A345340 -id:A345340 - cp's OEIS frontend

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

Peter Kagey, May 30 2021

nonn

			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.

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

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

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

(* 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 *)

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

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

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

A346905 Triangle read by rows: T(n,k) is the number of ways of choosing a k-dimensional cube from the vertices of an n-dimensional hypercube; 0 <= k <= n.

Original entry on oeis.org

1, 2, 1, 4, 6, 1, 8, 28, 6, 1, 16, 120, 36, 8, 1, 32, 496, 200, 40, 10, 1, 64, 2016, 1120, 280, 60, 12, 1, 128, 8128, 6272, 2240, 280, 84, 14, 1, 256, 32640, 35392, 15232, 2800, 448, 112, 16, 1, 512, 130816, 200832, 103936, 34272, 2016, 672, 144, 18, 1

Offset: 0

Peter Kagey, Aug 06 2021

			Table begins:
n\k |   0       1       2       3      4     5    6    7   8  9
----+----------------------------------------------------------
  0 |   1;
  1 |   2,      1;
  2 |   4,      6,      1;
  3 |   8,     28,      6,      1;
  4 |  16,    120,     36,      8,     1;
  5 |  32,    496,    200,     40,    10,    1;
  6 |  64,   2016,   1120,    280,    60,   12,   1;
  7 | 128,   8128,   6272,   2240,   280,   84,  14,   1;
  8 | 256,  32640,  35392,  15232,  2800,  448, 112,  16,  1;
  9 | 512, 130816, 200832, 103936, 34272, 2016, 672, 144, 18, 1
One of the T(7,3) = 2240 ways of choosing a 3-cube from the vertices of a 7-cube is the cube with the following eight points:
(0,0,0,1,1,1,0);
(1,1,0,1,1,1,0);
(0,0,1,1,1,0,0);
(0,0,0,1,0,1,1);
(1,1,1,1,1,0,0);
(1,1,0,1,0,1,1);
(0,0,1,1,0,0,1); and
(1,1,1,1,0,0,1).

Columns: A000079 (k=0), A006516 (k=1), A345340 (k=2).

Cf. A346906.

T[n_, 0] := 2^n
T[n_, k_] := 2^(n - k)*Sum[n!/(k!*(i!)^k*(n - i*k)!), {i, 1, n/k}]

T(n,0) = 2^n.
T(n,k) = 2^(n-k) * Sum_{i=1..floor(n/k)} n!/(k!*(i!)^k*(n-i*k)!).
T(n,k) = 2^(n-k) * A346906(n,k).

A362706 Number of squares formed by first n vertices of the infinite-dimensional hypercube.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 6, 6, 7, 9, 13, 16, 21, 27, 36, 36, 37, 40, 45, 50, 57, 66, 78, 85, 94, 106, 121, 136, 154, 175, 200, 200, 201, 205, 211, 219, 229, 242, 258, 271, 286, 305, 327, 351, 378, 409, 444, 463, 484, 510, 539, 571, 606, 646, 690, 729, 771, 819

Offset: 1

Hugo van der Sanden, Jun 22 2023

nonn

We can take the coordinates of a vertex to represent a binary number, so we define the n-th point to have coordinates represented by the binary expansion of n-1.
Let d(m) = a(m+1) - a(m) be the shifted first differences of a(n), so that d(m) represents the additional squares introduced by the (m+1)-th vertex. Then d(0) = d(2^x) = 0; when m = 2^x + 2^y, x > y, d(m) = A115990(x - 1, x - y - 1); generally, d(m) = sum d(k) for all k formed by selecting two 1's from the binary expansion of m. Thus d(7) = d(3) + d(5) + d(6).
a(n) is a lower bound for an infinite-dimensional extension of A051602. Peter Munn notes that it is not an upper bound: for example, the vertices of a regular {k-1}-simplex duplicated at unit distance in any orthogonal direction gives T_k squares from 2k+2 points, which exceeds a(n) at 6, 10 and 12 points.

			The 6 points (0,0,0), (1,0,0), (0,1,0), (1,1,0), (0,0,1), (1,0,1) give the squares (0,0,0), (1,0,0), (0,1,0), (1,1,0) and (0,0,0), (1,0,0), (0,0,1), (1,0,1). So a(6) = 2.

Cf. A345340, A115990, A051602.

a(2^k) = A345340(k).

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

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A346905 Triangle read by rows: T(n,k) is the number of ways of choosing a k-dimensional cube from the vertices of an n-dimensional hypercube; 0 <= k <= n.

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A362706 Number of squares formed by first n vertices of the infinite-dimensional hypercube.

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