OFFSET
0,2
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
FORMULA
G.f.: (3-z+q)*(1+z-q)^2/(64*q*z^2), where q = sqrt(1-6*z+z^2).
Recurrence: n*(n+2)*a(n) = (8*n^2 + 7*n - 3)*a(n-1) - (13*n^2 - 17*n - 6)*a(n-2) + 2*(n-3)*(n+1)*a(n-3). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ (1 + sqrt(2))^(2*n+3) / (2^(11/4) * sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012, simplified Dec 24 2017
a(n) = Sum_{j=0..n}((-1)^j*2^(n-j)*binomial(n+3,j)*binomial(2*n-j+2,n+2)). - Vladimir Kruchinin, Apr 08 2016
EXAMPLE
a(1)=4 because in the three dissections of a square we have altogether four triangles: no triangle in the "no-diagonals" dissection and two triangles in each of the dissections by one of the two diagonals of the square.
MATHEMATICA
Table[SeriesCoefficient[(3-x+Sqrt[1-6*x+x^2])*(1+x-Sqrt[1-6*x+x^2])^2/(64*Sqrt[1-6*x+x^2]*x^2), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 14 2012 *)
PROG
(PARI) x='x+O('x^66); q = sqrt(1-6*x+x^2); Vec((3-x+q)*(1+x-q)^2/(64*q*x^2)) \\ Joerg Arndt, May 10 2013
(Maxima)
a(n):=sum((-1)^j*2^(n-j)*binomial(n+3, j)*binomial(2*n-j+2, n+2), j, 0, n); /* Vladimir Kruchinin, Apr 08 2016 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Emeric Deutsch, Dec 28 2003
STATUS
approved