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

Author

Peter Kagey, May 30 2021

Keywords

• nonn

Examples

			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.
		

Links

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

Crossrefs

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

Programs

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 *)

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<Chai Wah Wu, Jan 04 2024

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

Extensions

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