A330197 Number of scalene triangles whose vertices are the vertices of a regular n-gon.

0, 0, 0, 12, 14, 32, 54, 80, 110, 168, 208, 280, 360, 448, 544, 684, 798, 960, 1134, 1320, 1518, 1776, 2000, 2288, 2592, 2912, 3248, 3660, 4030, 4480, 4950, 5440, 5950, 6552, 7104, 7752, 8424, 9120, 9840, 10668, 11438, 12320, 13230, 14168, 15134, 16224, 17248

Offset

3,4

Comments

The number of scalene triangles equals (number of triangles, i.e., binomial(n,3)) - (number of isosceles triangles).

The general formula is readily proved true by counting arguments.

Links

Colin Barker, Table of n, a(n) for n = 3..1000

Index entries for linear recurrences with constant coefficients, signature (0,2,2,-1,-4,-1,2,2,0,-1).

Formula

a(n) = binomial(n,3) - A320577(n).

a(n) = C(n,3)-n*(n-1)/2 if n mod 6 = 1 or 5; C(n,3)-n*(n-2)/2 if n mod 6 = 2 or 4; C(n,3)-n*(3*n-7)/6 if n mod 6 = 3; C(n,3)-n*(3*n-10)/6 otherwise [C(n,k) denoting binomial coefficients].

G.f.: 2*x^6*(2+x)*(3+x*(2+x))/((x-1)^4*(x+1)^2*(1+x+x^2)^2).

a(n) = 2*a(n-2) + 2*a(n-3) - a(n-4) - 4*a(n-5) - a(n-6) + 2*a(n-7) + 2*a(n-8) - a(n-10) for n>12. - Colin Barker, Jan 08 2020

Example

Trivial cases:
a(3)=0 since the only triangle formed by joining vertices is equilateral.
a(4)=a(5)=0 since all such triangles are isosceles.
For higher n, since a triangle is formed by choosing 3 vertices and joining them, there are C(n,3) such triangles. To obtain the number of scalene triangles, subtract the number of isosceles triangles (A320577).

Mathematica

a[n_] := If[Mod[n, 6]==1 || Mod[n, 6]==5, Binomial[n, 3]-Binomial[n, 2], If[Mod[n, 6]==2 || Mod[n, 6]==4, Binomial[n, 3]-n*(n-2)/2,
If[Mod[n, 6]==3, Binomial[n, 3]-n*(3*n-7)/6, Binomial[n, 3]-n*(3*n - 10)/6]]]; Array[a, 20, 3]
LinearRecurrence[{0, 2, 2, -1, -4, -1, 2, 2, 0, -1}, {0, 0, 0, 12, 14, 32, 54, 80, 110, 168}, 50] (* Harvey P. Dale, Aug 20 2021 *)

Prog

(Python)
from sympy import binomial
def a(n):
    assert (n>=3), "Sequence a(n) defined for n>=3"
    m = n % 6
    Cn3 = binomial(n, 3)
    if m in [1, 5]:   return Cn3 - (n*(n-1))//2
    elif m in [2, 4]: return Cn3 - (n*(n-2))//2
    elif m==3:       return Cn3 - (n*(3*n-7))//6
    else:            return Cn3 - (n*(3*n-10))//6
print([a(k) for k in range(3, 51)])
(PARI) concat([0, 0, 0], Vec(2*x^6*(2 + x)*(3 + 2*x + x^2) / ((1 - x)^4*(1 + x)^2*(1 + x + x^2)^2) + O(x^60))) \\ Colin Barker, Jan 08 2020

Crossrefs

Cf. A320577 (isosceles triangles).

Sequence in context: A238228 A214504 A140810 * A127401 A332876 A256786

Adjacent sequences: A330194 A330195 A330196 * A330198 A330199 A330200

Keyword

nonn,easy

Author

Adam Vellender, Dec 05 2019

Status

approved