A195582 Numerator of the average height of a binary search tree on n elements.

0, 1, 2, 8, 10, 19, 64, 1471, 3161, 3028, 6397, 27956, 58307, 168652, 190031, 794076401, 817191437, 57056556523, 65776878541, 112508501827291, 32836043478431, 24620974441660973, 30663050241335933, 280904716386831931, 1713934856212591039, 12438570098319186469

Offset: 0

Author

Alois P. Heinz, Sep 20 2011

Keywords

• nonn
• frac

Comments

Empty external nodes are counted in determining the height of a search tree.

Examples

			0/1, 1/1, 2/1, 8/3, 10/3, 19/5, 64/15, 1471/315, 3161/630, 3028/567, 6397/1134, 27956/4725, 58307/9450, 168652/26325, 190031/28665 ... = A195582/A195583
For n = 3 there are 2 permutations of {1,2,3} resulting in a binary search tree of height 2 and 4 permutations resulting in a tree of height 3.  The average height is (2*2+4*3)/3! = (4+12)/6 = 16/6 = 8/3.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..478
• B. Reed, The height of a random binary search tree, J. ACM, 50 (2003), 306-332.
• Wikipedia, Binary search tree
• Index entries for sequences related to rooted trees
• Index entries for sequences related to trees

Crossrefs

Denominators: A195583.

Cf. A000142, A195581, A195596, A195597, A195598, A195599, A195600, A195601, A244108, A316944.

Programs

Maple

b:= proc(n,k) option remember;
      if n=0 then 1
    elif n=1 then `if`(k>0, 1, 0)
    else add(binomial(n-1,r-1) *b(r-1,k-1) *b(n-r,k-1), r=1..n)
      fi
    end:
T:= (n, k)-> b(n, k)-`if`(k>0, b(n, k-1), 0):
a:= n-> add(T(n,k)*k, k=0..n)/n!:
seq(numer(a(n)), n=0..30);

Mathematica

b[n_, k_] := b[n, k] = If[n==0, 1, If[n==1, If[k>0, 1, 0], Sum[Binomial[n - 1, r-1]*b[r-1, k-1]*b[n-r, k-1], {r, 1, n}]]]; T[n_, k_] := b[n, k] - If[ k>0, b[n, k-1], 0]; a[n_] := Sum[T[n, k]*k, {k, 0, n}]/n!; Table[ Numerator[a[n]], {n, 0, 30}] (* Jean-François Alcover, Mar 01 2016, after Alois P. Heinz *)

Formula

A195582(n)/A195583(n) = 1/n! * Sum_{k=1..n} k * A195581(n,k).

A195582(n)/A195583(n) = alpha*log(n) - beta*log(log(n)) + O(1), with alpha = 4.311... (A195596) and beta = 1.953... (A195599).

A195582(n)/A195583(n) = A316944(n) / A000142(n).