A091868
a(n) = (n!)^(n+1).
Original entry on oeis.org
1, 1, 8, 1296, 7962624, 2985984000000, 100306130042880000000, 416336312719673760153600000000, 281633758444745849464726940024832000000000, 39594086612242519324387557078266845776303882240000000000
Offset: 0
Nicolau C. Saldanha (nicolau(AT)mat.puc-rio.br), Mar 10 2004
- Matvey Borodin, Eric Chen, Aidan Duncan, Tanya Khovanova, Boyan Litchev, Jiahe Liu, Veronika Moroz, Matthew Qian, Rohith Raghavan, Garima Rastogi, and Michael Voigt, Sequences of the Stable Matching Problem, arXiv:2201.00645 [math.HO], 2021.
- Peter J. Taylor, Determinant of matrix with Stirling numbers as elements
- Eric Weisstein's World of Mathematics, Fibonacci Polynomial
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[Factorial(n)^(n+1): n in [0..10]]; // Vincenzo Librandi, Nov 25 2015
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(n!)^(n+1);
a[0]:=1:for n from 1 to 20 do a[n]:=product(n!, k=0..n) od: seq(a[n], n=0..8); # Zerinvary Lajos, Jun 11 2007
seq(mul(mul(j,j=1..n), k=0..n), n=0..8); # Zerinvary Lajos, Sep 21 2007
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Table[(n!)^(n+1),{n,0,8}] (* Harvey P. Dale, Apr 30 2012 *)
A165643
Number of ways to assemble an n-cube from 2n labeled (n-1)-cubes with labeled vertices, where left-handed and right-handed counterparts are considered equivalent.
Original entry on oeis.org
1, 48, 3932160, 2958824445050880, 65878553108096586952810168320, 106856067198182393582849337977733120000000000000, 26299193579608484719502346688357440131301651853401391104000000000000000
Offset: 1
For n=2, we are constructing a square from 4 labeled line-segments with labeled endpoints. Solutions which differ by a rotation or a reflection are considered equivalent. There are 3 ways to order the line-segments, and each line-segment can be oriented in 2 ways, so the total number of solutions is 3 * 2^4 = 96. (equivalently, 6 * 2^3, which more closely resembles the formula provided.) For n=3, we are constructing a cube from 6 labeled squares with labeled vertices. Without loss of generality, we can pick one labeled square to serve as our face of reference. For this face, we do not care which side of the square will face the interior of the cube as this just translates into a reflection of the cube, nor we do not care about which rotation we pick as these just translate into rotations of the cube. From this reference square, there are 5! ways to assign the remaining squares to the faces of the cube, and each square can be oriented in 8 ways (we can pick which side of the square will face the interior of the cube, and we can pick from 4 rotations). This gives 8^5 * 5! solutions.
Cf.
A165642 (same idea, but reflections are distinct).
A165644 and
A091868 are the corresponding sequences for simplices instead of cubes.
A165644
Number of ways to assemble an n-simplex from n+1 labeled (n-1)-simplices with labeled vertices, where left-handed and right-handed counterparts are considered distinct.
Original entry on oeis.org
2, 16, 2592, 15925248, 5971968000000, 200612260085760000000, 832672625439347520307200000000, 563267516889491698929453880049664000000000
Offset: 1
For n=2, construct a triangle from 3 labeled line segments with labeled endpoints. Solutions which differ by a rotation are considered equivalent, but solutions which differ by a reflection are considered distinct (assume the triangle we are constructing will be embedded in a plane, so we cannot flip it over to convert a left-handed solution to a right-handed solution). There are 2 ways to order the line segments, and each line segment can be oriented in 2 ways, so the total number of solutions is 2 * 2^3 = 16.
For n=3, construct a tetrahedron from 4 labeled triangles with labeled vertices. Assume the tetrahedron is confined to 3-space where we cannot flip it around in 4-space to convert left-handed solutions to right-handed solutions, so we consider reflections to be distinct. Without loss of generality, we can pick one labeled triangle to serve as our face of reference. For this face, we must pick which side of the triangle will face the interior of the tetrahedron, but we do not care about which rotation we pick as these just translate into rotations of the tetrahedron. From this reference triangle, there are 6 (=3!) ways to assign the remaining triangles to the faces of the tetrahedron, and each triangle can be oriented in 6 (=3!) ways (we can pick which side of the triangle will face the interior of the tetrahedron, and we can pick from 3 rotations). This gives 2 * 6^4 solutions. The factor of "2" comes from the choice of which side of the reference triangle faces the interior (a choice that goes away if we consider reflections to be equivalent).
Cf.
A091868 (same idea, but reflections are equivalent).
A165642 and
A165643 are the corresponding sequences for cubes instead of simplices.
Showing 1-3 of 3 results.
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