OFFSET
3,2
FORMULA
EXAMPLE
....o-o..........o-o......
....| |..........|\ ......
....o-o..........o-o......
T(4,0)=3 because the graph on the left has 4 nodes and 0 nodes with degree 1. It has 3 labelings.
T(4,1)=12 because the graph on the right has 4 nodes and 1 node with degree 1. It has 12 labelings.
1,
3, 12,
12, 90, 120,
70, 600, 1800, 1200,
465, 4725, 19530, 31500, 12600,
3507, 42168, 211680, 529200, 529200, 141120,
30016, 414288, 2451456, 7902720, 13124160, 8890560, 1693440.
MATHEMATICA
nn=10; f[list_]:=Select[list, #>0&]; t=Sum[Sum[n!/k! StirlingS2[n-1, n-k]y^k x^n/n!, {k, 1, n}], {n, 0, nn}]; Map[Reverse, Map[f, Drop[Range[0, nn]!CoefficientList[Series[ Exp[Log[1/(1-t)]/2-t/2-t^2/4], {x, 0, nn}], {x, y}], 3]]]//Grid
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Geoffrey Critzer, Mar 23 2013
STATUS
approved