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A337083 Number of spanning trees of the 1-skeleton of the (n-1)-dimensional permutohedron.

%I A337083 #32 Jan 21 2021 07:15:08
%S A337083 1,1,6,101154816,
%T A337083 6187732257761496793412385090375984958331031826464768000000000
%N A337083 Number of spanning trees of the 1-skeleton of the (n-1)-dimensional permutohedron.
%C A337083 We have the factorizations:
%C A337083 a(4) = 2^15 * 3^2 * 7^3.
%C A337083 a(5) = 2^59 * 3^15 * 5^9 * 7^5 * 11^6 * 23^5 * 29^4 * 41^4.
%C A337083 a(6) = 2^215 * 3^178 * 5^47 * 7^15 * 11^39 * 13^10 * 19^16 * 23^15 * 29^16 * 41^16 * 61^5 * 67^9 * 71^5 * 1931^16 * 3253^9.
%H A337083 Richard Stanley, <a href="/A337083/b337083.txt">Table of n, a(n) for n = 1..6</a>
%H A337083 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BruhatGraph.html">Bruhat Graph</a>
%H A337083 Wikipedia, <a href="https://en.wikipedia.org/wiki/Permutohedron">Permutohedron</a>
%e A337083 For n=3 the permutohedron is a hexagon, which has six spanning trees.
%o A337083 (Python)
%o A337083 import sympy,itertools
%o A337083 def A337083(n):
%o A337083   p=tuple(itertools.permutations(range(n)))
%o A337083   m=len(p)
%o A337083   q={p[i]:i for i in range(m)}
%o A337083   Q=sympy.diag(*[n-1]*m)
%o A337083   for i in range(m):
%o A337083     for k in range(n-1):
%o A337083       Q[i,q[p[i][:k]+tuple(reversed(p[i][k:k+2]))+p[i][k+2:]]]=-1
%o A337083   return Q[:m-1,:m-1].det() # _Pontus von Brömssen_, Jan 18 2021
%Y A337083 Cf. A006237.
%K A337083 nonn
%O A337083 1,3
%A A337083 _Richard Stanley_, Aug 14 2020
%E A337083 a(1) prepended by _Pontus von Brömssen_, Jan 19 2021