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.

A277652 Numerators of factorial moments of order 2 for the number of parts in dissections of rooted and convex polygons.

Original entry on oeis.org

0, 0, 4, 40, 312, 2212, 14920, 97632, 626080, 3957448, 24747948, 153483720, 945638232, 5795135820, 35357242128, 214919392128, 1302250826880, 7869116134672, 47437683195220, 285373276253352, 1713562776624952, 10272384482513140, 61489533128765784, 367581030765071200
Offset: 0

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Author

Ricardo Gómez Aíza, Oct 26 2016

Keywords

Comments

a(n)/A001003(n) is the factorial moment of order two for the number of parts in a (uniform) random (rooted) dissection of a convex (n+2)-gon.

Examples

			A convex 3-gon is a triangle. There is only one dissection of a rooted triangle, with one single part. The factorial moment of order two is therefore 0 and hence a(1) = 0.
A convex 4-gon is a quadrilateral. There are three dissections of a rooted quadrilateral, two with two parts and one with one part. Then the expectation of the number of parts is 5/3, and the expectation of the number of parts squared is 9/3, hence the factorial moment of order two is 9/3 - 5/3 = 4/3. The second Schröder number is A001003(2) = 3, therefore a(2) = 4.
		

Crossrefs

Denominators are the Schröder numbers A001003.

Programs

  • Maple
    s := (z^2-6*z+1)^(1/2): g := z/s^3-(1/s-(z+1-s)/(4*z))/2: ser := series(g,z,30):
    seq(coeff(ser,z,n), n=0..23); # Peter Luschny, Nov 17 2016
  • Mathematica
    CoefficientList[Series[z/Sqrt[(z^2 - 6*z + 1)^3] - (1/Sqrt[z^2 - 6*z + 1] - (z + 1 - Sqrt[z^2 - 6*z + 1])/(4*z))/2, {z, 0, 20}], z]

Formula

a(n) = A002695(n) - A035029(n-1), n >= 1.
G.f.: (z/sqrt(z^2 - 6*z + 1)^3) - (1/sqrt(z^2 - 6*z + 1) - (z + 1 - sqrt(z^2 - 6*z + 1))/(4*z))/2.
D-finite with recurrence (-n^3-5*n^2-6*n)*a(n)+(6*n^3+27*n^2+35*n+12)*a(n+1)+(-n^3-4*n^2-3*n)*a(n+2) = 0. - Robert Israel, Nov 18 2016