A189765 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.

1, 4, -6, 12, 9, -36, 192, 30, -180, 180, 16, -120, 1200, 240, -2700, 6480, -140, 1680, -4200, 2800, 25, -300, 4800, 1050, -18900, 79380, -1400, 26880, -117600, 179200, 630, -12600, 56700, -88200, 44100, 36, -630, 14700, 3360, -88200, 564480, -7560, 211680

Offset

1,2

Comments

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).

The sums of select rows of the inverse matrix are sequences A002457, A002736, A002738, A007531, and A054559.

The largest magnitude in the matrix is A210356(n). - T. D. Noe, Mar 28 2012

The sum of the elements of the n-th matrix is n^2. - T. D. Noe, Apr 02 2012

Links

T. D. Noe, Rows n = 1..25, flattened

Eric Weisstein's World of Mathematics, Hilbert Matrix.

Formula

M(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.

Example

Row 3 is 9, -36, 192, 30, -180, 180 which corresponds to the inverse
    9  -36   30
  -36  192 -180
   30 -180  180

Mathematica

lowerTri[m_List] := Module[{n = Length[m]}, Flatten[Table[Take[m[[i]], i], {i, n}]]]; Flatten[Table[lowerTri[Inverse[HilbertMatrix[n]]], {n, 6}]]

Crossrefs

Cf. A002457, A002736, A002738, A005249 (determinant), A007531, A054559, A189766 (trace).

Sequence in context: A272063 A152678 A110758 * A074162 A365648 A374778

Adjacent sequences: A189762 A189763 A189764 * A189766 A189767 A189768

Keyword

sign,tabf

Author

T. D. Noe, May 02 2011

Status

approved