This question addresses a hierarchy of linear recurrences
which arise from an attempt to generalize the Nekrasov-Okounkov
formula to the Young-Fibonacci setting.
A related posting
extensions of the Nekrasov-Okounkov formula
asks how one might try to extend the Nekrasov-Okounkov formula
by replacing the Plancherel measure on the Young lattice $\Bbb{Y}$
with another ergodic, central measure.
In this discussion, I want to instead replace the Young lattice $\Bbb{Y}$
by the Young-Fibonacci lattice $\Bbb{YF}$ which comes equipped with its own Plancherel measure in virtue of being a $1$-differential poset.
Allow me to briefly review some basics of the Young-Fibonacci lattice
before I state the putative $\Bbb{YF}$-version of the Nekrasov-Okounkov partition function.
Young-Fibonacci Preliminaries:
Recall that a fibonacci word $u$ is a word formed
out of the alphabet $\{1,2\}$. As a set $\Bbb{YF}$ is the
collection of a (finite) fibonacci words and $\Bbb{YF}_n$
will denote the set of fibonacci words $u \in \Bbb{YF}$ of length
$|u|=n$ where
$|u|:= a_1 + \cdots + a_k$ and where $u=a_k \cdots a_1$ is the
parsing of $u$ into its digits $a_1, \dots, a_k \in \{1,2 \}$.
The adjective Fibonacci reflects the fact that the cardinality of $\Bbb{YF}_n$
is the $n$-th Fibonacci number. I will skip defining the poset structure
on $\Bbb{YF}$ and instead I point the readers to the Wikipedia page
https://en.wikipedia.org/wiki/Young–Fibonacci_lattice. Suffice it to
say that when endowed with an appropriate partial order $\unlhd$ the set $\Bbb{YF}$ becomes a ranked, modular (but not distributive), $1$-differential lattice. R. Stanley's $1$-differential property (see https://en.wikipedia.org/wiki/Differential_poset) is key here because it implies that the function
$\mu_\mathrm{P}: \Bbb{YF} \longrightarrow \Bbb{R}_{>0}$
defined by
\begin{equation}
\begin{array}{ll}
\mu_\mathrm{P}(u)
&\displaystyle := \ { \dim^2(u) \over {|u|!}} \quad \text{where} \\
\dim(u)
&\displaystyle := \
\# \left\{
\begin{array}{l}
\text{all saturated chains $(u_0 \lhd \cdots \lhd u_n)$ in $\Bbb{YF}$} \\
\text{starting with $u_0 = \emptyset$ and ending at $u_n =u$}
\end{array}
\right\}
\end{array}
\end{equation}
restricts to a positive probability distribution $\mu^{(n)}_\mathrm{P}$
on $\Bbb{YF}_n$ for each $n \geq 0$. In fact $\mu_\mathrm{P}$ satisfies a
stronger property known as coherence: The ratios
\begin{equation}
\tilde{\mu}_\mathrm{P}(u \lhd v) \ := \
{\mu_\mathrm{P}(v) \over {\mu_\mathrm{P}(u)}}
\end{equation}
restrict to a probability distribution $\tilde{\mu}_{\mathrm{P},u}$ on the set of
covering relations $u \lhd v$
(i.e. edges in the Hasse diagram of $\Bbb{YF}$)
for any fixed $u \in \Bbb{YF}_n$.
We refer to
$\mu^{(n)}_\mathrm{P}$ as the Plancherel
measure for $\Bbb{YF}_n$. If $S:\Bbb{YF} \longrightarrow \Bbb{R}_{\geq 0}$
is some statistic let $\langle S \rangle_n$ denote its expectation
value with respect to the Plancherel measure, i.e.
\begin{equation}
\langle S \rangle_n \ := \ \sum_{|u|=n} \, {\dim^2(u) \over {n!}} \, S(u)
\end{equation}
We may visualize a fibonacci word $u \in \Bbb{YF}$
using a profile of boxes
akin to the way one depicts a partition by its Young diagram.
The following example with $u = 12112211$
should illustrate the concept of a Young-Fibonacci diagram clearly. For emphasis
each digit of the fibonacci word $u$ is written directly underneath the corresponding column of boxes:
\begin{equation}
\begin{array}{cccccccc}
& \Box & & & \Box & \Box & & \\
\Box & \Box & \Box & \Box & \Box & \Box & \Box & \Box \\
1 & 2 & 1 & 1 & 2 & 2 & 1 & 1
\end{array}
\end{equation}
A Fibonacci word $u$ will be synonymous with its Young-Fibonacci diagram
and $\Box \in u$ will indicate membership of a box.
The hook length $\mathrm{h}(\Box)$ of a box $\Box \in u$
is defined to be $1$ whenever it is in the top row; otherwise $\mathrm{h}(\Box)$
equals $1$ plus the total number of boxes directly
above it and to its right. For example the hook lengths of the boxes of
$u = 12112211$ are indicated in the tableaux below:
\begin{equation}
\begin{array}{cccccccc}
& \boxed{1 \ \ } & & & \boxed{1 \ \ } & \boxed{1 \ \ } & & \\
\boxed{11} & \boxed{10} & \boxed{8 \ \ } & \boxed{7 \ \ }
& \boxed{6 \ \ } & \boxed{4 \ \ } & \boxed{2 \ \ } & \boxed{1 \ \ }
\end{array}
\end{equation}
These graphical conventions allows us to reformulate
the value of $\mu_\mathrm{P}(u)$ in terms of (the squares of)
the hook-lenghts of $u \in \Bbb{Y}$, i.e.
\begin{equation}
\mu_\mathrm{P}(u) \ = \ \prod_{\Box \, \in \, u} \, {|u|! \over
{\mathrm{h}^2(\Box)} }
\end{equation}
This is a non-trivial observation made by R. Stanley in the course
of his work examining differential posets.
The $\Bbb{YF}$-version of the Nekrasov-Okounkov partition function:
For a fibonacci words $u \in \Bbb{YF}$
define a $t$-statistic
$H_t(u) := \prod_{\Box \, \in \, u} \, \big(\mathrm{h}^2(\Box) – t \big)$ and the $\Bbb{YF}$-Nekrasov-Okounkov partition function as
\begin{equation}
\begin{array}{ll}
F(z;t)
&\displaystyle = \ \sum_{n \geq 0} {z^n \over {n!}}
\, \langle H_t \rangle_n \\
&\displaystyle = \ \sum_{n \geq 0} {z^n \over {n!}} \,
\sum_{|u|=n} \, {\dim^2(u) \over {n!}} \, H_t(u)
\end{array}
\end{equation}
Given a fibonacci word $u$ let $E_k(u)$ be the elementary symmetric polynomial in the square hook lengths
$\mathrm{h}^2(\Box)$ for $\Box \in u$ with the conventions
that $E_k(u) = 0$ whenever $k > |u|$ and that $E_0(u) = 1$
for all $u \in \Bbb{YF}$. Following a hint from Stanley's
notes "Partition Statistics with Respect to Plancherel Measure"
(http://www-math.mit.edu/~rstan/transparencies/plancherel.ps)
we will try to compute $F(z;t)$ by working out a recursion for the expectation values $\langle E_k \rangle_n$. It will be convenient to make a change of variable $z \mapsto -z$ and consider $F^\vee(z;t)
:= F(-z;t)$ instead; the effect of this sign-change is to
replace the statistic $H_t(u)$ by $H^\vee_t(u) := \prod_{\Box \, \in \, u} \, \big(t -\mathrm{h}^2(\Box) \big)$ in the definition of the partition
function. After expanding into elementary symmetric polynomials $E_k$ we
get
\begin{equation}
\begin{array}{ll}
\displaystyle H^\vee_t(u)
&\displaystyle = \ \sum_{k=0}^{|u|} \, (-1)^k \, E_{k}(u) \, t^{|u|-k} \\ \\
&\text{— and so —} \\ \\
\displaystyle F^\vee(z;t)
&\displaystyle = \
\sum_{n \geq 0} \, {z^n \over {n!}} \,
\langle H^\vee_t \rangle_n \\
&\displaystyle = \
\sum_{n \geq 0} \, {z^n \over {n!}} \,
\sum_{k=0}^n \, (-1)^k
\, \langle E_k \rangle_n \, t^{n-k} \\
&\displaystyle = \
\sum_{k \geq 0} \, (-t)^{-k} \,
\underbrace{\sum_{n \geq 0} \, (zt)^n \, {\langle E_k \rangle_n \over {n!}}}_{\text{$= \, F^\vee_k(zt)$ see below}}
\end{array}
\end{equation}
Evaluating expectation values:
Fibonacci words $u \in \Bbb{YF}_n$ with $n \geq 2$ can be separated into two
disjoint groups: Those of the form $u=1v$ for $v \in \Bbb{YF}_{n-1}$
and those of the form $u=2v$ for $v \in \Bbb{YF}_{n-2}$. Depending on
whether the prefix of $u$ is $1$ or $2$ we can write down a recursive
formula for the value of $E_k(u) := E_k \big( \mathrm{h}^2(\Box) \big)_{\Box \, \in \, u}$ by analyzing the hook length(s) of the box(es) in the left-most
column, specifically:
\begin{equation}
\begin{array}{lll}
E_k(1v)
&= E_k(v) + n^2E_{k-1}(v)
&\text{if} \ |v| = n-1 \\
E_k(2v)
&= E_k(v) + (n^2+1)E_{k-1}(v) + n^2E_{k-2}(v)
&\text{if} \ |v| = n-2
\end{array}
\end{equation}
Using the observation that $\dim(1v) = \dim(v)$ and
$\dim(2v) = (|v| + 1)^2 \dim(v)$ we may conclude
\begin{equation}
\langle E_k \rangle_n
= \left\{
\begin{array}{l}
\displaystyle {1 \over n} \langle E_k \rangle_{n-1}
\ + \ {n-1 \over n} \langle E_k \rangle_{n-2} \\ \\
\displaystyle + \ n \langle E_{k-1} \rangle_{n-1} \ + \
{(n-1)(n^2+1) \over n} \langle E_{k-1} \rangle_{n-2} \\ \\
\displaystyle + \ n(n-1) \langle E_{k-2} \rangle_{n-2}
\end{array}
\right.
\end{equation}
If we set $\sigma_k(n) := {1 \over {n!}} \, \langle E_k \rangle_n$ then
the above recursion can be rewritten as:
\begin{equation}
(\dagger) \ \
\left\{
\begin{array}{l}
\displaystyle
n^2\sigma_k(n) \ = \
\underbrace{\sigma_k(n-1) \ + \ \sigma_k(n-2)}_{\text{homogeneous part}}
\ + \ \gamma_{<k}(n) \quad \text{where} \\ \\
\displaystyle
\gamma_{<k}(n) \ = \
\underbrace{n^2\sigma_{k-1}(n-1)
\ + \ (n^2 +1)\sigma_{k-1}(n-2)
\ + \ n^2\sigma_{k-2}(n-2)}_{\text{inductive heap of inhomogeneous junk}}
\end{array} \right.
\end{equation}
all of which can be converted, using the usual yoga of generating functions, into the following second order inhomogeneous ODE for $F^\vee_k(x) := \sum_{n \geq 0} \sigma_k(n) x^n$.
\begin{equation}
\begin{array}{c}
\displaystyle x^2 \, {d^2 \over {dx^2}} F^\vee_k(x) \ + \
x {d \over {dx}} F^\vee_k(x) \ – \
\big(x^2 + x \big) F^\vee_k(x) \\
\displaystyle \ = \ \\
\displaystyle
G_{<k}(x) \ + \ \big( \sigma_k(1) – \sigma_k(0) \big)x
\end{array}
\end{equation}
which, after setting $F^\vee_k(x) := e^x J^\vee_k(x)$, can be rewritten as
\begin{equation}
(\dagger \dagger) \ \ \left\{
\begin{array}{c}
\displaystyle x {d^2 \over {dx^2}} J^\vee_k(x)
\ + \ \big(2x + 1 \big) {d \over {dx}} J^\vee_k(x) \\
= \\
\displaystyle {1 \over x} \Big[ G_{<k}(x) \ + \ \big( \sigma_k(1) – \sigma_k(0) \big)x \Big]
\end{array} \right.
\end{equation}
where the generating function
$G_{<k}(x) = \sum_{n \geq 2} \, \gamma_k(n) x^n$ will have been evaluated earlier by induction on $k \geq 0$.
The associated homogeneous ODE of $(\dagger \dagger)$
has two nice independent solutions $Y_1(x) = 1$ and $Y_2(x)= \int x^{-1} e^{-2x} dx$ whose Wronskian is $W=x^{-1} e^{-2x}$. One starts the inductive engine beginning with $F^\vee_0(x) = e^x$ or, equivalently with $J^\vee_0(x) = 1$. For $k=1$ clearly $\sigma_1(0)=0$ and $\sigma_1(1)=1$ while
\begin{equation}
\begin{array}{ll}
\displaystyle G_{<1}(x)
&\displaystyle = \ \sum_{n \geq 2} \, {n^3 + n -1 \over {(n-1)!}} \, x^n \\
&\displaystyle = \ \big(x + 8x^2 + 6x^3 + x^4 \big) \, e^x \ – \ x
\end{array}
\end{equation}
so the ODE for $J^\vee_1(x)$ becomes
\begin{equation}
\begin{array}{c}
\displaystyle x {d^2 \over {dx^2}} J^\vee_1(x) \ + \
\big(2x+1\big) {d \over {dx}} J^\vee_1(x) \\
\displaystyle = \\
\underbrace{\big(x + 8x^2 + 6x^3 + x^4 \big) \, e^x}_{G_{<1}(x) \ + \ x }
\end{array}
\end{equation}
By variation of parameters, a particular inhomogeneous solution is
\begin{equation}
\begin{array}{rl}
\displaystyle Y_\mathrm{particular}(x)
&\displaystyle = \ v_1(x) \cdot Y_1(x) \ + \ v_2(x) \cdot
Y_2(x) \\
\displaystyle v_1(x)
&\displaystyle = \ -\int xe^{2x} \, Y_2(x) \,
\Big(x + G_{<1}(x) \Big) \, dx \\
\displaystyle v_2(x)
&\displaystyle = \ \ \ \ \ \int xe^{2x} \, Y_1(x) \,
\Big(x + G_{<1}(x) \Big) \, dx
\end{array}
\end{equation}
After solving $J^\vee_1(x)$ (and thus for $F^\vee_1(x)$) we repeat the process for $k > 1$. At each stage we solve by variation of parameters, using the two homogeneous solutions $Y_1(x)$ and $Y_2(x)$, the $(\dagger \dagger)$-ODE
whose inhomogeneous term is itself computed
from the data obtained in the previous layer of computation.
Question.
Does anyone know how to solve either the $(\dagger)$-hierarchy of
linear recurrences, the $(\dagger \!\dagger)$-hierarchy of 2nd order inhomogeneous ODEs, or equivalently the $(\dagger \! \dagger \! \dagger)$-ODE explained in the second answer/response below? By solve I mean to express the solution in terms of elementary functions, continued fractions, or else by some nice class of special functions (e.g. hypergeometric).
thanks, ines.
Best Answer
I want to propose a (recursive) method to obtain $F_k(x)$ by exploiting a formula well known from the theory of Borel re-summation. In the present context it looks like this:
\begin{equation} \begin{array}{ll} \displaystyle \sum_{n \geq 0} \, \langle E_k \rangle_n \, z^n &\displaystyle = \ \int_0^\infty \, dt \, e^{-t} \sum_{n \geq 0} \, {\langle E_k \rangle_n \over {n!}} \, (zt)^n \\ &\displaystyle = \ \int_0^\infty \, dt \, e^{-t} \, F_k(zt) \\ &\displaystyle = \ \Big( {1 \over z} \Big) \cdot \,\text{$\frak{L}$} \big\{ F_k \big\} \Big({1 \over z} \Big) \quad \text{Laplace transform} \end{array} \end{equation}
I'll illustrate the approach for $k=1$. Let $E_1(x) := \sum_{n \geq 0} \, \langle E_1 \rangle_n \, x^n$. In this case the recurrence relation, as indicated the initial post, is
\begin{equation} \underbrace{n \langle E_1 \rangle_n}_{xE_1'(x) - x} \ = \ \left\{ \begin{array}{l} \displaystyle \underbrace{\langle E_1 \rangle_{n-1}}_{xE_1(x)} \ + \ \underbrace{(n-1) \langle E_1 \rangle_{n-2}}_{x^3E_1'(x) \ + \ x^2E_1(x)} \ + \ \\ \displaystyle \underbrace{\ n^2 \langle E_0 \rangle_{n-1} \ }_{\sum_{n \geq 2} n^2 x^n} \ + \ \underbrace{(n-1)(n^2+1)\langle E_0 \rangle_{n-2}}_{\sum_{n \geq 2} (n-1)(n^2+1) \, x^n} \end{array} \right. \end{equation}
for $n \geq 2$ with the initial values $\langle E_1\rangle_0 = 0$ and $\langle E_1 \rangle_1 = 1$. I've "underbraced" the terms of the recurrence by their respective contributions to the $1$-st order non-homogeneous ODE for the generating function $E_1(x)$ which is, upon simplification,
\begin{equation} E'_1(x) \ + \ {1 \over {x-1}} \, E_1(x) \ = \ {x^3 -x^2 + 5x +1 \over {(1+x)(1-x)^5}} \end{equation}
Mathematica tells me that the general solution is of the form
\begin{equation} \ {\text{const} \over {1-x}} \ + \ {1 \over 4} {x^2+x+2 \over {(1-x)^4}} \ + \ {3 \over 8} {1 \over {(1-x)}} \log \Big( {1-x \over {1+x}} \Big) \end{equation}
The value of the constant term is forced to be $\text{const} = - {1 \over 2}$ since $\langle E_1\rangle_0 = 0$. After performing the requisite inverse transforms we should get (excuse me for mimicking Wikipedia)
\begin{equation} \begin{array}{ll} \displaystyle F_k(zt) &\displaystyle = \ \sum_{n \geq 0} \, {\langle E_1 \rangle_n \over {n!}} \, (zt)^n \\ &\displaystyle = \ {1 \over {2\pi}} \, \int_{-\pi}^{\pi} \, d\theta \, E_1 \big(zt e^{-i\theta} \big) \exp \big( e^{i\theta} \big) \\ &\displaystyle \ \ \ \ \ \text{(Not sure what this is yet.)} \end{array} \end{equation}
ines.