A225692 Number of Dyck paths of semilength n avoiding the pattern U^(n-1) D^(n-1).

0, 0, 4, 25, 106, 392, 1380, 4797, 16714, 58685, 207890, 742755, 2674270, 9694648, 35357444, 129644533, 477638410, 1767262865, 6564120058, 24466266619, 91482563198, 343059613165, 1289904146794, 4861946400875, 18367353071526

Offset

2,3

Comments

It appears that for n>=0, a(n+2) represents twice the area of the n-gon created by the paired terms of Pascal's triangle interpreted as Cartesian point coordinates. The terms binomial(n,k) of the n-th row of the triangle define vertex coordinates (x_i,y_i) = (C(n,i),C(n,i+1)), i=0,...,n-1 of a non-self-intersecting closed polygon symmetric to the diagonal x=y. For example, row 4 contains the terms 1,4,6,4,1 which create the points (1,4),(4,6),(6,4),(4,1); these four points bound a four-sided figure with an area of 25/2 = a(6)/2 (see corresponding PARI script). - J. M. Bergot, May 20 2014

Links

Table of n, a(n) for n=2..26.

Axel Bacher, Antonio Bernini, Luca Ferrari, Benjamin Gunby, Renzo Pinzani, and Julian West, The Dyck pattern poset, Discrete Math. 321 (2014), 12--23. MR3154009.

A. Bernini, L. Ferrari, R. Pinzani and J. West, The Dyck pattern poset, arXiv preprint arXiv:1303.3785 [math.CO], 2013.

Formula

Conjecture: -10*(n+1)*(n-4)*a(n) +(51*n^2-196*n+80)*a(n-1) +(-45*n^2+134*n-99)*a(n-2) +2*(2*n-5)*(n-3)*a(n-3)=0. - R. J. Mathar, Jul 09 2013

Conjecture: a(n) ~ 4^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 30 2017

Maple

A009766 := proc(n, k)
    binomial(n+k, n)*(n-k+1)/(n+1) ;
end proc:
A225692d := proc(n, k)
    add( A009766(k-j, n-k+j)^2, j=1..k-n/2) ;
end proc:
A225692 := proc(n)
    A225692d(n, n-1) ;
end proc: # R. J. Mathar, Jul 09 2013

Mathematica

A009766[n_, k_] := Binomial[n + k, n]*(n - k + 1)/(n + 1);
A225692d[n_, k_] := Sum[A009766[k - j, n - k + j]^2, {j, 1, k - n/2}];
A225692[n_] := A225692d[n, n - 1];
Table[A225692[n], {n, 2, 26}] (* Jean-François Alcover, Nov 30 2017, after R. J. Mathar *)

Prog

(PARI) twice_area(n) = {vx = vector(n+1, i, binomial(n, i-1)); vy = vector(n+1, i, if (i<=n, vx[i+1])); vy[n+1] = vy[1]; abs(sum(k=1, #vx-1, vx[k]*vy[k+1] - vy[k]*vx[k+1])); } \\ using the Surveyor's Area Formula; Michel Marcus, May 22 2014

Crossrefs

Sequence in context: A167889 A329495 A042651 * A359524 A070764 A327646

Adjacent sequences: A225689 A225690 A225691 * A225693 A225694 A225695

Keyword

nonn

Author

N. J. A. Sloane, May 27 2013

Status

approved