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Problem $\text A6$ on the $2024$ William Lowell Putnam Mathematical Competition was as follows.

Let $c_0, c_1, c_2, \dots$ be the sequence defined so that $$ \frac{1 - 3x - \sqrt{1 - 14x + 9x^2}}{4} = \sum_{k=0}^{\infty} c_k x^k $$ for sufficiently small $x$. For a positive integer $n$, let $A$ be the $n$-by-$n$ matrix with $i, j$-entry $c_{i + j - 1}$ for $i$ and $j$ in $\{1, \dots, n\}$. Find the determinant of $A$.

In fact, the matrix looks like

$$A = \begin{bmatrix} c_1 & c_2 & c_3 & \cdots & c_n \\ c_2 & c_3 & c_4 & \cdots & c_{n+1} \\ c_3 & c_4 & c_5 & \cdots & c_{n+2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ c_n & c_{n+1} & c_{n+2} & \cdots & c_{2n-1} \end{bmatrix}$$

Related: Putnam $2024$ — For which real polynomial $p$ is there a real polynomial $q$ such that $ p\bigl(p(x)\bigr) - x = \bigl(p(x) - x\bigr)^2 q(x) $

Integreek
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Guy Fsone
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    It seems the answer to be $10^{\frac{n(n+1)}{2}}$ – yuanming luo Apr 25 '25 at 09:48
  • If we denote by $y$ your expression, it is a solution of quadratic equation $2y^2+(3x-1)y+x=0$ which can be expressed under the form of a continued fraction ; furthermore, there exist connections between continued fractions and Hankel determinants (see here)... – Jean Marie Apr 25 '25 at 10:43
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    Putnam 2024-A6: https://kskedlaya.org/putnam-archive/ – Catalin Zara Apr 25 '25 at 10:50
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    Take a look at OEIS A379103 and OEIS A110147 – Rodrigo de Azevedo Apr 25 '25 at 21:50
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    Nathaniel Johnston posted his solution on YouTube and in writing [PDF] – Rodrigo de Azevedo Apr 25 '25 at 22:21
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    Frankly, given that the Putnam is arguably the hardest math competition in the world, with a median score of roughly 0%, it is inconsiderate of you to post such a question without providing a source. – Rodrigo de Azevedo Apr 25 '25 at 23:59
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    Off the top of my head, given my experience with similar matrices such as Hilbert matrices, I'd look for some set of functions in some $L^2[a, b]$ space such that $\langle f_m, f_n \rangle = c_{m+n-1}$, so the matrix becomes a Grammian matrix, and then you can get to the answer if you can calculate the distance from $f_n$ to $\operatorname{span}(f_1, \ldots, f_{n-1})$. – Daniel Schepler Apr 26 '25 at 12:57
  • @Rodrigo de Azevedo : as well sequence OEIS A082148... – Jean Marie Apr 27 '25 at 21:37
  • @Daniel Schepler Your comment is extremely interesting. Could you give us a reference ? And how ncould be chosen the weight function $w(x)$ and the values of $a$ and $b$ in $\int_a^b f_m(x)f_n(x)w(x)dx$ ? How to connect it with $f(x)=\frac{1 - 3x - \sqrt{1 - 14x + 9x^2}}{4}$ knowing that $f$ can be considered as a moment generating function of the corresponding measure ? – Jean Marie Apr 30 '25 at 07:57
  • [ctd] ... knowing that there exist paths from MGF to underlying measure (see here). – Jean Marie Apr 30 '25 at 09:15
  • @JeanMarie As I hinted, I don't know whether that strategy could actually work in this case since I haven't put much more thought into it. But just as an example for how to apply the method to Hilbert matrices: $H_n$ is the Grammian of $1, x, x^2, \ldots, x^{n-1}$ in $L^2[0,1]$, and since those are linearly independent, the determinant is nonzero. – Daniel Schepler Apr 30 '25 at 13:36
  • @JeanMarie Also, since polynomials $\frac{d^j}{dx^j}(x^2-x)^j$ are orthogonal, $\frac{j!}{(2j)!} \frac{d^j}{dx^j}(x^2-x)^j$ is the component of $x^j$ orthogonal to the subspace spanned by $1, \ldots, x^{j-1}$ and you can compute the norm of that using iterated integration by parts. (So here, you get some help if you're familiar with the well-known sequence of polynomials $\frac{d^n}{dx^n}(x^2-1)^n$ whose name escapes me at the moment. Especially not sure if some similar trick would work for this problem.) – Daniel Schepler Apr 30 '25 at 13:38
  • @Daniel Schepler : Legendre polynomials. Thank you very much for this discussion. I am convinced that there must exist a more "enlightning" proof than the very technical one given by Nathaniel Johnston... – Jean Marie Apr 30 '25 at 15:53

1 Answers1

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The algorithm is

   detcl[n_] := 
    (cl = 
   CoefficientList[
    Normal@Series[
         1/4 (1 - 3 x - Sqrt[1 - 14 x + 9 x^2]), 
          {x, 0, 3 n}] , 
     x];
   Det[Array[(cl[[#1 ;; #1 + n]] &), n + 1,2] ])

The result

  Table[detcl[k], {k, 1, 7}]

$$\{10,1000,1000000,10000000000,1000000000000000,1000000000000000000000,10000000000000000000000000000\}$$

Roland F
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