Young-Fibonacci version of Nekrasov-Okounkov

Question

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.

Answer 1

This post is almost an answer to the question. It provides a method to directly calculate $F(z;t)$ and bypasses the obstacles related to computing the expectation values $\langle E_k \rangle_n$.

It's not hard to see that

\begin{equation} \begin{array}{lll} \displaystyle H_t(1v) &\displaystyle = \, \big(n^2 - t \big) H_t(v) &\text{if $|v|=n-1$} \\ \displaystyle H_t(2v) &\displaystyle = \, \big( 1 - t \big) \big( n^2 - t \big) H_t(v) &\text{if $|v|=n-2$} \end{array} \end{equation}

and consequently

\begin{equation} \langle H_t \rangle_n \ = \ \left\{ \begin{array}{ç} \displaystyle \ \ \ \, {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ \ + \\ \displaystyle {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{array} \right. \end{equation}

or, after setting $\Bbb{f}_t(n) := {1 \over {n!}} \langle H_t \rangle_n$

\begin{equation} n^2 \Bbb{f}_t(n) \ = \ \left\{ \begin{array}{c} \displaystyle \big(n^2 - t \big)\Bbb{f}_t(n-1) \\ + \\ \displaystyle \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{array} \right. \end{equation}

Of course $F(z;t) = \sum_{n \geq 0} \, \Bbb{f}_t(n) z^n$ and it will satisfy the following second order homogeneous ODE in light of the linear recurrence:

\begin{equation} \begin{array}{c} (\dagger \! \dagger \! \dagger) \ \ \displaystyle A(z) {d^2 \over {dz^2}} F(z;t) \ + \ B(z) {d \over {dz}} F(z;t) \ + \ C(z) F(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} A(z) \ = \ (1-t)z^3 + z^2 - z \\ B(z) \ = \ 5(1-t)z^2 + 3z - 1 \\ C(z) \ = \ (1-t)(4-t)z + (1-t) \end{array} \end{array} \end{equation}

I don't know how to solve this ODE for general values of the parameter $t$. Nevertheless, by construction, $F(z;0)$ must be the generating function for the Fibonacci numbers, i.e.

\begin{equation} \begin{array}{l} \displaystyle F(z;0) &\displaystyle = \ 1 + z + 2z^2 + 3z^3 + 5z^4 + \cdots \\ &\displaystyle = \ {1 \over {(1 - z -z^2)}} \end{array} \end{equation}

and one can check directly that $(1 - z -z^2)^{-1}$ is indeed a solution to the ODE when $t=0$. When $t=1$ the ODE's general solution is $c_1 + c_2 \int z^{-1} \, (z-1)^{-2} \, dz$ where $c_1, c_2$ are constants. However, any fibonacci word $u$ with $|u| > 0$ must contain a box $\Box \in u$ with $\mathrm{h}(\Box)=1$ and so the statistic $H_1(u)$ must vanish. Combinatorial realities thus force us to select the constant solution $F(z;1) = 1$.

We can eliminate the first order term in the $(*)$-ODE by introducing an appropriate phase in the solution $F(z;t) := e^{\varphi(x;t)} \, K(z;t)$ where $\varphi(z;t):= - \log \big[\sqrt{z} \, \big( (1-t)z^2 + z -1\big) \big]$ the $(\dagger \! \dagger \! \dagger)$-ODE becomes

\begin{equation} \begin{array}{c} \displaystyle (**) \ \ A(z) {d^2 \over {dz^2}} K(z;t) \ + \ D(z) \, K(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} \displaystyle A(z) \ = \ (1-t)z^3 + z^2 - z \\ \displaystyle D(z) \ = \ {(1-t)(4-t)z^2 + (1-4t)z - 1 \over {4z}} \\ \end{array} \end{array} \end{equation}

I'm not sure if this regauging will necessarily help; nevertheless I offer it as a potential avenue to crack the nut, so to speak.

The recurrence above is clearly implemented by tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ equals the initial $n \times n$ principal minor of the following semi-infinite tridiagonal matrix:

\begin{equation} \begin{pmatrix} 1-t & 1 - {1 \over 4} t & 0 & 0 & \\ t-1 & 1 - {1 \over 4} t & 1 - {1 \over 9} t & 0 & \\ 0 & t-1 & 1 - {1 \over 9} t & 1 - {1 \over 16}t & \\ 0 & 0 & t-1 & 1 - {1 \over 16}t & & \\ & & & & \ddots & \end{pmatrix} \end{equation}

So the problem of finding $F(z;t)$ can be viewed as a special case of the general problem of evaluating generating functions of the sort

\begin{equation} \sum_{n \geq 0} \, \det (T_n) \, z^n \end{equation}

where $T_n$ is the $n \times n$ leading, principal submatrix of a infinite $\Bbb{N} \times \Bbb{N}$ tridiagonal matrix $T$. If it were possible to show that the associated semi-infinite Hankel matrix

\begin{equation} H := \ \begin{pmatrix} \Bbb{f}_t(0) & \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \\ \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \\ \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \\ \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \Bbb{f}_t(6) & \\ & & & & \ddots \end{pmatrix} \end{equation}

were positive semi-definite (i.e. all finite, principal minors of $H$ are non-negative) then a result of Alan Sokal (see theorem 2 of https://arxiv.org/pdf/1804.04498.pdf) would imply that $F(z;t)$ has an expansion as $J$-type continued fraction

\begin{equation} {\alpha_0 \over {1 - z \gamma_0 \ - \ {\displaystyle z^2 \beta_1 \over {\displaystyle 1 - z \gamma_1 \ - \ {z^2 \beta_2 \over {\displaystyle 1 - z \gamma_2 \ - \ {z^2 \beta_3 \over {\ddots}}}}}}}} \end{equation}

where $\alpha_0 \geq 0$ is a real number and

\begin{equation} \begin{array}{ll} \displaystyle \underline{\beta} &\displaystyle = \ \big(\beta_1, \ \beta_2, \ \beta_3, \ \dots \big) \ \ \text{with $\beta_k \geq 0$ for all $k \geq 1$} \\ \displaystyle \underline{\gamma} &\displaystyle = \ \big(\gamma_0, \ \gamma_1, \ \gamma_2, \ \, \dots \big) \ \ \text{with $\gamma_k \in \Bbb{R}$ for all $k \geq 0$} \end{array} \end{equation}

When $t=0$ the Hankel matrix $H$ will consists of Fibonacci numbers

\begin{equation} H := \ \begin{pmatrix} 1 & 1 & 2 & 3 & \\ 1 & 2 & 3 & 5 & \\ 2 & 3 & 5 & 8 & \\ 3 & 5 & 8 & 13 & \\ & & & & \ddots \end{pmatrix} \end{equation}

which is clearly positive semi-definite. As WimC pointed out to me here (https://math.stackexchange.com/questions/4055848/what-is-the-j-type-continued-fraction-of-the-generating-function-of-the-fibonacc) the $J$-type continued fraction of $F(z;0) = \big(1 - z -z^2 \big)^{-1}$ is itself, i.e.

\begin{equation} {1 \over {1 - 1 \cdot z \ - \ {\displaystyle 1 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\ddots}}}}}}}} \end{equation}

Initial calculations using Mathematica endorse the claim that $H$ is positive semi-definite for generic values of $t$, but the initial parameters $\beta_1, \beta_2, \beta_3$ and $\gamma_0, \gamma_1, \gamma_2$ in the $J$-fraction expansion
do not have a particularly enlightening shape from which to discern a pattern.

Any help in these matters would be greatly appreciated.

ines.

Answer 2

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.