OFFSET
0,3
COMMENTS
From Emeric Deutsch, Dec 15 2007: (Start)
Row 0 has 1 term; row n (n >= 1) has ceiling(n/2) terms.
Row sums yield the Catalan numbers (A000108).
Column 0 yields A135307.
T(2n+1, n) = binomial(2n,n)/(n+1) (the Catalan numbers, A000108). (End)
LINKS
Alois P. Heinz, Rows n = 0..200, flattened
A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
FORMULA
From Emeric Deutsch, Dec 15 2007: (Start)
T(n,k) = (1/n)*binomial(n,k)*Sum_{j=k..floor((n-1)/2)} (-1)^(j-k)*binomial(n-k, j-k)*binomial(2n-3j, n-j+1).
G.f.: G = G(t,z) satisfies z*G^3 - ((1-t)*z+1)*G^2 + (1+2*(1-t)*z)*G - (1-t)*z = 0. (End)
EXAMPLE
Triangle begins:
1;
1;
2;
4, 1;
9, 5;
23, 17, 2;
63, 54, 15;
178, 177, 69, 5;
514, 594, 273, 49;
1515, 1997, 1056, 280, 14;
4545, 6698, 4077, 1308, 168;
...
T(4,1) = 5 because we have U(UDDU)DUD, U(UDDU)UDD, UU(UDDU)DD, UDU(UDDU)D and UUD(UDDU)D (the UDDU's are shown between parentheses).
MAPLE
A135306 := proc(n, k) if n =0 then 1 ; else add((-1)^(j-k)*binomial(n-k, j-k)*binomial(2*n-3*j, n-j+1), j=k..floor((n-1)/2)) ; %*binomial(n, k)/n ; fi ; end: for n from 0 to 20 do for k from 0 to max(0, (n-1)/2) do printf("%a, ", A135306(n, k)) ; od: od: # R. J. Mathar, Dec 08 2007
T:=proc(n, k) options operator, arrow: binomial(n, k)*(sum((-1)^(j-k)*binomial(n-k, j-k)*binomial(2*n-3*j, n-j+1), j=k..floor((1/2)*n-1/2)))/n end proc: 1; for n to 13 do seq(T(n, k), k=0..ceil((n-2)*1/2)) end do; # yields sequence in triangular form; Emeric Deutsch, Dec 15 2007
MATHEMATICA
T[n_, k_] := Binomial[n, k]*Sum[(-1)^(j-k)*Binomial[n-k, j-k]*Binomial[2*n - 3*j, -j+n+1], {j, k, (n-1)/2}]/n; T[0, 0] = 1; Table[T[n, k], {n, 0, 13}, {k, 0, If[n == 0, 0, Quotient[n-1, 2]]}] // Flatten (* Jean-François Alcover, Nov 27 2014, after Emeric Deutsch *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
N. J. A. Sloane, Dec 07 2007
EXTENSIONS
More terms from R. J. Mathar and Emeric Deutsch, Dec 08 2007
STATUS
approved