%I A108759
%S A108759 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,
%T A108759 120,1386,2640,715,1,165,2772,8580,5005,273,1,220,5148,24024,25025,4368,
%U A108759 1,286,9009,60060,100100,37128,1428,1,364,15015,137280,340340,222768
%N A108759 Triangle read by rows: T(n,k)=binomial(3k,k)*binomial(n+k,3k)/(2k+1) 
               (0<=k<=floor(n/2)).
%C A108759 Row n has 1+floor(n/2) terms. Row sums are the Catalan numbers (A000108). 
               T(2n,n)=binomial(3n,n)/(2n+1) (A001764).
%C A108759 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 non-leaf non-root vertex. - David Callan (callan(AT)stat.wisc.edu), 
               Jul 25 2005
%C A108759 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 (callan(AT)stat.wisc.edu), Jul 22 2008
%D A108759 H. Niederhausen, Catalan traffic at the beach, The Electronic Journal 
               of Combinatorics, 9 (2002), #R33 (p.12).
%H A108759 David Callan, <a href="http://front.math.ucdavis.edu/math.CO/0502532">
               Some Identities for the Catalan and Fine Numbers</a>
%H A108759 Yidong Sun, <a href="http://front.math.ucdavis.edu/search?a=&t=0805.1279&q=&c=math.CO&n=25&s=Listings">
               A simple bijection between binary trees and colored ternary trees</
               a>.
%F A108759 T(n, k) = binom[n+1, 2k+1]binom[n+k, k]/(n+1).
%e A108759 Table begins
%e A108759 \ k..0....1....2....3....4
%e A108759 n\
%e A108759 0 |..1
%e A108759 1 |..1
%e A108759 2 |..1....1
%e A108759 3 |..1....4
%e A108759 4 |..1...10....3
%e A108759 5 |..1...20...21
%e A108759 6 |..1...35...84...12
%e A108759 7 |..1...56..252..120
%e A108759 8 |..1...84..630..660...55
%e A108759 The ordered trees on 3 edges with 1 node adjacent to a leaf are (drawn 
               down
%e A108759 from the root)
%e A108759 /\..../\....|
%e A108759 |......|..../\
%e A108759 together with the path of 3 edges; so T(3,1)=4. (Example reworked by 
               David Callan (callan(AT)stat.wisc.edu), Oct 08 2005)
%p A108759 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
%Y A108759 Sequence in context: A138775 A121529 A006370 this_sequence A158824 A039806 
               A030320
%Y A108759 Adjacent sequences: A108756 A108757 A108758 this_sequence A108760 A108761 
               A108762
%K A108759 nonn,tabf
%O A108759 0,6
%A A108759 Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 24 2005