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User: Adam Vellender

Adam Vellender has authored 2 sequences.

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

Adam Vellender, Dec 05 2019

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.

			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).

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).

Cf. A320577 (isosceles triangles).

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 *)

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

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)])

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

A309000 Number of strings of length n from a 3-symbol alphabet (A,B,C, say) containing at least one "A" and at least two "B"s.

Original entry on oeis.org

3, 22, 105, 416, 1491, 5034, 16365, 51892, 161799, 498686, 1524705, 4635528, 14037627, 42391378, 127763925, 384536924, 1156232175, 3474201510, 10434138825, 31326533680, 94029932643, 282194655482, 846802070205, 2540859195396, 7623517110231, 22872497487694

Offset: 3

Adam Vellender, Jul 04 2019

This sequence can be thought of as the number of ways of rolling n 3-sided dice (with sides "A", "B", and "C") and obtaining at least one A and at least two B's.

The general formula is readily proved true by counting arguments.

			Suppose three-sided dice each have sides labeled A,B,C.
If there are three dice, then ABB, BAB, and BBA are the three strings resulting from rolling the dice satisfying the property of at least one A and at least two B's, hence a(3)=3 [Note a(0)=a(1)=a(2)=0].
If there are four such dice, there are 22 such permutations, hence a(4)=22: AABB, ABAB, ABBA, ABBB, ABBC, ABCB, ACBB, BAAB, BABA, BABB, BABC, BACB, BBAA, BBAB, BBAC, BBBA, BBCA, BCAB, BCBA, CABB, CBAB, CBBA.

Index entries for linear recurrences with constant coefficients, signature (9,-31,51,-40,12).

Cf. A269914, A269915, A269916, A269917, A186244, A186314.

[3^n-2^(n+1)-n*2^(n-1)+n+1: n in [3..40]]; // Vincenzo Librandi, Jul 05 2019

Array[3^# - 2^(# + 1) - # 2^(# - 1) + # + 1 &, 27, 3] (* or *)
CoefficientList[Series[(-3 + 5 x)/((-1 + 3 x) (1 - 3 x + 2 x^2)^2), {x, 0, 26}], x] (* Michael De Vlieger, Jul 04 2019 *)

[3**n-2**(n+1)-n*2**(n-1)+n+1 for n in range(3,20)]

a(n) = 3^n - 2^(n+1) - n*2^(n-1) + n + 1.
G.f.: x^3*(-3 + 5*x)/((-1 + 3*x)*(1 - 3*x + 2*x^2)^2). - Michael De Vlieger, Jul 04 2019.
a(n) = 9*a(n-1) - 31*a(n-2) + 51*a(n-3) - 40*a(n-4) + 12*a(n-5) for n > 7. - Stefano Spezia, Jul 05 2019

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User: Adam Vellender

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

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