A185423 Ordered forests of k increasing plane unary-binary trees with n nodes.

1, 1, 2, 3, 6, 6, 9, 30, 36, 24, 39, 150, 270, 240, 120, 189, 918, 1980, 2520, 1800, 720, 1107, 6174, 16254, 25200, 25200, 15120, 5040, 7281, 47070, 142884, 266616, 327600, 272160, 141120, 40320, 54351, 394470, 1369710, 2948400, 4331880

Offset: 1

Author

Peter Bala, Jan 28 2011

Keywords

• nonn
• easy
• tabl

Comments

An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A080635 enumerates increasing plane (ordered) unary-binary trees with n nodes with labels drawn from the set {1,2,...,n}. The entry T(n,k) of the present table counts ordered forests of k increasing plane unary-binary trees with n nodes. See below for an example. See A185421 for the corresponding table in the non-plane case. See A185422 for the table for unordered forests of increasing plane unary-binary trees.

Examples

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....1
..2|....1......2
..3|....3......6......6
..4|....9.....30.....36.....24
..5|...39....150....270....240....120
..6|..189....918...1980...2520...1800....720
..7|.1107...6174..16254..25200..25200..15120...5040
..
Examples of recurrence relation:
T(5,3) = 270 = 3*(T(4,2) + T(4,3) + T(4,4)) = 3*(30+36+24);
T(6,1) = 1*(T(5,0) + T(5,1) + T(5,2)) = 39+150.
Examples of ordered forests:
T(4,2) = 30. The 15 unordered forests consisting of two plane increasing unary-binary trees on 4 nodes are shown below. We can order the trees in a forest in two ways giving rise to 30 ordered forests.
......... ......... ...3.....
.2...3... .3...2... ...|.....
..\./.... ..\./.... ...2.....
...1...4. ...1...4. ...|.....
......... ......... ...1...4.
.
......... ......... ...4.....
.2...4... .4...2... ...|.....
..\./.... ..\./.... ...2.....
...1...3. ...1...3. ...|.....
......... ......... ...1...3.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...1...2. ...1...2. ...|.....
......... ......... ...1...2.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...2...1. ...2...1. ...|.....
......... ......... ...2...1.
.
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... ..........
		

Crossrefs

Cf. A019538, A080635, A185421, A185422.

Programs

Maple

#A185423
P := proc(n,x)
description 'polynomial sequence P(n,x) A185415'
if n = 0 return 1
else
return x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
end proc:
with combinat:
T:=(n,k) -> add ((-1)^(k-j)*binomial(k,j)*P(n,j),j = 0..k):
for n from 1 to 10 do
seq(T(n,k),k = 1..n);
end do;

PARI

{T(n,k)=if(n<1||k<1||k>n,0,if(n==1,if(k==1,1,0),k*(T(n-1,k-1)+T(n-1,k)+T(n-1,k+1))))}

Formula

TABLE ENTRIES

(1)... T(n,k) = Sum_{j = 0..k} (-1)^j*binomial(k,j)*P(n,j),

where P(n,x) are the polynomials described in A185415.

Compare (1) with the formula for the number of ordered set partitions

(2)... A019538(n,k) = Sum_{j = 0..k} (-1)^j*binomial(k,j)*j^n.

RECURRENCE RELATION

(3)... T(n+1,k) = k*(T(n,k-1) + T(n,k) + T(n,k+1))

GENERATING FUNCTION

Let E(t) = 1/2 + (sqrt(3)/2)*tan(sqrt(3)/2*t + Pi/6) be the e.g.f. for A080635.

The e.g.f. for the present triangle is

(4)... 1/(1 + x*(1 - E(t))) = Sum {n >= 0} R(n,x)*t^n/n!

= 1 + x*t + (x + 2*x^2)*t^2/2! + (3*x + 6*x^2 + 6*x^3)*t^3/3! + ....

ROW POLYNOMIALS

The ordered Bell polynomials OB(n,x) are the row polynomials of A019538 given by

(5)... OB(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.

By comparing the e.g.f.s for A019538 and the present table we obtain the surprising identity

(6)... (-i*sqrt(3))^(n-1)*OB(n,x)/x = R(n,y)/y,

where i = sqrt(-1) and x = (sqrt(3)*i/3)*y + (-1/2+sqrt(3)*i/6).

RELATIONS WITH OTHER SEQUENCES

Column 1: A080635.

A185422(n,k) = T(n,k)/k!.

Setting y = 0 in (6) gives

(7)... A080635(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k) * (-1/2+sqrt(3)*i/6)^(k-1).

The row polynomials have their zeros on the vertical line Re(z) = -1/2 in the complex plane (this follows from the above connection with the ordered Bell polynomials, which have negative real zeros). - Peter Bala, Oct 23 2023