A212796
Square array read by antidiagonals: T(m,n) = number of spanning trees in C_m X C_n.
Original entry on oeis.org
1, 2, 2, 3, 32, 3, 4, 294, 294, 4, 5, 2304, 11664, 2304, 5, 6, 16810, 367500, 367500, 16810, 6, 7, 117600, 10609215, 42467328, 10609215, 117600, 7, 8, 799694, 292626432, 4381392020, 4381392020, 292626432, 799694, 8, 9, 5326848, 7839321861, 428652000000, 1562500000000, 428652000000, 7839321861, 5326848, 9
Offset: 1
Array begins:
1, 2, 3, 4, 5, 6 7, ...
2, 32, 294, 2304, 16810, 117600, 799694, ...
3, 294, 11664, 367500, 10609215, 292626432, 7839321861, ...
4, 2304, 367500, 42467328, 4381392020, 428652000000, 40643137651228, ...
...
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Digits:=200;
T:=(m,n)->round(Re(evalf(simplify(expand(
m*n*mul(mul( 4*sin(h*Pi/m)^2+4*sin(k*Pi/n)^2, h=1..m-1), k=1..n-1))))));
-
default(realprecision, 120);
{T(n, k) = round(n*k*prod(a=1, n-1, prod(b=1, k-1, 4*sin(a*Pi/n)^2+4*sin(b*Pi/k)^2)))} \\ Seiichi Manyama, Jan 13 2021
A175243
Array read by antidiagonals: total number of spanning trees R_n(m) of the complete prism K_m X C_n.
Original entry on oeis.org
1, 2, 1, 3, 12, 3, 4, 75, 294, 16, 5, 384, 11664, 16384, 125, 6, 1805, 367500, 5647152, 1640250, 1296, 7, 8100, 10609215, 1528823808, 6291456000, 259200000, 16807, 8, 35287, 292626432, 380008339280, 18911429680500, 13556617751088, 59549251454
Offset: 1
The array starts in row n=1 as:
1, 1, 3, 16, 125
2, 12, 294, 16384, 1640250
3, 75, 11664, 5647152, 6291456000
4, 384, 367500, 1528823808,
5, 1805, 10609215,
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A175243 := proc(n,m) n*2^(m-1)/m*( orthopoly[T](n,1+m/2)-1)^(m-1) ; end proc:
for d from 2 to 10 do for m from 1 to d-1 do n := d-m ; printf("%d,",A175243(n,m)) ; end do: end do:
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r[n_, m_] := n*2^(m-1)*(ChebyshevT[n, 1+m/2]-1)^(m-1)/m; Table[r[n-m, m], {n, 2, 9}, {m, 1, n-1}] // Flatten (* Jean-François Alcover, Jan 10 2014 *)
Original entry on oeis.org
4, 2304, 367500, 42467328, 4381392020, 428652000000, 40643137651228, 3771854305099776, 344499209234302500, 31074298464967845120, 2774871814779003772844, 245741556726521856000000, 21611621448116558812137652, 1889376666754339457990201088, 164334311374716912516773437500
Offset: 1
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Table[2^(6*n-4)*n*Product[Sin[j*Pi/4]^2 + Sin[k*Pi/n]^2, {j,1,3}, {k,1,n-1}], {n,1,20}]//Round (* Vaclav Kotesovec, Feb 26 2021 *)
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# Using graphillion
from graphillion import GraphSet
def make_CnXCk(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))
grids.append((i + (n - 1) * k, i))
for i in range(1, k * n, k):
for j in range(1, k):
grids.append((i + j - 1, i + j))
grids.append((i + k - 1, i))
return grids
def A212799(n):
if n == 1: return 4
if n == 2: return 2304
universe = make_CnXCk(4, n)
GraphSet.set_universe(universe)
spanning_trees = GraphSet.trees(is_spanning=True)
return spanning_trees.len()
print([A212799(n) for n in range(1, 8)]) # Seiichi Manyama, Nov 22 2020
Showing 1-3 of 3 results.