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A128727
Triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n having k DDU and LDU's.
0
1, 1, 3, 9, 1, 27, 9, 81, 54, 2, 243, 270, 30, 729, 1215, 270, 5, 2187, 5103, 1890, 105, 6561, 20412, 11340, 1260, 14, 19683, 78732, 61236, 11340, 378, 59049, 295245, 306180, 85050, 5670, 42, 177147, 1082565, 1443420, 561330, 62370, 1386, 531441
OFFSET
0,3
COMMENTS
A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of steps in it.
Row n has ceiling(n/2) terms (n >= 1).
Row sums yield A002212.
Apparently a(n) = A126177(n-1). - Georg Fischer, Oct 28 2018
LINKS
E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203
FORMULA
T(n,0) = 3^(n-1).
T(2k+1,k) = binomial(2k,k)/(k+1) (the Catalan numbers, A000108).
T(2k,k-1) = 3binomial(2k-1,k) = A003409(k).
Sum_{k>=0} k*T(n,k) = A026377(n-1).
T(n,k) = (1/n)*3^(n-1-2k)*binomial(n,k)*binomial(n-k,k+1).
G.f.: G = G(t,z) satisfies tzG^2 - (1 - 3z + 2tz)G + 1 - 2z + tz = 0.
EXAMPLE
T(5,2)=2 because we have UU(DDU)U(DDU)D and UUU(DDU)(DDU)D (the 2 subwords are shown between parentheses).
Triangle starts:
1;
1;
3;
9, 1;
27, 9;
81, 54, 2;
243, 270, 30;
729, 1215, 270, 5;
MAPLE
T:=(n, k)->3^(n-1-2*k)*binomial(n, k)*binomial(n-k, k+1)/n: 1; for n from 1 to 13 do seq(T(n, k), k=0..floor((n-1)/2)) od; # yields sequence in triangular form
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Mar 31 2007
STATUS
approved