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.

A005036 Number of nonequivalent dissections of a polygon into n quadrilaterals by nonintersecting diagonals up to rotation and reflection.

Original entry on oeis.org

1, 1, 2, 5, 16, 60, 261, 1243, 6257, 32721, 175760, 963900, 5374400, 30385256, 173837631, 1004867079, 5861610475, 34469014515, 204161960310, 1217145238485, 7299007647552, 44005602441840
Offset: 1

Views

Author

Keywords

Comments

Closed formula is given in my paper linked below. - Nikos Apostolakis, Aug 01 2018
Number of unoriented polyominoes composed of n square cells of the hyperbolic regular tiling with Schläfli symbol {4,oo}. A stereographic projection of this tiling on the Poincaré disk can be obtained via the Christensson link. For unoriented polyominoes, chiral pairs are counted as one. - Robert A. Russell, Jan 20 2024

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=4 of A295260.
Polyominoes: A005034 (oriented), A369315 (chiral), A047749 (achiral), A385149 (asymmetric), A001764 (rooted), A000207 {3,oo}, A005040 {5,oo}, A005038 {5,oo} (oriented).

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