0,3

Every tree in the forest must have at least 2 nodes, i.e. at least one more node besides the root. - N. J. A. Sloane, Nov 26 2010

First, T(n), the number of rooted trees with n+1 nodes A000081(n+1) can be computed using partitions of n as follows: let n = (q1*1 + q2*2 + q3*3 + ... + qn*n) be a nonnegative integer partition of n (the "q"s are the multiplicities of the part sizes), and define a^b to be (a+b-1)! / (a-1)! / b! (the number of ways to color b identical items with a colors), then compute the sum of T(0)^q1 * T(1)^q2 * ... * T(n-1)^qn over all such partitions of n.

Then F(n), the number of forests of rooted trees containing N nodes not counting the roots, can be similarly computed as the sum of T(1)^q1 * T(2)^q2 * ... * T(n)^qn over all such partitions of n.

These are the diagonal sums of the triangle in A174135. - N. J. A. Sloane, Nov 26 2010.

N. J. A. Sloane and Alois P. Heinz, Table of n, a(n) for n = 0..2133

A. Mansuy, Preordered forests, packed words and contraction algebras, J. Algebra 411 (2014) 259-311, section 4.1

R. J. Mathar, Topologically Distinct Sets of Non-intersecting Circles in the Plane, arXiv:1603.00077 [math.CO], 2016.

a(n) ~ c * d^n / n^(3/2), where d = 2.955765285651994974714817524... is the Otter's rooted tree constant (see A051491), and c = 10.088029891871277227771831767... . - Vaclav Kotesovec, May 09 2014

a(n) = A033185(2n, n). - Alois P. Heinz, Feb 15 2016

a(n) = A033185(2n+k, n+k) for all n, k >= 0. - Michael Somos, Aug 20 2018

Trees for example (leaving out the "^0" factors for clarity):

T(0) = 1, T(1) = 1

T(2) = T(1)^1 + T(0)^2 = 2,

T(3) = T(2)^1 + T(1)^1*T(0)^1 + T(0)^3 = 4,

T(4) = T(3)^1 + T(2)^1*T(0)^1 + T(1)^2 + T(1)^1*T(0)^2 +T(0)^4 = 9,

T(5) = T(4)^1 + T(3)^1*T(0)^1 + T(2)^1*T(1)^1 + T(2)^1*T(0)^2 + T(1)^2*T(0)^1 + T(1)^1*T(0)^3 + T(0)^5 = 20.

Forests for example (leaving out the "^0" factors for clarity):

F(2) = T(2)^1 + T(1)^2 = 3,

F(3) = T(3)^1 + T(2)^1*T(1)^1 + T(1)^3 = 7,

F(4) = T(4)^1 + T(3)^1*T(1)^1 + T(2)^2 + T(2)*T(1)^2 + T(1)^4 = 19,

F(5) = T(5)^1 + T(4)^1*T(1)^1 + T(3)^1*T(2)^1 + T(3)^1*T(1)^2 + T(2)^2*T(1)^1 + T(2)^1*T(1)^3 + T(1)^5 = 47.

{Examples of this a^b definition:

2^1 = 2, 2^2 = 3, 2^3 = 4, 2^4 = 5,

3^1 = 3, 3^2 = 6, 3^3 = 10, 3^4 = 15, (triangular numbers)

4^1 = 4, 4^2 = 10, 4^3 = 20, 4^4 = 35, (tetrahedral numbers)

equivalently a^b = (b == 0 ? 1 : (a == 1 || b == 1 ? a : (a * (a+1)^(b-1) / b))) }

# First read 110 terms of A000081 into array b1

M:=100;

t1:=1/mul((1-x*y^i)^b1[i+1], i=2..M):

t2:=series(t1, y, M):

t3:=series(t2, x, M):

a:=(n, k)->coeff(coeff(t3, x, k), y, n);

g:=n->add(a(n-1+i, i), i=1..n-1);

[seq(g(n), n=1..48)]; # N. J. A. Sloane, Nov 26 2010

# Alternative:

g:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,

add(binomial(T(i-1)+j-1, j) *g(n-i*j, i-1), j=0..n/i)))

end:

T:= n-> g(n, n):

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,

add(binomial(T(i)+j-1, j) *b(n-i*j, i-1), j=0..n/i)))

end:

a:= n-> b(n, n):

seq(a(n), n=0..40); # Alois P. Heinz, Jul 20 2012

# Alternative:

g:= proc(n) option remember; `if`(n<=1, n, (add(add(d*

g(d), d=numtheory[divisors](j))*g(n-j), j=1..n-1))/(n-1))

end:

a:= proc(n) option remember; `if`(n=0, 1, add(add(d*

g(d+1), d=numtheory[divisors](j))*a(n-j), j=1..n)/n)

end:

seq(a(n), n=0..40); # Alois P. Heinz, Sep 19 2017

g[n_, i_] := g[n, i] = If[n==0, 1, If[i<1, 0, Sum[Binomial[T[i-1]+j-1, j]*g[n-i*j, i-1], {j, 0, n/i}]]]; T[n_] := g[n, n]; b[n_, i_] := b[n, i] = If[n==0, 1, If[i<1, 0, Sum[Binomial[T[i]+j-1, j]*b[n-i*j, i-1], {j, 0, n/i}]]]; a[n_] := b[n, n] // FullSimplify; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Mar 30 2015, after Alois P. Heinz *)

Cf. A000081 (rooted trees).

Cf. A093637 (products of partition numbers).

Cf. A174135, A033185.

Sequence in context: A110014 A026581 A151535 * A001372 A179467 A308153

Adjacent sequences: A181357 A181358 A181359 * A181361 A181362 A181363

nonn,changed

Peter A. Lawrence, Oct 15 2010

a(0)=1 prepended by Alois P. Heinz, Sep 19 2017

approved