OFFSET
0,2
COMMENTS
This sequence counts dissections of a convex 3n+3-sided polygon into one triangle and n pentagons, modulo a simple equivalence relation. This equivalence relation is defined by moving the triangle according to a simple rule (not detailed here).
(The equivalence relation is not defined by a group, but by local moves. Consider the hexagon formed by a pentagon adjacent to the triangle. The local move is half-rotation of such hexagons.)
The terms seem to be odd exactly for indices in A002450. - F. Chapoton Mar 08 2020
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
FORMULA
a(n) = binomial(4*n+1,n-1)*(n+2)/n = binomial(4*n+1,n)*(n+2)/(3*n+2).
a(n) = binomial(n+2,2) * A000260(n). - F. Chapoton Feb 22 2024
EXAMPLE
For n=0, there is just one triangle, so that a(0)=1. For n=1, one can dissect an hexagon in 6 ways into a pentagon and a triangle. In this case, the equivalence relation just relates every such dissection to its half rotated image, so that a(1)=3.
MAPLE
f := RootOf(81*x - 8 - (75*x - 8)*_Z + (288*x^2 - 30*x)*_Z^2 + (256*x^3 - 27*x^2)*_Z^4): seq(coeff(series(f, x, 20), x, n), n = 0..19); # Peter Luschny, Apr 06 2023
MATHEMATICA
Table[Binomial[4*n + 1, n]*(n + 2)/(3*n + 2), {n, 0, 50}] (* G. C. Greubel, Jun 23 2017 *)
PROG
(Sage)
def A185113(n):
return binomial(4*n+1, n) * (n+2) / (3*n+2)
(PARI) for(n=0, 25, print1(binomial(4*n+1, n)*(n+2)/(3*n+2), ", ")) \\ G. C. Greubel, Jun 23 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
F. Chapoton, Feb 03 2011
STATUS
approved