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A332959
Triangle read by rows: T(n,k) is the number of labeled forests with n trees and 2n nodes and with the largest tree having exactly k nodes, (n >= 1, 2 <= k <= n+1).
3
1, 3, 12, 15, 180, 240, 105, 5040, 6720, 7000, 945, 151200, 352800, 315000, 272160, 10395, 6029100, 21067200, 20790000, 17962560, 13311144, 135135, 276215940, 1387386000, 1765764000, 1471133664, 1211314104, 787218432, 2027025, 14983768800, 105945840000, 165225060000, 146023637760, 121131410400, 94466211840, 54717165360
OFFSET
1,2
COMMENTS
The first formula is based on Kolchin's formula (1.4.2) [see the Kolchin reference].
REFERENCES
V. F. Kolchin, Random Graphs. Encyclopedia of Mathematics and Its Applications 53. Cambridge Univ. Press, Cambridge, 1999, pp 30-31.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)
FORMULA
T(n,k) = ((2*n)!/n!) * Sum_{compositions p_1 + ... + p_n = 2*n, 1 <= p_i <= k}
Product_{j=1..n} f(p_j) / p_j!, where f(p_j) = A000272(p_j) = p_j^(p_j-2).
EXAMPLE
Triangle T(n,k) begins:
1;
3, 12;
15, 180, 240;
105, 5040, 6720, 7000;
945, 151200, 352800, 315000, 272160;
10395, 6029100, 21067200, 20790000, 17962560, 13311144;
...
The graphs for T(2,2) and T(2,3) are illustrated below:
o---o : o o
: |
o---o : o---o
T(2,2) = 3 since the graph on the left has 3 labelings.
T(2,3) = 12 since the graph on the right has 12 labelings.
PROG
(PARI)
T(n, k) = { my(S = 0);
forpart(a = 2*n,
if(a[n] == k,
my(D = Set(a));
my(Pr = prod(j=1, #D, my(p = D[j], c = #select(x->x==p, Vec(a))); p^((p-2)*c) / (p!^c*c!)));
S += n!*Pr )
, [1, k], [n, n]); (2*n)! / n! * S };
(PARI)
B(n, k)={my(p=sum(j=1, k, j^(j-2)*x^j/j!)); (2*n)!*polcoef( polcoef( exp(y*p + O(x*x^(2*n))), 2*n, x), n, y)}
T(n, k)={B(n, k)-B(n, k-1)} \\ Andrew Howroyd, May 08 2020
CROSSREFS
Columns k=2..3 are A001147, A332960.
Row sums give A302112.
Main diagonal is A332958.
Sequence in context: A361925 A096854 A013191 * A009783 A341013 A192786
KEYWORD
nonn,tabl,easy
AUTHOR
Washington Bomfim, Apr 13 2020
STATUS
approved