A344854 The number of equilateral triangles with vertices from the vertices of the n-dimensional hypercube.

0, 0, 0, 8, 64, 320, 2240, 17920, 121856, 831488, 6215680, 46069760, 333639680, 2468257792, 18538397696, 138630955008, 1038902624256, 7848847736832, 59474614157312, 451122104369152, 3432752856694784, 26200670667276288, 200322520455315456, 1534319564383322112

Offset

0,4

Links

Table of n, a(n) for n=0..23.

Albert Stadler, Problems and Solutions, Problem 12261, The American Mathematical Monthly, 128:6 (2021), 563.

Formula

a(n) = 2^n*Sum_{k=1..floor(n/3)}n!/(6*(n - 3*k)!*k!^3). - Drake Thomas, May 30 2021

a(n) = 2^n*(hypergeom([-n/3, (1 - n)/3, (2 - n)/3], [1, 1], -27) - 1) / 6. - derived from Drake Thomas's formula by Peter Luschny, May 31 2021

From Vaclav Kotesovec, Jun 01 2021: (Start)

E.g.f.: exp(2*x)*(-1 + hypergeom([], [1, 1], 8*x^3))/6.

Recurrence: (n-3)*n^2*a(n) = 2*(4*n^3 - 15*n^2 + 13*n - 4)*a(n-1) - 4*(n-1)*(6*n^2 - 21*n + 16)*a(n-2) + 8*(n-2)*(n-1)*(31*n-90)*a(n-3) - 448*(n-3)*(n-2)*(n-1)*a(n-4).

a(n) ~ 8^n / (3^(3/2)*Pi*n). (End)

Let Exp(x, m) = Sum_{k>=0} (x^k / k!)^m, then the above e.g.f. can be stated as:

a(n) = (n!/3!) * [x^n] Exp(2*x, 1)*(Exp(2*x, 3) - 1). - Peter Luschny, Jun 01 2021

Example

For n = 3, the a(3) = 8 equilateral triangles are
  (0,0,0), (1,1,0), and (1,0,1);
  (0,0,0), (1,1,0), and (0,1,1);
  (0,0,0), (1,0,1), and (0,1,1);
  (1,0,0), (0,1,0), and (0,0,1);
  (1,0,0), (0,1,0), and (1,1,1);
  (1,0,0), (0,0,1), and (1,1,1);
  (0,1,0), (0,0,1), and (1,1,1); and
  (1,1,0), (1,0,1), and (0,1,1).
For n = 6, the a(6) = 2240 equilateral triangles are
  (0,0,0,0,0,0),(0,0,0,0,1,1),(0,0,0,1,0,1); and
  (0,0,0,0,0,0),(0,0,1,1,1,1),(1,1,0,0,1,1); and all of the equilateral triangles that can be generated by mapping these under the 2^6*6! symmetries of the 6-cube.

Maple

a := n -> 2^n*(hypergeom([-n/3, (1 - n)/3, (2 - n)/3], [1, 1], -27) - 1) / 6:
seq(simplify(a(n)), n = 0..23); # Peter Luschny, May 31 2021

Mathematica

(* Based on Drake Thomas's formula *)
A344854[n_] := 2^n*Sum[n!/(6*(n - 3 k)!*(k!)^3), {k, 1, Floor[n/3]}]
nmax = 20; CoefficientList[Series[E^(2*x)*(-1 + HypergeometricPFQ[{}, {1, 1}, 8*x^3])/6, {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Jun 01 2021 *)

Prog

(Python)
from sympy import hyperexpand, Rational
from sympy.functions import hyper
def A344854(n): return (hyperexpand(hyper((Rational(-n, 3), Rational(1-n, 3), Rational(2-n, 3)), (1, 1), -27))-1)//3<<n-1 if n else 0 # Chai Wah Wu, Jan 04 2024

Crossrefs

Cf. A016283 (rectangles), A345340 (squares).

Sequence in context: A189065 A189190 A188875 * A223843 A356703 A267189

Adjacent sequences: A344851 A344852 A344853 * A344855 A344856 A344857

Keyword

nonn

Author

Peter Kagey, May 30 2021

Extensions

a(9)-a(23) from Drake Thomas, May 30 2021

Status

approved