A333667 Triangle T(n,k), n >= 2, 0 <= k <= floor(n^2/2)-2*n+2, read by rows, where T(n,k) is the number of 2*(k+2*n-2)-cycles in the n X n grid graph which pass through NW and SE corners ((0,0),(n-1,n-1)).

1, 3, 20, 16, 6, 175, 420, 562, 456, 186, 1764, 8064, 21224, 39500, 55376, 57248, 37586, 10260, 1072, 19404, 138600, 569768, 1717152, 4151965, 8371428, 14126846, 19364732, 20241450, 14759356, 6998166, 1927724, 230440

Offset

2,2

Links

Seiichi Manyama, Rows n = 2..9, flattened

Formula

T(n,0) = A000891(n-2).

Example

T(3,0) = 3;
   +--*--*   +--*--*   +--*
   |     |   |     |   |  |
   *--*  *   *     *   *  *--*
      |  |   |     |   |     |
      *--+   *--*--+   *--*--+
Triangle starts:
=======================================================================
n\k|      0        1         2 ...      4 ...   8 ...    12 ...     18
---|-------------------------------------------------------------------
2  |      1;
3  |      3;
4  |     20,      16,        6;
5  |    175,     420,      562, ... , 186;
6  |   1764,    8064,    21224, .......... , 1072;
7  |  19404,  138600,   569768, .................. , 230440;
8  | 226512, 2265120, 12922446, ............................ , 4638576;

Prog

(Python)
# Using graphillion
from graphillion import GraphSet
import graphillion.tutorial as tl
def A333667(n):
    universe = tl.grid(n - 1, n - 1)
    GraphSet.set_universe(universe)
    cycles = GraphSet.cycles().including(1).including(n * n)
    return [cycles.len(2 * k).len() for k in range(2 * n - 2, n * n // 2 + 1)]
print([i for n in range(2, 8) for i in A333667(n)])

Crossrefs

Row sums give A333323.

Cf. A003763, A302337, A333651, A333652, A333668.

Sequence in context: A115280 A212995 A081849 * A169642 A222482 A022129

Adjacent sequences: A333664 A333665 A333666 * A333668 A333669 A333670

Keyword

nonn,tabf

Author

Seiichi Manyama, Apr 01 2020

Status

approved