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%I #49 Jan 14 2024 15:05:25
%S 0,0,4,25,106,392,1380,4797,16714,58685,207890,742755,2674270,9694648,
%T 35357444,129644533,477638410,1767262865,6564120058,24466266619,
%U 91482563198,343059613165,1289904146794,4861946400875,18367353071526
%N Number of Dyck paths of semilength n avoiding the pattern U^(n-1) D^(n-1).
%C 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
%H Axel Bacher, Antonio Bernini, Luca Ferrari, Benjamin Gunby, Renzo Pinzani, and Julian West, <a href="http://dx.doi.org/10.1016/j.disc.2013.12.011">The Dyck pattern poset</a>, Discrete Math. 321 (2014), 12--23. MR3154009.
%H A. Bernini, L. Ferrari, R. Pinzani and J. West, <a href="http://arxiv.org/abs/1303.3785">The Dyck pattern poset</a>, arXiv preprint arXiv:1303.3785 [math.CO], 2013.
%F 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
%F Conjecture: a(n) ~ 4^n / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Nov 30 2017
%p A009766 := proc(n, k)
%p binomial(n+k, n)*(n-k+1)/(n+1) ;
%p end proc:
%p A225692d := proc(n,k)
%p add( A009766(k-j,n-k+j)^2,j=1..k-n/2) ;
%p end proc:
%p A225692 := proc(n)
%p A225692d(n,n-1) ;
%p end proc: # _R. J. Mathar_, Jul 09 2013
%t A009766[n_, k_] := Binomial[n + k, n]*(n - k + 1)/(n + 1);
%t A225692d[n_, k_] := Sum[A009766[k - j, n - k + j]^2, {j, 1, k - n/2}];
%t A225692[n_] := A225692d[n, n - 1];
%t Table[A225692[n], {n, 2, 26}] (* _Jean-François Alcover_, Nov 30 2017, after _R. J. Mathar_ *)
%o (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
%K nonn
%O 2,3
%A _N. J. A. Sloane_, May 27 2013
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