OFFSET
1,2
COMMENTS
The first formula is based on Kolchin's formula (1.4.2) [see the Kolchin reference].
Let S be the set of labeled forests with n trees and 2n nodes.
We know that the largest trees in S have n+1 nodes. It follows from line n=6 of the triangle that more than 33% of the forests in S do not have trees with more than 4/7*(n+1) nodes.
The percentages goes to 61%, 83%, and 100% respectively for (5/7)*(n+1) nodes, (6/7)*(n+1) nodes, and n+1 nodes.
REFERENCES
V. F. Kolchin, Random Graphs. Encyclopedia of Mathematics and Its Applications 53. Cambridge Univ. Press, Cambridge, 1999, pp 30-31.
FORMULA
EXAMPLE
Triangle T(n,k) begins:
1;
3, 15;
15, 195, 435;
105, 5145, 11865, 18865;
945, 152145, 504945, 819945, 1092105;
10395, 6039495, 27106695, 47896695, 65859255, 79170399;
...
The graphs for T(2,2) and T(2,3) are illustrated below:
o---o : o---o o o
: |
o---o : o---o o---o
T(2,2) = 3 since the first graph on the left has 3 labelings.
T(2,3) = 15 since the first graph has 3 labelings, and the second has 12 labelings.
PROG
(PARI) T(n, k) = { my(S = 0, D, p, c);
forpart(a = 2*n, D = Set(a);
S += prod(j=1, #D, p=D[j]; c=#select(x-> x==p, Vec(a)); (p^(p-2)/p!)^c /c!)
, [1, k], [n, n]); (2*n)! * S };
CROSSREFS
KEYWORD
AUTHOR
Washington Bomfim, May 11 2020
STATUS
approved
