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A114502
Triangle read by rows: T(n,k) is number of ordered trees with n edges and having exactly k vertices all of whose children are leaves (1<=k<=floor(n/2) for n>=2).
0
1, 2, 5, 13, 1, 34, 8, 89, 42, 1, 233, 183, 13, 610, 717, 102, 1, 1597, 2622, 624, 19, 4181, 9134, 3275, 205, 1, 10946, 30691, 15473, 1650, 26, 28657, 100284, 67684, 11020, 366, 1, 75025, 320466, 279106, 64553, 3716, 34, 196418, 1005630, 1098402, 342867
OFFSET
1,2
COMMENTS
Row 1 has one term; row n (n>=2) has floor(n/2) terms. Row sums are the Catalan numbers (A000108). Column 1 yields the Fibonacci numbers with odd index (A001519). Sum(kT(n,k),k=1..floor(n/2))=[1+sum(binomial(2j,j),j=0..n-1)]/2 (A024718).
LINKS
Jean-Luc Baril, Pamela E. Harris, Kimberly J. Harry, Matt McClinton, and José L. Ramírez, Enumerating runs, valleys, and peaks in Catalan words, arXiv:2404.05672 [math.CO], 2024. See p. 14.
FORMULA
G.f.: G=G(t, z) satisfies z(1-z)G^2-(1-z-z^2+tz^2)G+1-2z+tz=0.
G.f. G(t,z) can be derived easily from the symbolic decomposition of an ordered tree according to the degree of the root; one obtains G = 1 + z*(G-1+t) + z^2*(G^2-1+t) + z^3*(G^3-1+t) + ... . - Emeric Deutsch, Feb 12 2015
EXAMPLE
T(4,2)=1 because we have the tree with two paths of length two, rab and rcd, emanating from the root r; a and b are vertices all of whose children are leaves.
Triangle starts:
1;
2;
5;
13,1;
34,8;
89,42,1;
233,183,13;
610,717,102,1;
...
MAPLE
G:=1/2/(z^2-z)*(-1+z+z^2-t*z^2+sqrt(1-6*z+11*z^2-2*t*z^2-6*z^3+2*z^3*t+z^4-2*z^4*t+t^2*z^4)): Gser:=simplify(series(G, z=0, 18)): for n from 1 to 15 do P[n]:=coeff(Gser, z^n) od: 1; for n from 1 to 15 do seq(coeff(P[n], t^j), j=1..floor(n/2)) od; # yields sequence in triangular form
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Dec 02 2005
STATUS
approved