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A339257 Number of spanning trees in the n X 5 king graph.

%I A339257 #17 Feb 16 2025 08:34:01
%S A339257 1,27648,146356224,698512774464,3271331573452800,15258885095892902976,
%T A339257 71111090441547013886784,331335100372867196224868352,
%U A339257 1543757070688065237574186369344,7192607774929149127350811889484864,33511424900308657559195109303117533184,156134620449573478209362729027690283037248
%N A339257 Number of spanning trees in the n X 5 king graph.
%H A339257 Seiichi Manyama, <a href="/A339257/b339257.txt">Table of n, a(n) for n = 1..200</a>
%H A339257 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/KingGraph.html">King Graph</a>
%H A339257 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SpanningTree.html">Spanning Tree</a>
%F A339257 Empirical g.f.: -x*(218700000000*x^8 - 2040471000000*x^7 + 538526880000*x^6 + 311791396500*x^5 - 17462695797*x^4 - 80280747*x^3 + 10513308*x^2 - 21759*x - 1) / (656100000000*x^8 - 4293081000000*x^7 + 4819127400000*x^6 - 930215250900*x^5 + 51621632181*x^4 - 1033572501*x^3 + 5949540*x^2 - 5889*x + 1). - _Vaclav Kotesovec_, Dec 09 2020
%o A339257 (Python)
%o A339257 # Using graphillion
%o A339257 from graphillion import GraphSet
%o A339257 def make_nXk_king_graph(n, k):
%o A339257     grids = []
%o A339257     for i in range(1, k + 1):
%o A339257         for j in range(1, n):
%o A339257             grids.append((i + (j - 1) * k, i + j * k))
%o A339257             if i < k:
%o A339257                 grids.append((i + (j - 1) * k, i + j * k + 1))
%o A339257             if i > 1:
%o A339257                 grids.append((i + (j - 1) * k, i + j * k - 1))
%o A339257     for i in range(1, k * n, k):
%o A339257         for j in range(1, k):
%o A339257             grids.append((i + j - 1, i + j))
%o A339257     return grids
%o A339257 def A338029(n, k):
%o A339257     if n == 1 or k == 1: return 1
%o A339257     universe = make_nXk_king_graph(n, k)
%o A339257     GraphSet.set_universe(universe)
%o A339257     spanning_trees = GraphSet.trees(is_spanning=True)
%o A339257     return spanning_trees.len()
%o A339257 def A339257(n):
%o A339257     return A338029(n, 5)
%o A339257 print([A339257(n) for n in range(1, 15)])
%Y A339257 Column 5 of A338029.
%Y A339257 Cf. A003779.
%K A339257 nonn
%O A339257 1,2
%A A339257 _Seiichi Manyama_, Nov 29 2020