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%I #15 Dec 18 2015 18:17:25
%S 1,-1,-8,1296,7962624,-2985984000000,-100306130042880000000,
%T 416336312719673760153600000000,
%U 281633758444745849464726940024832000000000
%N Determinant of n X n matrix whose (i,j)-th element is (i+j)^(n-1)
%C a(n) = A057077(n)*A091868(n-1), signed variant of A091868.
%C Concerning the general case det((u(i) + v(j))^(n-1) for 1<=i,j<=n and u(1), ...,u(n), v(1), ..., v(n) integers, the reference give the proof that det((u(i) + v(j))^(n-1) = (1/x) *(((n-1)!)^n)*y*z with the following results :
%C x = 1! * 2!*...(n-1)! ;
%C y = (u(1) - u(2))*((u(1) - u(3))*....*(u(n-1) - u(n)) ;
%C z = (v(n) - v(n-1))*((v(n) - v(n-2))*....*(v(2) - v(1)).
%C If the (i,j)-th element is (i+j)^(n-1), then u(i) = i and v(j) = n+1-j. Finally, det(n X n) = ((-1)^p)* (n!)^(n+1) with n = 2p + 1 or n=2p.
%D J. M. Monier, Algebre & geometrie, Dunod (1996), p.216.
%F a(n) = (-1)^floor(n/2)* ((n-1)!)^n.
%e a(3) = determinant(M_3) = -8 where M_3 is the matrix
%e [4 9 16]
%e [9 16 25]
%e [16 25 36]
%t a[n_] = ((-1)^((n - Mod[n, 2])/2))*((n - 1)!)^n; Table[a[n], {n, 9}] (* from Jean-François Alcover, Aug 31 2011 *)
%Y Cf. A091868.
%K sign
%O 1,3
%A _Michel Lagneau_, Apr 08 2010
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