The matrix I am inquiring about here is the $n \times n$ matrix where the entry $A_{ij}$ is $\frac{1}{(i+j1)^2}$. The $2 \times 2$ matrix looks like $$ \begin{pmatrix} 1 & 1/4 \\ 1/4 & 1/9 \end{pmatrix}. $$ The $3 \times 3$ case looks like $$ \begin{pmatrix} 1 & 1/4 & 1/9 \\ 1/4 & 1/9 & 1/16 \\ 1/9 & 1/16 & 1/25 \end{pmatrix}. $$ The $n \times n$ matrix looks like $$ \begin{pmatrix} 1 & 1/4 & \cdots & 1/(n^2) \\ 1/4 & 1/9 & \cdots & 1/{(n+1)^2} \\ \vdots & \vdots & \cdots & \vdots \\ 1/(n^2) & 1/{(n+1)^2} & \cdots & 1/{(2n1)^2} \end{pmatrix}. $$ If anyone has any information or knows of any papers talking about this matrix please let me know. Thanks!

4$\begingroup$ I think this could stand some motivation. $\endgroup$ – Todd Trimble♦ Jul 10 '13 at 17:40

10$\begingroup$ It's the Hadamard product of the Hilbert matrix with itself, so it is positive definite and its determinant is positive. $\endgroup$ – Clément de Seguins Pazzis Jul 10 '13 at 17:46

1$\begingroup$ You can give a formula for the determinant using the JacobiTrudi identities $\endgroup$ – John WiltshireGordon Jul 10 '13 at 18:01

4$\begingroup$ When $n=5$, the numerator of the determinant is divisible by the large prime $179357$. This implies that, unlike the Hilbert matrix, there is no simple formula for the determinant as a ratio of products of factorials. $\endgroup$ – Neil Strickland Jul 10 '13 at 18:02
Not an answer, but an amusing observation: The determinant of the matrix usually (but not always) has denominator a perfect square (empirically), while the numerator always seems to have a huge prime factor.
EDIT Actually, it seems that the expectation of $\log(P(n))/\log(n),$ where $P(n)$ is the largest prime factor of $n$ approaches the GolombDickman constant, which is about 0.62. In view of this, the largest prime factor of the numerator is pretty much par for the course, or at least, not obviously NOT par for the course.
n=1: det = 1/1
n=2: det=7/(12^2)
n=3 det = 647/(2160^2)
n=4 det = (19 * 571)/(672000^2)
n=5 det = (179 * 179357)/(7*4233600000^2)
n=6 det = (97 * 157 * 384191938531)/(186313420339200000^2)
n=7 det = (23 * 1280587616051046200369)/(2067909047925770649600000^2)
n=8 det = (317 * 6337 * 25997 * 87403 * 511645991608091)/(365356847125734485878112256000000^2)
(and some more. Note that mysterious 7 which keeps popping up in the denominator, for n=5,9,11 )
n=9 det = (55091 * 7731550926975871647518813143593349)/(7 * 146968826339795671126721851844198400000000^2)
n=10 det = (257 * 47360083 * 530916328215423816923887043836865928533)/(15402297982638230438765209613012092908994560000000000^2)
n=11 det = (31 * 1193 * 2647 * 538580971 * 957346850101 * 71222443011485886519799225151)/(7 * 175251348661711183890804992735665222783492739007774720000000000^2))

$\begingroup$ These seem to be related to the numerators and denominators of permanents of the Hilbert matrices: oeis.org/A101812 oeis.org/A101811 oeis.org/A092326 $\endgroup$ – Steve Huntsman Jul 10 '13 at 18:58

$\begingroup$ Cool! There may be a theorem lurking here... $\endgroup$ – Igor Rivin Jul 10 '13 at 19:09

4$\begingroup$ Indeed there is. See (3.9) of arxiv.org/abs/math/9902004 $\endgroup$ – Steve Huntsman Jul 10 '13 at 19:13

$\begingroup$ Very cool! Still doesn't explain the humongous prime factors, though :( $\endgroup$ – Igor Rivin Jul 10 '13 at 19:21

$\begingroup$ Is there some reason you didnt factor the denominator for the $n=4$ case into its squareful and squarefree parts like the others? Note that its squarefree part also contains 7 (along with 3 and 5)... $\endgroup$ – ARupinski Jul 11 '13 at 0:34
The determinant formula for $\det (1/(x_i+y_j)^2)_{i,j}$ is due to Borchardt, see Krattenthaler's article given in Steve's answer, which contains a $(q)$deformation of it as well. I want to mention, that Emmanuel Preissmann conjectured also a formula for the generalized Cauchy determinant identity
$$\det \left(\frac{1}{(x_i+y_j)^k}\right)_{i,j}$$ for every $k\ge 1$, see http://ipg.epfl.ch/~leveque/Conjectures/cauchy.pdf.
Further generalizations of Cauchy's determinant identity are given in Okada's paper
http://www.math.kobeu.ac.jp/publications/rlm18/9.pdf, where $(1.6)$ is again Borchardt's identity.
See (3.9) and (3.10)/Theorem 25 in Krattenthaler's "Advanced determinant calculus", http://arxiv.org/abs/math/9902004

$\begingroup$ Alas, a cursory web search reveals nothing about the permanent of Hilbert matrices... $\endgroup$ – Steve Huntsman Jul 10 '13 at 19:23

$\begingroup$ It is given in Theorem $25$, for $q=1$. $\endgroup$ – Dietrich Burde Jul 10 '13 at 19:29

$\begingroup$ @DietrichBurde: the paper states that "equation (3.9) results immediately from (3.10) by setting $q = 1$." Am I missing an explicit formula for the permanent? $\endgroup$ – Steve Huntsman Jul 10 '13 at 19:46

$\begingroup$ Sorry, I just meant that the comparison of $(3.9)$ and $(3.10),q=1$ yields a formula for the permanent of the matrix $(1/(x_iy_j)$. $\endgroup$ – Dietrich Burde Jul 10 '13 at 20:04

$\begingroup$ Ah, I see. I had hoped that the specific instance of the Hilbert matrix might have a simpler form. $\endgroup$ – Steve Huntsman Jul 10 '13 at 20:35
Just a couple observations, building upon and explaining why Igor Rivin found that the denominators are often perfect squares or close to being such.
Let $A$ be the matrix with components $A_{ij} = 1/(x_i + y_j)^2$. Expressing its determinant as a sum over permutations and putting everything on the same denominator gives $$ \det A = \frac{P(\{x_i\}, \{y_i\})}{\prod_{i,j} (x_i + y_j)^2} \,, $$ for some polynomial $P$ with integer coefficients. Also, if two $x_i$ or two $y_i$ are equal, then the corresponding rows/columns are equal, and the determinant vanishes. Hence $$ \det A = \frac{Q(\{x_i\}, \{y_i\}) \prod_{i\neq j} (x_i  x_j) (y_i  y_j)}{\prod_{i,j} (x_i + y_j)^2} \,. $$ Actually, as pointed out in other answers, the Cauchy formula provides $Q$ as the permanent of some matrix, but all I need here is to know $Q$ has integer coefficients.
In your particular case, the products give $$ \frac{\prod_{i\neq j} (i  j) ((i  1)  (j  1))}{\prod_{i,j} (i + j  1)^2} = \frac{\prod_{i < j} (j  i)^4}{\prod_{i,j} (i + j  1)^2} = \prod_{j=1}^N \left[(j  1)!^4 \frac{(j  1)!^2}{(j + N  1)!^2} \right] = \frac{\prod_{j=0}^{N1} j!^8}{\prod_{j=0}^{2N1} j!^2} $$ Note that $j!^8$ divides $(2j)!(2j+1)!$, so the numerator divides the denominator. All in all, $$ \det A = \frac{Q}{\prod_{j=0}^{2N1} j!^2 \bigg/ \prod_{j=0}^{N1} j!^8} $$ for some integer $Q$. The denominator appearing in this formula is typically rather close to the ones given in Igor Rivin's answer, and are trivially squares: $1$, $12^2$, $2160^2$, $6048000^2 = 672000^2 \cdot 9^2$, etc. The discrepancy is of course accounted for by cancellations between $Q$ and the product of factorials. Again using Igor Rivin's data, I find (indices denote the size of the matrix) $$ \begin{aligned} Q_1 &= 1 \\ Q_2 &= 7 \\ Q_3 &= 647 = 2^3 \cdot 3^4  1 = 8 \cdot (16 \cdot (6  1) + 1)  1 \\ Q_4 &= 878769 = 16 \cdot (12 \cdot (32 \cdot (144  1) + 1)  1) + 1 \\ Q_5 &= 18203480001 = 40000 \cdot ( 48 \cdot ( 120 \cdot (80  1) + 1)  1) + 1 \\ Q_6 &= 5850859031888599 \,. \end{aligned} $$ Here I've included some expressions which may or may not generalize. In any case, it is certain that $Q$ will not be expressed as a (simple) product, as it involves some very large prime factors.