A210356 -id:A210356 - cp's OEIS frontend

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

T. D. Noe, May 02 2011

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

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

T. D. Noe, Rows n = 1..25, flattened
Eric W. Weisstein, MathWorld: Hilbert Matrix

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

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

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.

A210357 Location of the maximum modulus in the inverse of Hilbert's matrix.

Original entry on oeis.org

1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8, 9, 10, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 21, 22, 22, 23, 24, 24, 25, 26, 27, 27, 28, 29, 29, 30, 31, 31, 32, 33, 34, 34, 35, 36, 36, 37, 38, 39, 39, 40, 41, 41, 42, 43, 44, 44, 45, 46, 46, 47, 48, 48

Offset: 1

T. D. Noe, Mar 28 2012

nonn

The maximum value always occurs on the diagonal. These numbers are close to n/sqrt(2).

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A210356 (largest element in the inverse of Hilbert's matrix).

Table[im = Inverse[HilbertMatrix[n]]; pos = Position[im, Max[Abs[Flatten[im]]]]; If[Length[pos] > 1, Print[{n, pos}]; 0, pos[[1, 1]]], {n, 70}]
Table[t = Table[(2*i-1) Binomial[n+i-1, n-i]^2 * Binomial[2*i-2, i-1]^2, {i, n}]; Position[t, Max[t]][[1, 1]], {n, 100}]

A061065 For n <= 6, entry of maximal modulus in the inverse of the n-th Hilbert matrix. For n >= 3, this is the (n-1,n-1)-th entry.

Original entry on oeis.org

1, 12, 192, 6480, 179200, 4410000, 100590336, 2175421248, 45229916160, 912328045200, 17965673440000, 346945899203904, 6592659294154752, 123580568462478400, 2289795064260480000, 42003815644116000000

Offset: 1

Roger Cuculière, May 28 2001

Incorrect version of the largest element in the inverse of Hilbert's matrix. See A210356 for the correct version. See A210357 for the location of the maximal value. - T. D. Noe and Clark Kimberling, Mar 28 2012

Harry J. Smith, Table of n, a(n) for n = 1..100

Cf. A005249, A000515, A210356, A210357.

{ for (n=1, 100, if (n>2, a=((2*n-2)^2)*(2*n-3)*binomial(2*n-4, n-2)^2, if (n==1, a=1, a=12)); write("b061065.txt", n, " ", a) ) } \\ Harry J. Smith, Jul 17 2009

For n >= 3, a(n) = ((2n-2)^2)*(2n-3)*C(2n-4, n-2)^2. - David Wasserman, Jun 08 2002

More terms from David Wasserman, Jun 08 2002

A348419 Triangular table read by rows: T(n,k) is the k-th entry of the main diagonal of the inverse Hilbert matrix of order n.

Original entry on oeis.org

1, 4, 12, 9, 192, 180, 16, 1200, 6480, 2800, 25, 4800, 79380, 179200, 44100, 36, 14700, 564480, 3628800, 4410000, 698544, 49, 37632, 2857680, 40320000, 133402500, 100590336, 11099088, 64, 84672, 11430720, 304920000, 2134440000, 4249941696, 2175421248, 176679360

Offset: 1

Jianing Song, Oct 18 2021

			The inverse Hilbert matrix of order 4 is given by
  [  16  -120   240  -140]
  [-120  1200 -2700  1680]
  [ 240 -2700  6480 -4200]
  [-140  1680 -4200  2800].
Hence the 4th row is 16, 1200, 6480, 2800.
The first 8 rows of the table are:
  1,
  4, 12,
  9, 192, 180,
  16, 1200, 6480, 2800,
  25, 4800, 79380, 179200, 44100,
  36, 14700, 564480, 3628800, 4410000, 698544,
  49, 37632, 2857680, 40320000, 133402500, 100590336, 11099088,
  64, 84672, 11430720, 304920000, 2134440000, 4249941696, 2175421248, 176679360,
  ...

Jianing Song, Table of n, a(n) for n = 1..5050 (first 100 rows)
Eric Weisstein's World of Mathematics, Hilbert Matrix

Cf. A189766 (row sums), A189765, A005249.
A210356 gives the maximum value of each row and A210357 gives the positions of the maximum values.
Main diagonal gives A000515(n-1).

T:= n-> (M-> seq(M[i, i], i=1..n))(1/LinearAlgebra[HilbertMatrix](n)):
seq(T(n), n=1..8);  # Alois P. Heinz, Jun 19 2022

T[n_, k_] := Inverse[HilbertMatrix[n]][[k, k]]; Table[T[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* Amiram Eldar, Oct 18 2021 *)

PARI
```
T(n,k) = (1/mathilbert(n))[k,k]
```

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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.

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A210357 Location of the maximum modulus in the inverse of Hilbert's matrix.

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A061065 For n <= 6, entry of maximal modulus in the inverse of the n-th Hilbert matrix. For n >= 3, this is the (n-1,n-1)-th entry.

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A348419 Triangular table read by rows: T(n,k) is the k-th entry of the main diagonal of the inverse Hilbert matrix of order n.

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