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%I #50 Jan 04 2024 14:02:29
%S 0,0,0,8,64,320,2240,17920,121856,831488,6215680,46069760,333639680,
%T 2468257792,18538397696,138630955008,1038902624256,7848847736832,
%U 59474614157312,451122104369152,3432752856694784,26200670667276288,200322520455315456,1534319564383322112
%N The number of equilateral triangles with vertices from the vertices of the n-dimensional hypercube.
%H Albert Stadler, <a href="https://doi.org/10.1080/00029890.2021.1901495">Problems and Solutions, Problem 12261</a>, The American Mathematical Monthly, 128:6 (2021), 563.
%F a(n) = 2^n*Sum_{k=1..floor(n/3)}n!/(6*(n - 3*k)!*k!^3). - _Drake Thomas_, May 30 2021
%F 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
%F From _Vaclav Kotesovec_, Jun 01 2021: (Start)
%F E.g.f.: exp(2*x)*(-1 + hypergeom([], [1, 1], 8*x^3))/6.
%F 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).
%F a(n) ~ 8^n / (3^(3/2)*Pi*n). (End)
%F Let Exp(x, m) = Sum_{k>=0} (x^k / k!)^m, then the above e.g.f. can be stated as:
%F a(n) = (n!/3!) * [x^n] Exp(2*x, 1)*(Exp(2*x, 3) - 1). - _Peter Luschny_, Jun 01 2021
%e For n = 3, the a(3) = 8 equilateral triangles are
%e (0,0,0), (1,1,0), and (1,0,1);
%e (0,0,0), (1,1,0), and (0,1,1);
%e (0,0,0), (1,0,1), and (0,1,1);
%e (1,0,0), (0,1,0), and (0,0,1);
%e (1,0,0), (0,1,0), and (1,1,1);
%e (1,0,0), (0,0,1), and (1,1,1);
%e (0,1,0), (0,0,1), and (1,1,1); and
%e (1,1,0), (1,0,1), and (0,1,1).
%e For n = 6, the a(6) = 2240 equilateral triangles are
%e (0,0,0,0,0,0),(0,0,0,0,1,1),(0,0,0,1,0,1); and
%e (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.
%p a := n -> 2^n*(hypergeom([-n/3, (1 - n)/3, (2 - n)/3], [1, 1], -27) - 1) / 6:
%p seq(simplify(a(n)), n = 0..23); # _Peter Luschny_, May 31 2021
%t (* Based on Drake Thomas's formula *)
%t A344854[n_] := 2^n*Sum[n!/(6*(n - 3 k)!*(k!)^3), {k, 1, Floor[n/3]}]
%t 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 *)
%o (Python)
%o from sympy import hyperexpand, Rational
%o from sympy.functions import hyper
%o 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
%Y Cf. A016283 (rectangles), A345340 (squares).
%K nonn
%O 0,4
%A _Peter Kagey_, May 30 2021
%E a(9)-a(23) from _Drake Thomas_, May 30 2021
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