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.

A114848 Triangle read by rows T(n,k) = the number of Dyck paths of semilength n with k UUDDU's, 0<=k<=[(n-1)/2].

Original entry on oeis.org

1, 1, 2, 4, 1, 10, 4, 28, 13, 1, 82, 44, 6, 248, 153, 27, 1, 770, 536, 116, 8, 2440, 1889, 486, 46, 1, 7858, 6696, 1992, 240, 10, 25644, 23849, 8042, 1180, 70, 1, 84618, 85276, 32124, 5552, 430, 12, 281844, 305933, 127287, 25306, 2430, 99, 1, 946338, 1100692
Offset: 0

Views

Author

I. Tasoulas (jtas(AT)unipi.gr), Feb 20 2006

Keywords

Comments

Row sums are Catalan numbers A000108.

Examples

			T(4,1) = 4 because there exist 4 Dyck paths with one occurrence of UUDDU : UDUUDDUD, UUDDUDUD, UUDDUUDD, UUUDDUDD.
Triangle begins:
:  0 :     1;
:  1 :     1;
:  2 :     2;
:  3 :     4,     1;
:  4 :    10,     4;
:  5 :    28,    13,     1;
:  6 :    82,    44,     6;
:  7 :   248,   153,    27,    1;
:  8 :   770,   536,   116,    8;
:  9 :  2440,  1889,   486,   46,   1;
: 10 :  7858,  6696,  1992,  240,  10;
: 11 : 25644, 23849,  8042, 1180,  70,  1;
: 12 : 84618, 85276, 32124, 5552, 430, 12;
		

Crossrefs

Programs

  • Maple
    b:= proc(x, y, t) option remember; `if`(y<0 or y>x, 0,
         `if`(x=0, 1, expand(b(x-1, y+1, [2, 3, 3, 2, 2][t])
          *`if`(t=5, z, 1) +b(x-1, y-1, [1, 1, 4, 5, 1][t]))))
        end:
    T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0, 1)):
    seq(T(n), n=0..15);  # Alois P. Heinz, Jun 10 2014
  • Mathematica
    For[n = 1, n <= 20, n++, For[k = 0, k <= Floor[(n - 1)/2], k++, Print[Sum[(-1)^j * Binomial[n - 1 - (j + k), j + k] * Binomial[j + k, k] * Binomial[2(n - 2(j + k)), n - 2(j + k)]/(n - 2(j + k) + 1), {j, 0, Floor[(n - 1)/2] - k}]]]]

Formula

T(n,k) = Sum((-1)^j * binomial(n-1-(j+k), j+k) * binomial(j + k, k) * A000108(n-2(j+k)), j=0..[(n-1)/2]-k).
G.f. G = G(t,z) satisfies G = C(z/(z^2(1-t)+1)), where C(z) is g.f. of Catalan numbers.