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A143911
Triangle read by rows: T(n,k) = number of forests on n labeled nodes, where k is the maximum of the number of edges per tree (n>=1, 0<=k<=n-1).
2
1, 1, 1, 1, 3, 3, 1, 9, 12, 16, 1, 25, 60, 80, 125, 1, 75, 330, 480, 750, 1296, 1, 231, 1680, 3920, 5250, 9072, 16807, 1, 763, 9408, 33600, 49000, 72576, 134456, 262144, 1, 2619, 56952, 254016, 598500, 762048, 1210104, 2359296, 4782969, 1, 9495, 348120
OFFSET
1,5
LINKS
FORMULA
See program.
EXAMPLE
T(4,1) = 9, because 9 forests on 4 labeled nodes have 1 as the maximum of the number of edges per tree:
.1-2. .1.2. .1.2. .1.2. .1.2. .1.2. .1-2. .1.2. .1.2.
..... ...|. ..... .|... ..\.. ../.. ..... .|.|. ..X..
.4.3. .4.3. .4-3. .4.3. .4.3. .4.3. .4-3. .4.3. .4.3.
Triangle begins:
1;
1, 1;
1, 3, 3;
1, 9, 12, 16;
1, 25, 60, 80, 125;
1, 75, 330, 480, 750, 1296;
MAPLE
A:= (n, k)-> coeff(series(exp(add(j^(j-2) *x^j/j!, j=1..k)), x, n+1), x, n)*n!: T:= (n, k)-> A(n, k+1)-A(n, k): seq(seq(T(n, k), k=0..n-1), n=1..11);
MATHEMATICA
A[n_, k_] := SeriesCoefficient[Exp[Sum[j^(j-2)*x^j/j!, {j, 1, k}]], {x, 0, n}]*n!; T[n_, k_] := A[n, k+1] - A[n, k];
Table[T[n, k], {n, 1, 11}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, May 31 2016, translated from Maple *)
CROSSREFS
Columns k=0-1 give: A000012, A001189.
Row sums give A001858.
Rightmost diagonal gives A000272.
Cf. A138464.
Sequence in context: A260301 A216916 A157401 * A185422 A131889 A292386
KEYWORD
nonn,tabl
AUTHOR
Alois P. Heinz, Sep 04 2008
STATUS
approved