cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A114608 Triangle read by rows: T(n,k) is the number of bicolored Dyck paths of semilength n and having k peaks of the form ud (0 <= k <= n). A bicolored Dyck path is a Dyck path in which each up-step is of two kinds: u and U.

Original entry on oeis.org

1, 1, 1, 3, 4, 1, 11, 19, 9, 1, 45, 96, 66, 16, 1, 197, 501, 450, 170, 25, 1, 903, 2668, 2955, 1520, 365, 36, 1, 4279, 14407, 18963, 12355, 4165, 693, 49, 1, 20793, 78592, 119812, 94528, 41230, 9856, 1204, 64, 1, 103049, 432073, 748548, 693588, 372078, 117054
Offset: 0

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Author

Emeric Deutsch, Dec 15 2005

Keywords

Comments

Row sums yield A052701. Column 0 yields the little Schroeder numbers (A001003). Sum_{k=0..n} k*T(n,k) = A069720(n).
Triangle T(n,k), 0 <= k <= n, read by rows; given by [1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 23 2005

Examples

			T(3,2)=9 because we have (ud)(ud)Ud, (ud)Ud(ud), Ud(ud)(ud), (ud)u(ud)d,
(ud)U(ud)d, u(ud)d(ud), U(ud)d(ud), u(ud)(ud)d and U(ud)(ud)d (the ud peaks are shown between parentheses).
Triangle starts:
   1;
   1,  1;
   3,  4,  1;
  11, 19,  9,  1;
  45, 96, 66, 16,  1;
		

Crossrefs

Programs

  • Maple
    T:=proc(n,k) if k<=n-1 then (1/n)*binomial(n,k)*sum(2^j*binomial(n,j+1)*binomial(n-k,j),j=0..n-k) elif k=n then 1 else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
  • Mathematica
    T[n_, k_] := If[k <= n-1, (1/n)*Binomial[n, k]*Sum[2^j*Binomial[n, j+1]* Binomial[n-k, j], {j, 0, n-k}], If[k == n, 1, 0]];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 11 2018, from Maple *)

Formula

T(n,k) = (1/n)*binomial(n,k)*Sum_{j=0..n-k} 2^j*binomial(n, j+1)*binomial(n-k, j) (k <= n-1); T(n, n)=1.
G.f. = G = G(t, z) satisfies G = 1 + z*(G-1+t)*G + z*G^2.