|
| |
|
|
A102402
|
|
Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n having k ascents of length 2.
|
|
5
|
|
|
|
1, 1, 1, 1, 2, 3, 6, 6, 2, 17, 15, 10, 46, 51, 30, 5, 128, 175, 91, 35, 372, 568, 336, 140, 14, 1109, 1827, 1296, 504, 126, 3349, 5980, 4785, 2010, 630, 42, 10221, 19833, 17215, 8415, 2640, 462, 31527, 66078, 61908, 34210, 11385, 2772, 132, 98178, 220649, 223444, 134706, 50908, 13299, 1716
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,5
|
|
|
COMMENTS
|
T(n,k) is the number of Łukasiewicz paths of length n having k steps (1,1). A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(3,0)=2 because we have HHH and U(2)DD, where H=(1,0), U(2)=(1,2) and D=(1,-1). Row n has 1+floor(n/2) terms. Row sums yield the Catalan numbers (A000108). T(2n,n)=A000108(n). Column 0 is A102403
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f.: G=G(t,z) satisfies z^3*(1-t)G^3+z(1-z+tz)G^2-G+1=0.
|
|
|
EXAMPLE
|
T(4,2) = 2 because we have UUDDUUDD and UUDUUDDD, where U=(1,1) and D=(1,-1).
Triangle begins:
1;
1;
1, 1;
2, 3;
6, 6, 2;
17, 15, 10;
|
|
|
MATHEMATICA
|
m = 14; G[_, _] = 0;
Do[G[t_, z_] = 1 + G[t, z]^2 z + G[t, z]^2 t z^2 - G[t, z]^2 z^2 + G[t, z]^3 z^3 - G[t, z]^3 t z^3 + O[t]^m + O[z]^m, {m}];
CoefficientList[#, t]& /@ Take[CoefficientList[G[t, z], z], m] // Flatten (* Jean-François Alcover, Oct 05 2019 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,tabf
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
