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A092098
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Number of regions that the line segments in A091908(n) cut the equilateral triangle into.
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24
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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
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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Note that 3 divides a(2k) and a(2k+1)-1. - T. D. Noe, Jun 29 2005
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EXAMPLE
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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.
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MAPLE
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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
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MATHEMATICA
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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}],
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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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
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STATUS
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approved
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