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%I #52 Jun 22 2022 23:35:38
%S 1,1,8,1296,7962624,2985984000000,100306130042880000000,
%T 416336312719673760153600000000,
%U 281633758444745849464726940024832000000000,39594086612242519324387557078266845776303882240000000000
%N a(n) = (n!)^(n+1).
%C 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
%C From _Andrew Weimholt_, Sep 23 2009: (Start)
%C 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.
%C 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.
%C Cf. A165644 (same idea, but reflections are distinct). A165642 and A165643 are the corresponding sequences for cubes instead of simplices. (End)
%C 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
%C 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
%H Matvey Borodin, Eric Chen, Aidan Duncan, Tanya Khovanova, Boyan Litchev, Jiahe Liu, Veronika Moroz, Matthew Qian, Rohith Raghavan, Garima Rastogi, and Michael Voigt, <a href="https://arxiv.org/abs/2201.00645">Sequences of the Stable Matching Problem</a>, arXiv:2201.00645 [math.HO], 2021.
%H Peter J. Taylor, <a href="https://mathoverflow.net/a/423415/136218">Determinant of matrix with Stirling numbers as elements</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial</a>
%F a(n) = (n!)^(n+1) = a(n-1) * n^n * n!.
%F a(n) = A000178(n)*A002109(n), i.e., product of superfactorials and hyperfactorials. - _Henry Bottomley_, Nov 13 2009
%F 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
%p (n!)^(n+1);
%p 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
%p seq(mul(mul(j,j=1..n), k=0..n), n=0..8); # _Zerinvary Lajos_, Sep 21 2007
%t Table[(n!)^(n+1),{n,0,8}] (* _Harvey P. Dale_, Apr 30 2012 *)
%o (Magma) [Factorial(n)^(n+1): n in [0..10]]; // _Vincenzo Librandi_, Nov 25 2015
%Y Cf. A036740.
%K easy,nonn,nice
%O 0,3
%A Nicolau C. Saldanha (nicolau(AT)mat.puc-rio.br), Mar 10 2004
%E Edited by _N. J. A. Sloane_, Oct 24 2009 at the suggestion of _R. J. Mathar_
%E a(9) from _Vincenzo Librandi_, Nov 25 2015
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