A101974 Triangle read by rows: number of Dyck paths of semilength n with k peaks before the first return (1<= k

1, 2, 4, 1, 9, 4, 1, 23, 11, 7, 1, 65, 27, 28, 11, 1, 197, 66, 87, 62, 16, 1, 626, 170, 239, 250, 122, 22, 1, 2056, 471, 627, 829, 630, 219, 29, 1, 6918, 1398, 1656, 2448, 2553, 1419, 366, 37, 1, 23714, 4381, 4554, 6803, 8813, 6979, 2917, 578, 46, 1, 82500, 14282

Offset: 1

Author

Emeric Deutsch, Dec 22 2004

Keywords

• nonn
• tabf

Examples

			T(4,2)=4 because we have U(UD)(UD)D|UD, U(UD)U(UD)DD|, UU(UD)D(UD)D| and
UU(UD)(UD)DD|, where U=(1,1), D=(1,-1) (the peaks before the first return | are shown between parentheses).
     1
       2
     4      1
     9      4      1
    23     11      7      1
    65     27     28     11      1
   197     66     87     62     16      1
   626    170    239    250    122     22      1
  2056    471    627    829    630    219     29      1
  6918   1398   1656   2448   2553   1419    366     37      1
 23714   4381   4554   6803   8813   6979   2917    578     46      1
 82500  14282  13231  18571  27362  28364  17206   5567    872     56      1
		

References

• E. Deutsch, Dyck path enumeration, Discrete Math., 204, 1999, 167-202.

Crossrefs

Cf. A000108 (row sums), A014137 (column k=1), A014151 (column k=2), A101975.

Programs

Maple

c:=n->binomial(2*n,n)/(n+1):
T:=proc(n,k) if k=1 then sum(c(i),i=0..n-1) else sum(c(j)*binomial(n-1-j,k-1)*binomial(n-1-j,k)/(n-1-j),j=0..n-2) fi end proc:
T(1,1);
for n from 1 to 12 do seq(T(n,k),k=1..n-1) od; # yields the sequence in triangular form

Formula

T(n, 1)=sum(c(i), i=0..n-1), T(n, k)=sum(c(j)*binomial(n-1-j, k-1)*binomial(n-1-j, k)/(n-1-j), j=0..n-2) for k>1, where c(i)=binomial(2i, i)/(i+1) (i=0, 1, ...) are the Catalan numbers (A000108);

G.f.=1+tzC(z)[1+r(t, z)], where C(z)=1+zC(z)^2 is the Catalan function and r(t, z)=z[1+r(t, z)][1+tr(t, z)] is the Narayana function.