

A108759


Triangle read by rows: T(n,k) = binomial(3k,k)*binomial(n+k,3k)/(2k+1) (0<=k<=floor(n/2)).


1



1, 1, 1, 1, 1, 4, 1, 10, 3, 1, 20, 21, 1, 35, 84, 12, 1, 56, 252, 120, 1, 84, 630, 660, 55, 1, 120, 1386, 2640, 715, 1, 165, 2772, 8580, 5005, 273, 1, 220, 5148, 24024, 25025, 4368, 1, 286, 9009, 60060, 100100, 37128, 1428, 1, 364, 15015, 137280, 340340, 222768
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OFFSET

0,6


COMMENTS

Row n has 1+floor(n/2) terms. Row sums are the Catalan numbers (A000108). T(2n,n)=binomial(3n,n)/(2n+1) (A001764).
Triangle read by rows: number of ordered trees counted by number of interior vertices adjacent to a leaf. T(n,k) = number of ordered trees on n edges (A000108) containing k nodes adjacent to a leaf, where a node is a nonleaf nonroot vertex.  David Callan, Jul 25 2005
T(n,k) counts full binary trees on 2n edges by the value k of the following statistic X. Delete all right edges leaving the left edges in place. This partitions the left edges into line segments of lengths say ell(1),ell(2),...,ell(t), with Sum[ell(i),i=1..t] = n. Then X = Sum[Floor(ell(i)/2),i=1..t]. This result is implicit in the Sun reference. Also, there is a standard bijection from full binary trees on 2n edges to Dyck paths of length 2n: draw tree up from root; walk clockwise around the tree starting at the root; process in turn each edge *that has not previously been traversed*: a left edge becomes an upstep and a right edge becomes a downstep. Translated to Dyck paths using this walkaround bijection, the statistic X becomes the sum, taken over all the ascents A, of Floor(Length(A)/2). (An ascent is a maximal sequence of contiguous upsteps. Its length is the number of upsteps in it).  David Callan, Jul 22 2008


LINKS

Table of n, a(n) for n=0..54.
David Callan, Some Identities for the Catalan and Fine Numbers, arXiv:math/0502532 [math.CO], 2005.
Heinrich Niederhausen, Catalan Traffic at the Beach, Electronic Journal of Combinatorics, Volume 9 (2002), #R33 (p.12).
Yidong Sun, A simple bijection between binary trees and colored ternary trees, arXiv:0805.1279 [math.CO], 2008.


FORMULA

T(n,k) = binomial(n+1,2k+1) * binomial(n+k,k) / (n+1).
Sum_{k=0..floor(n/2)} T(n,k)*2^k = A049171(n+1).  Philippe Deléham, Dec 08 2009


EXAMPLE

Table begins
\ k..0....1....2....3....4
n\
0 ..1
1 ..1
2 ..1....1
3 ..1....4
4 ..1...10....3
5 ..1...20...21
6 ..1...35...84...12
7 ..1...56..252..120
8 ..1...84..630..660...55
The ordered trees on 3 edges with 1 node adjacent to a leaf are (drawn down
from the root)
/\..../\....
........../\
together with the path of 3 edges; so T(3,1)=4. (Example reworked by David Callan, Oct 08 2005)


MAPLE

T:=(n, k)>binomial(3*k, k)*binomial(n+k, 3*k)/(2*k+1): for n from 0 to 14 do seq(T(n, k), k=0..floor(n/2)) od; # yields sequence in triangular form


MATHEMATICA

Flatten[Table[Binomial[3k, k] Binomial[n+k, 3k]/(2k+1), {n, 0, 20}, {k, 0, Floor[n/2]}]] (* Harvey P. Dale, May 08 2012 *)


CROSSREFS

Cf. A000108, A001764, A049171.
Sequence in context: A121529 A006370 A262370 * A158824 A039806 A030320
Adjacent sequences: A108756 A108757 A108758 * A108760 A108761 A108762


KEYWORD

nonn,tabf


AUTHOR

Emeric Deutsch, Jun 24 2005


STATUS

approved



