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.

A114626 Triangle read by rows: T(n,k) is the number of hill-free Dyck paths of semilength n and having k peaks at level 2; 0<= k<=n-1, n>=2 (a Dyck path is said to be hill-free if it has no peaks at level 1).

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 2, 1, 1, 6, 6, 3, 2, 1, 19, 17, 12, 5, 3, 1, 61, 56, 36, 20, 8, 4, 1, 202, 185, 120, 66, 31, 12, 5, 1, 683, 624, 409, 224, 110, 46, 17, 6, 1, 2348, 2144, 1408, 784, 385, 172, 66, 23, 7, 1, 8184, 7468, 4920, 2760, 1380, 624, 257, 92, 30, 8, 1, 28855, 26317
Offset: 2

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Author

Emeric Deutsch, Dec 18 2005

Keywords

Comments

Row n has n terms (n>=2). Row sums yield the Fine numbers (A000957). T(n,0)=A114627(n-3). Sum(kT(n,k),k=0..n-1)=A114495(n).

Examples

			T(5,2)=3 because we have U(UD)(UD)UUDDD, UUUDD(UD)(UD)D and U(UD)UUDD(UD)D, where U=(1,1), D=(1,-1) (the peaks at level 2 are shown between parentheses).
Triangle begins:
0,1;
1,0,1;
2,2,1,1;
6,6,3,2,1;
19,17,12,5,3,1;
		

Crossrefs

Programs

  • Maple
    C:=(1-sqrt(1-4*z))/2/z: G:=(1+z-t*z-z*C)/(1+z+z^2-t*z-t*z^2-z*(1+z)*C): Gser:=simplify(series(G,z=0,15)): for n from 2 to 12 do P[n]:=coeff(Gser,z^n) od: for n from 2 to 12 do seq(coeff(t*P[n],t^j),j=1..n) od; # yields sequence in triangular form

Formula

G.f.=(1+z-tz-zC)/[1+z+z^2-tz-tz^2-z(1+z)C], where C=[1-sqrt(1-4z)]/(2z) is the Catalan function.