cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A210356 Maximum modulus in the inverse of Hilbert's matrix.

Original entry on oeis.org

1, 12, 192, 6480, 179200, 4410000, 133402500, 4249941696, 122367445200, 3480673996800, 117643011932160, 3659449159080000, 106518477825760000, 3521767173114190000, 114708987924290760000, 3525270042097046880000, 110552468520163390156800
Offset: 1

Views

Author

T. D. Noe, Mar 28 2012

Keywords

Comments

This sequence corrects A061065, which was wrong for n > 6. These values appear at the A210357(n) position on the diagonal. Thanks to Clark Kimberling for verifying this sequence.

Crossrefs

Cf. A061065, A189765 (inverse matrix), A210357, A210358 (minimum).

Programs

  • Mathematica
    Table[im = Inverse[HilbertMatrix[n]]; Max[Abs[Flatten[im]]], {n, 20}]
  • PARI
    for(n=1,17, my(h=1/mathilbert(n),s=0); for(j=1,n, for(k=1,n, s=max(s,h[j,k]))); print1(s,", ")) \\ Hugo Pfoertner, Feb 11 2020

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

Views

Author

Roger Cuculière, May 28 2001

Keywords

Comments

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

Crossrefs

Programs

  • PARI
    { 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

Formula

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

Extensions

More terms from David Wasserman, Jun 08 2002

A210358 Minimum value in the inverse of Hilbert's matrix.

Original entry on oeis.org

1, -6, -180, -4200, -117600, -3969000, -115259760, -3030051024, -106051785840, -3363975014400, -98227319533200, -3110531785218000, -103492384705710000, -3211851661880141280, -96355549856404238400, -3303308595002047632000, -106681852579497947388000
Offset: 1

Views

Author

T. D. Noe, Apr 02 2012

Keywords

Crossrefs

Cf. A061065, A189765 (inverse matrix), A210357.

Programs

  • Mathematica
    Table[im = Inverse[HilbertMatrix[n]]; Min[Flatten[im]], {n, 20}]
  • PARI
    a(n)=vecmin(1/mathilbert(n)) \\ Hugo Pfoertner, Feb 11 2020

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

Views

Author

Jianing Song, Oct 18 2021

Keywords

Examples

			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,
  ...
		

Crossrefs

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).

Programs

  • Maple
    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
  • Mathematica
    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]
Showing 1-4 of 4 results.