A005036 Number of nonequivalent dissections of a polygon into n quadrilaterals by nonintersecting diagonals up to rotation and reflection.
1, 1, 2, 5, 16, 60, 261, 1243, 6257, 32721, 175760, 963900, 5374400, 30385256, 173837631, 1004867079, 5861610475, 34469014515, 204161960310, 1217145238485, 7299007647552, 44005602441840
Offset: 1
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- T. D. Noe, Table of n, a(n) for n = 1..100
- Nikos Apostolakis, Non-crossing trees, quadrangular dissections, ternary trees, and duality preserving bijections, arXiv:1807.11602 [math.CO], July 2018.
- Malin Christensson, Make hyperbolic tilings of images, web page, 2019.
- F. Harary, E. M. Palmer, and R. C. Read, On the cell-growth problem for arbitrary polygons, computer printout, circa 1974
- F. Harary, E. M. Palmer and R. C. Read, On the cell-growth problem for arbitrary polygons, Discr. Math. 11 (1975), 371-389.
- E. V. Konstantinova, A survey of the cell-growth problem and some its variations, Com 2 MaC-KOSEF, 2001. (Archived link.)
- Index entries for "core" sequences
Crossrefs
Programs
-
Mathematica
p=4; Table[(Binomial[(p-1)n, n]/(((p-2)n+1)((p-2)n+2)) + If[OddQ[n], If[OddQ[p], Binomial[(p-1)n/2, (n-1)/2]/n, (p+1)Binomial[((p-1)n-1)/2, (n-1)/2]/((p-2)n+2)], 3Binomial[(p-1)n/2, n/2]/((p-2)n+2)]+Plus @@ Map[EulerPhi[ # ]Binomial[((p-1)n+1)/#, (n-1)/# ]/((p-1)n+1)&, Complement[Divisors[GCD[p, n-1]], {1, 2}]])/2, {n, 1, 20}] (* Robert A. Russell, Dec 11 2004 *) Table[(3Binomial[3n,n]/(2n+1)-Binomial[3n+1,n]/(n+1)-If[OddQ[n],-10Binomial[(3n-1)/2,(n-1)/2]-If[1==Mod[n,4],4Binomial[(3n-3)/4,(n-1)/4],0],-6Binomial[3n/2,n/2]]/(n+1))/8,{n,0,30}] (* Robert A. Russell, Jun 19 2025 *)
Formula
a(n) ~ 3^(3*n + 1/2) / (sqrt(Pi) * n^(5/2) * 2^(2*n + 4)). - Vaclav Kotesovec, Mar 13 2016
a(n) = A005034(n) - A369315(n) = (A005034(n) + A047749(n)) / 2 = A369315(n) + A047749(n). - Robert A. Russell, Jan 19 2024
G.f.: (3*G(z) - G(z)^2 + 6*G(z^2) + 5z*G(z^2)^2 + 2z*G(z^4)) / 8, where G(z)=1+z*G(z)^3 is the g.f. for A001764. - Robert A. Russell, Jun 19 2025
Extensions
More terms from Sascha Kurz, Oct 13 2001
Name edited by Andrew Howroyd, Nov 20 2017
Comments