A307026 Number of (undirected) paths in the m X n king graph (triangle read by rows with m = 1..n and n = 1..).
0, 1, 30, 3, 235, 5148, 6, 1448, 96956, 6014812, 10, 7909, 1622015, 329967798, 57533191444, 15, 40674, 25281625, 16997993692, 9454839968415, 4956907379126694, 21, 202719, 375341540, 834776217484, 1482823362091281, 2480146959625512771, 3954100866385811897908
Offset: 1
Examples
0; 1, 30; 3, 235, 5148; 6, 1448, 96956, 6014812; 10, 7909, 1622015, 329967798, 57533191444; 15, 40674, 25281625, 16997993692, ...;
Links
- Eric Weisstein's World of Mathematics, Graph Path
- Eric Weisstein's World of Mathematics, King Graph
Crossrefs
Programs
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Python
# Using graphillion from graphillion import GraphSet def make_nXk_king_graph(n, k): grids = [] for i in range(1, k + 1): for j in range(1, n): grids.append((i + (j - 1) * k, i + j * k)) if i < k: grids.append((i + (j - 1) * k, i + j * k + 1)) if i > 1: grids.append((i + (j - 1) * k, i + j * k - 1)) for i in range(1, k * n, k): for j in range(1, k): grids.append((i + j - 1, i + j)) return grids def A(start, goal, n, k): universe = make_nXk_king_graph(n, k) GraphSet.set_universe(universe) paths = GraphSet.paths(start, goal) return paths.len() def A307026(n, k): m = k * n s = 0 for i in range(1, m): for j in range(i + 1, m + 1): s += A(i, j, n, k) return s print([A307026(n, k) for n in range(1, 8) for k in range(1, n + 1)]) # Seiichi Manyama, Dec 15 2020
Formula
T(1, n) = binomial(n, 2).
T(n, n) = A288033(n).
Extensions
a(20)-a(28) from Seiichi Manyama, Dec 15 2020
Comments