A091868 a(n) = (n!)^(n+1).

1, 1, 8, 1296, 7962624, 2985984000000, 100306130042880000000, 416336312719673760153600000000, 281633758444745849464726940024832000000000, 39594086612242519324387557078266845776303882240000000000

Offset: 0

Author

Nicolau C. Saldanha (nicolau(AT)mat.puc-rio.br), Mar 10 2004

Keywords

• easy
• nonn
• nice

Comments

Let f(x) be a monic polynomial of degree n. Let u be any number and let m be the matrix of values f(u+i-j) for i,j=1..n. Then the determinant of m is a(n). - T. D. Noe, Aug 23 2008

From Andrew Weimholt, Sep 23 2009: (Start)

Also, 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 equivalent.

For n=2, we are constructing a triangle from 3 labeled line-segments with labeled endpoints. Solutions which differ by a rotation or a reflection are considered equivalent. Because reflections are equivalent, there is only 1 way to order the line-segments, and each line-segment can be oriented in 2 ways, so the total number of solutions is 2^3 = 8. For n=3, we are constructing a tetrahedron from 4 labeled triangles with labeled vertices. Without loss of generality, we can pick one labeled triangle to serve as our face of reference. For this face, we do not care which side of the triangle will face the interior of the tetrahedron as this just translates into a reflection of the tetrahedron, nor do we 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 6^4 solutions.

Cf. A165644 (same idea, but reflections are distinct). A165642 and A165643 are the corresponding sequences for cubes instead of simplices. (End)

a(n) is the number of preference profiles in the stable marriage problem with n women and n men, where all the men rank women in the same order. Given such a profile, the Gale-Shapley men-proposing algorithm ends in n rounds. Equivalently, this is the number of preference profiles where all the women rank men in the same order. - Tanya Khovanova and MIT PRIMES STEP Senior group, May 23 2021

a(n-1) is the determinant of the n X n matrix with elements m(i,j) = s(n+i-1,j), 1 <= i <= n, 1 <= j <= n, where s(x,y) are the unsigned Stirling numbers of the first kind. - Fabio Visonà, May 22 2022

Links

• 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

Crossrefs

Cf. A036740.

Programs

Magma

[Factorial(n)^(n+1): n in [0..10]]; // Vincenzo Librandi, Nov 25 2015

Maple

(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

Mathematica

Table[(n!)^(n+1),{n,0,8}] (* Harvey P. Dale, Apr 30 2012 *)

Formula

a(n) = (n!)^(n+1) = a(n-1) * n^n * n!.

a(n) = A000178(n)*A002109(n), i.e., product of superfactorials and hyperfactorials. - Henry Bottomley, Nov 13 2009

a(n) ~ (2*Pi)^((n+1)/2) * n^((n+1)*(2*n+1)/2) / exp(n^2 + n - 1/12). - Vaclav Kotesovec, Jul 10 2015

Extensions

Edited by N. J. A. Sloane, Oct 24 2009 at the suggestion of R. J. Mathar

a(9) from Vincenzo Librandi, Nov 25 2015