A092098 Number of regions that the line segments in A091908(n) cut the equilateral triangle into.

1, 6, 19, 30, 61, 78, 127, 150, 217, 246, 331, 366, 469, 510, 625, 678, 817, 870, 1027, 1080, 1261, 1326, 1519, 1566, 1801, 1878, 2107, 2190, 2437, 2520, 2791, 2886, 3169, 3270, 3559, 3678, 3997, 4110, 4447, 4548, 4921, 5034, 5419, 5550, 5899, 6078, 6487

Offset: 1

Author

Hugo Pfoertner, Feb 19 2004

Keywords

• nonn

Comments

Number of chambers in an n-sected triangle. That is, n sectors are extended from each vertex to the opposite edge of the triangle. - Eric Gottlieb, Jun 26 2005

How many chambers does the edge n-sected simplex with m vertices have? We have given just the first few terms of the case m = 3. This question is natural in the context of central hyperplane arrangements as it generalizes the braid arrangement. Mike Ackerman, Sul-Young Choi, Peter Coughlin, Japheth Wood and I originally encountered the question in the context of voting theory, where we were exploring ways to tabulate votes when voters' preferences are partially ordered. Unfortunately, it turns out that the chambers of the 3-sected simplex with n vertices are not in correspondence with the set of posets on n letters as the chain with three elements and a fourth incomparable element illustrates. - Eric Gottlieb, Jun 26 2005

"Equilateral" is not needed: the sequence counts regions correctly for any triangle with n-sected sides. Ceva's Theorem is used to deduct vanishing regions from the naive count. The first deduction is at n=15 for n odd and n=20 for n even. - Len Smiley and Brian Wick (mathclub(AT)math.uaa.alaska.edu), Jul 04 2005

Examples

			E.g. the number of chambers in the bisected triangle is six, the number of permutations on 3 letters. The number of chambers in the trisected triangle is equal to 19, the number of posets on 3 elements. - _Eric Gottlieb_, Jun 26 2005
a(2)=6: The 3 line segments cut the equilateral triangle into 6 triangles.
a(3)=19: The 3*2 line segments form 12 triangles, 3 quadrilaterals, 3 pentagons and 1 central non-regular hexagon. See pictures at Pfoertner link.
		

Links

• Hugo Pfoertner, Table of n, a(n) for n = 1..1000
• Hugo Pfoertner, Visualization of diagonal intersections in an equilateral triangle.
• Hugo Pfoertner, Visualization of diagonal intersections in an equilateral triangle. [Local copy]
• Scott R. Shannon, Image for n = 6.
• Scott R. Shannon, Image for n = 7.
• Scott R. Shannon, Image for n = 12.
• Scott R. Shannon, Image for n = 13.
• Scott R. Shannon, Image for n = 100.
• Scott R. Shannon, Image for n = 101.
• Sequences formed by drawing all diagonals in regular polygon

Crossrefs

Cf. A091908 (number of intersections), A091910 (radial locations of intersection points), A006533.

Programs

Maple

regions:=proc(n::nonnegint)
   local j,k,l,a;
   a:=0;
   if (n mod 2<>0) then
      a:=3*n^2-3*n+1
   else
      a:=3*n^2-6*n+6
   fi;
   for l from 1 to floor(n/2)-1 do
      for k from 1 to floor(n/2)-1 do
         for j from 1 to floor(n/2)-1 do
            if((n-k)*l*j=k*(n-l)*(n-j)) then
               a:=a-6
            fi
         od
      od
   od;
   return a
end proc;
seq(regions(i),i=1..100);  # Len Smiley and Brian Wick, Jun 30 2005

Mathematica

regions[n_]:=
If[Mod[n,2] == 0, 3n^2-6n+6, 3n^2-3n+1]-
   6*Count[
  Flatten@
   Table[
    Abs[(n-k)l*j - k(n-l)(n-j)],
    {j,1,Floor[n/2]-1},
    {k,1,Floor[n/2]-1},
    {l,1,Floor[n/2]-1}],
  0] (* Ethan Beihl, Oct 13 2016 *)

PARI

for(n=1,100,regions=0;if(n%2!=0,regions=3*n^2-3*n+1,regions=3*n^2-6*n+6);for(l=1,floor(n/2)-1,for(k=1,floor(n/2)-1,for(j=1,floor(n/2)-1,if((n-k)*l*j==k*(n-l)*(n-j),regions-=6))));print1(regions,",")) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 27 2006

Formula

Note that 3 divides a(2k) and a(2k+1)-1. - T. D. Noe, Jun 29 2005

Extensions

More terms from T. D. Noe, Jun 29 2005

Further terms from Brian Wick (mathclub(AT)math.uaa.alaska.edu), Jun 30 2005

More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 27 2006