A344854 -id:A344854 - cp's OEIS frontend

A344559 a(n) = (1/6) * 2^(-n) * n! * [x^n] Exp(2x, 1)(Exp(2*x, 3) - 1), where Exp(x, m) = Sum_{k>=0} (x^k / k!)^m.

0, 0, 0, 1, 4, 10, 35, 140, 476, 1624, 6070, 22495, 81455, 301301, 1131494, 4230681, 15852396, 59881956, 226877648, 860447129, 3273728234, 12493453344, 47760610689, 182905145214, 701883651799, 2697952583635, 10385325566785, 40033903418860, 154534663044346

Offset: 0

Peter Luschny, Jun 01 2021

nonn

Cf. A344854.

Exp := (x, m) -> sum((x^k / k!)^m, k=0..infinity):
gf := Exp(2*x, 1)*(Exp(2*x, 3) - 1): ser := series(gf, x, 34):
seq((1/6)*2^(-n)*n!*simplify(coeff(ser, x, n)), n = 0..28);

a[n_] := (1/6) (HypergeometricPFQ[{-n/3, (1 - n)/3, (2 - n)/3}, {1, 1}, -27] - 1);
Table[a[n], {n, 0, 28}]

from sympy import hyperexpand, Rational
from sympy.functions import hyper
def A344559(n): return (hyperexpand(hyper((Rational(-n,3),Rational(1-n,3),Rational(2-n,3)),(1,1),-27))-1)//6 # Chai Wah Wu, Jan 04 2024

a(n) = A344854(n) / 2^n.

a(n) = (1/6)*(hypergeom([-n/3, (1 - n)/3, (2 - n)/3], [1, 1], -27) - 1).

A345340 The number of squares with vertices from the vertices of the n-dimensional hypercube.

Original entry on oeis.org

0, 0, 1, 6, 36, 200, 1120, 6272, 35392, 200832, 1145856, 6566912, 37779456, 218050560, 1262030848, 7322034176, 42570760192, 247970693120, 1446799212544, 8453937692672, 49463868522496, 289761061240832, 1699288462655488, 9975342691254272, 58611909535989760

Offset: 0

Peter Kagey, Jun 14 2021

nonn

			For n = 4, there are a(4) = 36 such squares, nine of which contain the origin:
(0,0,0,0),(0,0,0,1),(0,0,1,0),(0,0,1,1);
(0,0,0,0),(0,0,0,1),(0,1,0,0),(0,1,0,1);
(0,0,0,0),(0,0,0,1),(1,0,0,0),(1,0,0,1);
(0,0,0,0),(0,0,1,0),(0,1,0,0),(0,1,1,0);
(0,0,0,0),(0,0,1,0),(1,0,0,0),(1,0,1,0);
(0,0,0,0),(0,1,0,0),(1,0,0,0),(1,1,0,0);
(0,0,0,0),(0,0,1,1),(1,1,0,0),(1,1,1,1);
(0,0,0,0),(0,1,0,1),(1,0,1,0),(1,1,1,1); and
(0,0,0,0),(0,1,1,0),(1,0,0,1),(1,1,1,1).

Cf. A097861, A362706.

Cf. A001788 (2-dimensional faces), A016283 (rectangles), A344854 (equilateral triangles).

a(n) = 2^(n-2) * Sum_{k=1..floor(n/2)} n!/(2*k!*k!*(n-2*k)!). - Drake Thomas, Jun 14 2021

a(n) = 2^(n-2) * A097861(n).

cp's OEIS Frontend

A344559 a(n) = (1/6) * 2^(-n) * n! * [x^n] Exp(2x, 1)(Exp(2*x, 3) - 1), where Exp(x, m) = Sum_{k>=0} (x^k / k!)^m.

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A345340 The number of squares with vertices from the vertices of the n-dimensional hypercube.

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Author

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Formula

cp's OEIS Frontend

A344559 a(n) = (1/6) * 2^(-n) * n! * [x^n] Exp(2*x, 1)*(Exp(2*x, 3) - 1), where Exp(x, m) = Sum_{k>=0} (x^k / k!)^m.

Views

Author

Keywords

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A345340 The number of squares with vertices from the vertices of the n-dimensional hypercube.

Views

Author

Keywords

Examples

Crossrefs

Formula

A344559 a(n) = (1/6) * 2^(-n) * n! * [x^n] Exp(2x, 1)(Exp(2*x, 3) - 1), where Exp(x, m) = Sum_{k>=0} (x^k / k!)^m.