OFFSET
1,2
COMMENTS
Let d(m) be the number of interior triple concurrencies when each vertex of an equilateral triangle emits m equal-angle cevians (angles split into m+1 equal parts).
Because d(2n)=0 (see proof below), the sequence gives the odd cases: a(n) = d(2n-1).
For every odd m there is at least one concurrency (the three angle bisectors), so a(n) >= 1 for all n.
The closed form in FORMULA is conjectural but matches computations for n <= 200.
Theorem (even parameter): For the full sequence d(m) one has d(m) = 0 for all even m. Sketch proof: Put theta = Pi/(3*(m+1)) and R(x) = 2*tan(x)/(sqrt(3)-tan(x)). Trigonometric Ceva says that i,j,k concur iff R(i*theta) * R(j*theta) * R(k*theta)=1. We have R(x) * R(pi/3-x) = 1 and, for 0 < u <= v < Pi/6, the strict inequality R(u)*R(v) < R(u+v). If m is even, at least two of i,j,k are <= m/2, so R(i*theta) * R(j*theta) < R((i+j)*theta); with the complement identity the whole product is < 1, a contradiction.
LINKS
Alexander Bogomolny, Trigonometric Form of Ceva's Theorem.
Constantinos Kourouzides, equal-angle-cevian-concurrency (project: code, tests, b-files).
Constantinos Kourouzides, Python program to compute a(n) and list solution triples.
Jim Propp and Adam Propp-Gubin, Counting Triangles in Triangles, arXiv:2409.17117 [math.CO], 2024.
FORMULA
a(n) = d(2*n-1).
Conjecture: a(n) = 6*n - 5 if n mod 5 != 0, and a(n) = 6*n+7 if n mod 5 = 0.
Conjectured g.f.: 6*x/(1-x)^2 - 5*x/(1-x) + 12*x^5/(1-x^5).
EXAMPLE
n = 1 (m = 1): the three medians concur at one interior point, so a(1) = 1.
n = 2 (m = 3): with 3 cevians per vertex there are 7 interior triple intersections, so a(2) = 7.
n = 3 (m = 5): with 5 cevians per vertex there are 13 interior triple intersections, so a(3) = 13.
CROSSREFS
KEYWORD
nonn
AUTHOR
Constantinos Kourouzides, Aug 30 2025
STATUS
approved