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
Examples
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;
Links
- Seiichi Manyama, Rows n = 2..9, flattened
Programs
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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)])
Formula
T(n,0) = A000891(n-2).