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A108759 Triangle read by rows: T(n,k) = binomial(3k,k)*binomial(n+k,3k)/(2k+1) (0 <= k <= floor(n/2)).

%I A108759 #39 Feb 26 2024 11:01:08
%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,
%U A108759 4368,1,286,9009,60060,100100,37128,1428,1,364,15015,137280,340340,222768
%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_, 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_{i=1..t} ell(i) = n. Then X = Sum_{i=1..t} floor(ell(i)/2). 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
%H A108759 Michael De Vlieger, <a href="/A108759/b108759.txt">Table of n, a(n) for n = 0..10200</a> (rows 0 <= n <= 200, flattened)
%H A108759 David Callan, <a href="https://arxiv.org/abs/math/0502532">Some Identities for the Catalan and Fine Numbers</a>, arXiv:math/0502532 [math.CO], 2005.
%H A108759 Wenqin Cao, Emma Yu Jin, and Zhicong Lin, <a href="https://doi.org/10.1016/j.dam.2019.01.035">Enumeration of inversion sequences avoiding triples of relations</a>, Discrete Applied Mathematics (2019); see also <a href="http://www.emmayujin.at/Pubs/CaoJinLin19.pdf">author's copy</a>
%H A108759 Thomas Einolf, Robert Muth, and Jeffrey Wilkinson, <a href="https://arxiv.org/abs/2107.13417">Injectively k-colored rooted forests</a>, arXiv:2107.13417 [math.CO], 2021.
%H A108759 Sergey Kitaev and Philip B. Zhang, <a href="https://arxiv.org/abs/2310.17236">Non-overlapping descents and ascents in stack-sortable permutations</a>, arXiv:2310.17236 [math.CO], 2023.
%H A108759 Heinrich Niederhausen, <a href="https://doi.org/10.37236/1649">Catalan Traffic at the Beach</a>, Electronic Journal of Combinatorics, Volume 9 (2002), #R33 (p.12).
%H A108759 Yidong Sun, <a href="https://arxiv.org/abs/0805.1279">A simple bijection between binary trees and colored ternary trees</a>, arXiv:0805.1279 [math.CO], 2008.
%H A108759 Tad White, <a href="https://arxiv.org/abs/2401.01462">Quota Trees</a>, arXiv:2401.01462 [math.CO], 2024. See p. 20.
%F A108759 T(n,k) = binomial(n+1,2k+1) * binomial(n+k,k) / (n+1).
%F A108759 Sum_{k=0..floor(n/2)} T(n,k)*2^k = A049171(n+1). - _Philippe Deléham_, Dec 08 2009
%e A108759 Table begins
%e A108759   n\k..0....1....2....3....4
%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 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_, 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
%t A108759 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 *)
%Y A108759 Cf. A000108, A001764, A049171.
%K A108759 nonn,tabf
%O A108759 0,6
%A A108759 _Emeric Deutsch_, Jun 24 2005