A074148, A302337, A329633.

Programs

Python

# 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)])