Following Richard Stanley's pointers regarding my earlier MO question, I decided to "scale-down" the problem and add a slight "twist" to it.

Consider the one-line partition $\lambda_n=(n)$ and its corresponding Young diagram $Y_n$, which is a $1\times n$ straight bar of $n$ cells. Now, start tiling $Y_n$ using monomers ($1\times1$ squares) and dimers ($1\times2$ rectangles). There are $F_{n+1}$ (Fibonacci number) different ways of doing so.

Next, insert the hook-lengths $h(\square)$ into each cell $\square\in Y_n$. Associate a weight: a monomer at $\square$ receives $h(\square)$, a dimer sitting on $\square$ and $\square'$ gets the product $h(\square)\cdot h(\square')$. Each tiling $T$ will have weight assigned as the sum of the weights of its monomers and dimers. Let $b_n$ be the entire sum of the weights of all possible tiltings of $Y_n$. For example, if $n=3$ then we get $$b_3=(3+2+1)+(3\cdot2+1)+(3+2\cdot1)=6+7+5=18.$$ The first few values are: $b_1=1, b_2=5, b_3=18, b_4=59, b_5=162$, this is not listed on OEIS.

QUESTION. If $G(x)=\sum_nb_nx^n$ is a generating function of $b_n$, is $G(x)$ a rational function?

Remark. I might upgrade the question later depending on responses.

POSTSCRIPT. Fedor's comment leads to $G(x)=\frac{x(1+x+5x^3-3x^4)}{(1-x-x^2)^4}$.

  • 1
    $\begingroup$ it is rational, since $b_{n}=nF_n+b_{n-1}+n(n-1)F_{n-1}+b_{n-2}$ (look at the first cell). $\endgroup$ – Fedor Petrov Apr 6 at 18:08

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