A191521 Triangle read by rows: T(n,k) is the number of left factors of Dyck paths of length n that have k valleys (i.e., a (1,-1)-step followed by a (1,1)-step).

1, 1, 2, 2, 1, 3, 3, 3, 6, 1, 4, 12, 4, 4, 18, 12, 1, 5, 30, 30, 5, 5, 40, 60, 20, 1, 6, 60, 120, 60, 6, 6, 75, 200, 150, 30, 1, 7, 105, 350, 350, 105, 7, 7, 126, 525, 700, 315, 42, 1, 8, 168, 840, 1400, 840, 168, 8, 8, 196, 1176, 2450, 1960, 588, 56, 1, 9, 252, 1764, 4410, 4410, 1764, 252, 9

Offset: 0

Author

Emeric Deutsch, Jun 05 2011

Keywords

• nonn
• tabf

Comments

Row n>=1 contains ceiling(n/2) entries.

Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).

Sum_{k>=0} k*T(n,k) = A191522(n).

Examples

			T(4,1)=3 because we have U(DU)D, U(DU)U, and UU(DU), where U=(1,1) and D=(1,-1) (the valleys are shown between parentheses).
Triangle starts:
  1;
  1;
  2;
  2,  1;
  3,  3;
  3,  6,  1;
  4, 12,  4;
  4, 18, 12,  1;
  ...
		

Links

• Alois P. Heinz, Rows n = 0..300, flattened

Crossrefs

Cf. A001405, A124428, A191522.

Programs

Maple

Q := sqrt(((1-z)^2-t*z^2)*((1+z)^2-t*z^2)): G := (1+t*z^2-z^2-Q)/(t*z*(t*z^2-1+2*z-z^2+Q)): Gser := simplify(series(G, z = 0, 19)): for n from 0 to 16 do P[n] := sort(coeff(Gser, z, n)) end do: 1; for n to 16 do seq(coeff(P[n], t, k), k = 0 .. ceil((1/2)*n)-1) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(x, y, t) option remember; expand(`if`(x=0, 1,
     `if`(y>0, b(x-1, y-1, z), 0)+b(x-1, y+1, 1)*t))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0, 1)):
seq(T(n), n=0..30);  # Alois P. Heinz, Mar 29 2017

Mathematica

T[n_, m_] := If [n == 0 && m == 0, 1, If[n == 0, 0, If[OddQ[n-1], (2* Binomial[n/2, m]*Binomial[n/2, m+1]*(n/2 + 1))/n, Binomial[(n+1)/2, m+1]*Sum[(-1)^(k-1)*Binomial[(n+1)/2, m-k+1], {k, 1, (n+1)/2}]]]];
Table[T[n, m], {n, 0, 16}, {m, 0, If[n <= 2, 0, Quotient[n-1, 2]]}] // Flatten (* Jean-François Alcover, Feb 16 2021, after Vladimir Kruchinin *)

Maxima

T(n,m):=if n=0 and m=0 then 1 else if n=0 then 0 else if oddp(n-1) then (2*binomial(n/2,m)*binomial(n/2,m+1)*(n/2+1))/n else binomial((n+1)/2,m+1)*sum((-1)^(k-1)*binomial((n+1)/2,m-k+1),k,1,(n+1)/2);
/* Vladimir Kruchinin, Jul 24 2019 */

Formula

G.f.: G(t,z) = (1+t*z^2-z^2-Q)/(t*z*(t*z^2-1+2*z-z^2+Q)), where Q = sqrt(((1-z)^2-t*z^2)*((1+z)^2-t*z^2)).

T(n,k) = 2*C(n/2,k)*C(n/2,k+1)*(n/2+1)/n, for even n, C((n+1)/2,k+1)*Sum_{j=1..(n+1)/2} (-1)^(j-1)*C((n+1)/2,k-j+1), for odd n, T(0,0)=1. - Vladimir Kruchinin, Jul 24 2019