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.

Previous Showing 11-12 of 12 results.

A361240 Number of nonequivalent noncrossing triangular cacti with n triangles up to rotation and reflection.

Original entry on oeis.org

1, 1, 1, 4, 19, 124, 931, 7801, 68685, 630850, 5966610, 57808920, 571178751, 5737672339, 58455577800, 602859484608, 6283968796705, 66119472527814, 701526880303315, 7498841163925819, 80696081185766970, 873654670250482120, 9510760874015305314, 104056578392127906720
Offset: 0

Views

Author

Andrew Howroyd, Mar 06 2023

Keywords

Crossrefs

Column 3 of A361239.

Formula

a(2*n) = (A361237(2*n) + A002294(n))/2; a(2*n+1) = (A361237(2*n+1) + A118970(n))/2.

A367872 Number of dissections of a convex (4n+4)-sided polygon into n hexagons and one square (up to equivalence).

Original entry on oeis.org

1, 4, 30, 272, 2695, 28080, 302064, 3321120, 37095201, 419276660, 4782798020, 54960207120, 635339153865, 7380876649216, 86101923008160, 1007980225327680, 11836181297108565, 139353762142502100
Offset: 0

Views

Author

F. Chapoton, Feb 22 2024

Keywords

Comments

This sequence counts dissections of a convex 4n+4-sided polygon into one square and n hexagons, modulo a simple equivalence relation. The equivalence relation is not defined by a group, but by local moves. Consider the octagon formed by a hexagon adjacent to the square. The local move is half-rotation of such octagons.
It seems that a(n) is divisible by n+1.

Examples

			For n=0, there is just one square, so that a(0)=1. For n=1, one can dissect an octagon in 8 ways into a hexagon and a square. In this case, the equivalence relation just relates every such dissection to its half rotated image, so that a(1)=4.
		

Crossrefs

Cf. A174687, A185113 (similar), A118970 (related).

Programs

  • Mathematica
    Table[Binomial[5*n + 2, n]*(n + 3)/(4*n + 3), {n, 0, 50}]
  • PARI
    for(n=0,25, print1(binomial(5*n+2,n)*(n+3)/(4*n+3), ", "))
  • Sage
    def A367872(n):
        return binomial(5*n+2, n) * (n+3) / (4*n+3)
    

Formula

a(n) = binomial(5*n+2,n)*(n+3)/(4*n+3).
Previous Showing 11-12 of 12 results.