I am looking for the determinant

$$ \det(I_n + H_n) $$

where $I_n$ is the $n \times n$ identity matrix and $H_n$ is the $n \times n$ Hilbert matrix, whose entries are given by

$$ [H_n]_{ij} = \frac{1}{i+j-1}, \qquad\qquad 1 \le i,j \le n $$

Is anything known about this determinant for finite $n$ or about its asymptotic behaviour for $n \rightarrow \infty$?

More generally, are there results about the determinant of "identity plus Hankel" matrices or their asymptotic behaviour?

  • $\begingroup$ On the other hand, there is a nice formula for the determinant of a Cauchy matrix plus $J$, where $J$ is the $n \times n$-matrix whose all entries (not just the diagonal ones) are $1$. $\endgroup$ – darij grinberg Mar 8 '18 at 4:48

One thing you can say is that your determinant is the sum of determinants of the Cauchy matrices $C_S$ for subsets $S$ of $\{1,\ldots, n\}$, where $(C_S)_{i,j} = 1/(S_i + S_j - 1)$ for $1 \le i,j \le |S|$ (in the case $S = \emptyset$ we take the determinant to be $1$). This means $$ \det(I_n + H_n) = \sum_{S \subseteq \{1\ldots n\}} \frac{\prod_{1 \le i < j \le |S|} (S_i - S_j)^2}{\prod_{i,j=1}^{|S|} (S_i + S_j -1)}$$ For $n=1$ to $8$ your determinants are $$ 2,{\frac{29}{12}},{\frac{2927}{1080}},{\frac{659251}{224000}},{\frac{ 46508430817}{14817600000}},{\frac{616473989937916861}{ 186313420339200000}},{\frac{3577562384224548869428843}{ 1033954523962885324800000}},{\frac{1314142513507030576449489451528961} {365356847125734485878112256000000}} $$ I don't see an apparent pattern, nor does Maple's gfun package. The numerators and denominators don't seem to be in the OEIS.

EDIT: They are now. A295426 and A295427

  • $\begingroup$ The prime decompositions of denominators are kind of nice, although it is hard to see a pattern. The primes are very small compared to the denominators themselves. $\endgroup$ – Mark Sapir Feb 10 '18 at 1:26
  • 3
    $\begingroup$ Of course, all primes in the denominator for $\det(I_n + H_n)$ are at most $2n-1$. $\endgroup$ – Robert Israel Feb 11 '18 at 2:48

If you look at $H=\frac 1{(i+j-1)\pi}$ instead, then $\det(1 + H)\sim n^{3/8}$, as $n\to\infty$. This is basically proven in arXiv:1808.08009, arXiv:1905.03154


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