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.

A191525 Triangle read by rows: T(n,k) is the number of left factors of Dyck paths of length n and having k hills; a hill is a (1,1)-step starting at level 0 and followed by a (1,-1)-step.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 4, 1, 1, 7, 2, 1, 13, 5, 1, 1, 24, 8, 2, 1, 46, 16, 6, 1, 1, 86, 28, 9, 2, 1, 166, 58, 19, 7, 1, 1, 314, 103, 32, 10, 2, 1, 610, 211, 71, 22, 8, 1, 1, 1163, 382, 121, 36, 11, 2, 1, 2269, 781, 261, 85, 25, 9, 1, 1, 4352, 1432, 456, 140, 40, 12, 2, 1, 8518, 2920, 976, 316, 100, 28, 10, 1, 1
Offset: 0

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Author

Emeric Deutsch, Jun 06 2011

Keywords

Comments

Row n contains 1+floor(n/2) entries.
Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).
T(n,0)=A191526(n).
Sum(k*T(n,k), k>=0) = A037952(n-1).

Examples

			T(5,1)=2 because we have (UD)UUD and (UD)UUU, where U=(1,1) and D=(1,-1) (the hills are shown between parentheses).
Triangle starts:
  1;
  1;
  1,1;
  2,1;
  4,1,1;
  7,2,1;
  13,5,1,1;
		

Crossrefs

Programs

  • Maple
    c := ((1-sqrt(1-4*z^2))*1/2)/z^2: G := 1/((1-z*c)*(1-z^4*c^2-t*z^2)): Gser := simplify(series(G, z = 0, 19)): for n from 0 to 16 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 16 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form

Formula

G.f.: G(t,z)=1/((1-z*c)*(1-z^4*c^2-t*z^2)), where c=(1-sqrt(1-4*z^2))/(2*z^2) is the Catalan function with argument z^2.