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%I #17 Sep 07 2013 14:24:39
%S 1,4,-6,12,9,-36,192,30,-180,180,16,-120,1200,240,-2700,6480,-140,
%T 1680,-4200,2800,25,-300,4800,1050,-18900,79380,-1400,26880,-117600,
%U 179200,630,-12600,56700,-88200,44100,36,-630,14700,3360,-88200,564480,-7560,211680
%N Triangle in which row n has the n(n+1)/2 elements of the lower triangular part of the inverse of the n-th order Hilbert matrix.
%C The n-th order Hilbert matrix has elements h(i,j) = 1/(i+j-1) for 1 <= i,j <=n. Only the lower triangular matrix is shown because the Hilbert matrix and its inverse are symmetric. The n-th row begins with n^2 and ends with A000515(n+1).
%C The sums of select rows of the inverse matrix are sequences A002457, A002736, A002738, A007531, and A054559.
%C The largest magnitude in the matrix is A210356(n). - T. D. Noe, Mar 28 2012
%C The sum of the elements of the n-th matrix is n^2. - T. D. Noe, Apr 02 2012
%H T. D. Noe, <a href="/A189765/b189765.txt">Rows n = 1..25, flattened</a>
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/HilbertMatrix.html">MathWorld: Hilbert Matrix</a>
%F a(n,i,j) = (-1)^(i+j) (i+j-1) binomial(n+i-1, n-j) binomial(n+j-1, n-i) binomial(i+j-2, i-1)^2 is the (i,j) element of the inverse of the n-th Hilbert matrix.
%e Row 3 is 9, -36, 192, 30, -180, 180 which corresponds to the inverse
%e 9 -36 30
%e -36 192 -180
%e 30 -180 180
%t lowerTri[m_List] := Module[{n = Length[m]}, Flatten[Table[Take[m[[i]], i], {i, n}]]]; Flatten[Table[lowerTri[Inverse[HilbertMatrix[n]]], {n, 6}]]
%Y Cf. A002457, A002736, A002738, A005249 (determinant), A007531, A054559, A189766 (trace).
%K sign,tabf
%O 1,2
%A _T. D. Noe_, May 02 2011
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