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
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
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
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, May 27 2013
STATUS
approved