A128727 - OEIS

%I #21 Oct 30 2018 09:40:38

%S 1,1,3,9,1,27,9,81,54,2,243,270,30,729,1215,270,5,2187,5103,1890,105,

%T 6561,20412,11340,1260,14,19683,78732,61236,11340,378,59049,295245,

%U 306180,85050,5670,42,177147,1082565,1443420,561330,62370,1386,531441

%N Triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n having k DDU and LDU's.

%C 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.

%C Row n has ceiling(n/2) terms (n >= 1).

%C Row sums yield A002212.

%C Apparently a(n) = A126177(n-1). - _Georg Fischer_, Oct 28 2018

%H E. Deutsch, E. Munarini, S. Rinaldi, <a href="http://dx.doi.org/10.1016/j.jspi.2010.01.015">Skew Dyck paths</a>, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203

%F T(n,0) = 3^(n-1).

%F T(2k+1,k) = binomial(2k,k)/(k+1) (the Catalan numbers, A000108).

%F T(2k,k-1) = 3binomial(2k-1,k) = A003409(k).

%F Sum_{k>=0} k*T(n,k) = A026377(n-1).

%F T(n,k) = (1/n)*3^(n-1-2k)*binomial(n,k)*binomial(n-k,k+1).

%F G.f.: G = G(t,z) satisfies tzG^2 - (1 - 3z + 2tz)G + 1 - 2z + tz = 0.

%e T(5,2)=2 because we have UU(DDU)U(DDU)D and UUU(DDU)(DDU)D (the 2 subwords are shown between parentheses).

%e Triangle starts:

%e     1;

%e     1;

%e     3;

%e     9,    1;

%e    27,    9;

%e    81,   54,    2;

%e   243,  270,   30;

%e   729, 1215,  270,    5;

%p 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

%Y Cf. A000108, A002212, A003409, A026377, A126177.

%K nonn,tabf

%O 0,3

%A _Emeric Deutsch_, Mar 31 2007