PROOFS OF THREE GEODE CONJECTURES

TEWODROS AMDEBERHAN AND DORON ZEILBERGER

Abstract. In the May 2025 issue of the Amer. Math. Monthly, Norman J. Wildberger and Dean Rubine intoduced a new kind of multiindexed numbers, that they call ‘Geode numbers’, obtained from the hyper-Catalan numbers. They posed three intriguing conjectures about them, that are proved in this note.

1. Introduction

In a recent captivating Monthly article [2], by Norman J. Wildberger and Deane Rubine, the authors utilize a generating series to solve the general univariate polynomial equation. They also explored a “curious factorization" of this hyper-Catalan generating series, and in the penultimate section, they made three conjectures about this algebraic object that they termed the Geode array. In this note, we prove these three conjectures. At least as interesting as the actual statements of the conjectures (now theorems) is how we proved them, using several important tools of the trade.

The first tool is the multinomial theorem

(1.1)  (x_{1} + ⋯ + x_{r})^{n} = ∑_{\begin{matrix}m_{1},....m_{r}≥0 \\ m_{1}+··· +m_{r}=n \end{matrix}} \binom{n}{m_{1}, …, m_{r}}x_{1}^{m_{1}}⋯ x_{r}^{m_{r}}.

The second tool is constant-term extraction, the third is Wilf-Zeilberger (WZ) algorithmic proof theory[3] and the last-but-not-least tool is Lagrange Inversion [4] that states that: if u(t) and Φ(t) are formal Laurent series starting at t^{1} and t^{0}, respectively, then u(t) = tΦ(u(t)) implies

[t^{n}]u(t)= \frac{1}{n}[z^{n−1}]Φ(z)^{n}. (1.2)

Here [z^{n}]F(z) means the coefficient of z^{n} is the Laurent expansion of F(z). We shall use the notation CT CT CT_{z}F(z) for the constant-term of F(z). We now bring in the relevant notation adopted in [2] with a caveat that indices are shifted slightly. Consider the equation 0 = 1 − α + ∑ _{k≥1} t_{k}α^{k+1} and denote its series solution by α = S S S[t_{1}, t_{2}, …]. Letting S S S_{1} = t_{1}+ t_{2}+ ⋯, Wildberger-Rubine proved [2, Theorem 12] the existence of a (remarkable!) factorization S S S − 1 = S S S_{1}G G Gand the factor G G G[t_{1}, t_{2}, …] (that they dubbed the Geode series). Furthermore, we opt to use G[m_{1}, m_{2}, …] for the coefficient of t_{1}^{m_{1}}t_{2}^{m_{2}}⋯in the polyseries G G G[t_{1}, t_{2}, …]. We are now ready to state and prove the three conjectures from [2, p. 399]. For the sake of clarity, let’s describe the first of these in some detail. Suppose we are solving the polynomial equation 0 = 1 − α + t_{1}α^{2} + t_{2}α^{3} through the formal power series

Date: June 18, 2025.

2000 Mathematics Subject Classification. Primary, Secondary.

α = S S S[t_{1}, t_{2}] = ∑_{m_{1},m_{2}≥0}C[m_{1}, m_{2}] t_{1}^{m_{1}}t_{2}^{m_{2}}.

Consequently, the corresponding Geode series becomes G G G[t_{1}, t_{2}] = \frac{S S S[t_{1},t_{2}]−1}{t_{1}+t_{2}}. We follow closely [4] to engage the LagrangeInversion in the extraction of the coefficients C[m_{1}, m_{2}] satisfying n = m_{1} + m_{2}. Then, the amalgamation of such monomials is given by (1.2) in the form of

∑_{m_{1}+m_{2}=n}C[m_{1}, m_{2}] t_{1}^{m_{1}}t_{2}^{m_{2}}= [Y^{n}] (∑_{k=1}^{3n+1} \frac{1}{k} [z^{k−1}] (1 + Yt_{1}z^{2} + Yt_{2}z^{3})^{k})

= [Y^{n}] ∑_{m_{1},m_{2}≥0} \frac{\binom{1+2m_{1}+3m_{2}}{m_{1},m_{2},1+m_{1}+2m_{2}}}{1 + 2m_{1} + 3m_{2}}Y^{m_{1}+m_{2}}t_{1}^{m_{1}}t_{2}^{m_{2}}

= ∑_{\begin{matrix}m_{1},m_{2}≥0 \\ m_{1}+m_{2}=n \end{matrix}} \frac{\binom{1+2m_{1}+3m_{2}}{m_{1},m_{2},1+m_{1}+2m_{2}}}{1 + 2m_{1} + 3m_{2}}t_{1}^{m_{1}}t_{2}^{m_{2}}

= ∑_{m_{2}=0}^{n} \frac{\binom{1+2n+m_{2}}{n−m_{2},m_{2},1+n+m_{2}}}{1 + 2n + m_{2}}t_{1}^{n−m_{2}}t_{2}^{m_{2}}

= ∑_{k=0}^{n} \frac{\binom{n}{k}\binom{2n+1+k}{n+1+k}}{2n + 1 + k}t_{1}^{n−k}t_{1}^{k}.

For example, the following reveal both coefficients C[m_{1}, m_{2}] and G[m_{1}, m_{2}]:

∑_{m_{1}+m_{2}=3}C[m_{1}, m_{2}] t_{1}^{m_{1}}t_{2}^{m_{2}}= (t_{1} + t_{2})(5t_{1}^{2} + 16t_{1}t_{2} + 12t_{2}^{2}),

∑_{m_{1}+m_{2}=4}C[m_{1}, m_{2}] t_{1}^{m_{1}}t_{2}^{m_{2}}= (t_{1} + t_{2})(14t_{1}^{3} + 70t_{1}^{2}t_{2} + 110t_{1}t_{2}^{2} + 55t_{2}^{3}).

As a first step, we reprove that the linear term t_{2} + t_{3} divides the polynomial

P_{n}(t_{1}, t_{2}) := ∑_{k=0}^{n} \frac{\binom{n}{k}\binom{2n+1+k}{n+1+k}}{2n + 1 + k}t_{1}^{n−k}t_{2}^{k}.

This is equivalent to proving that P_{n}(−t_{2}, t_{2}) = 0, which, in turn, is equivalent to the following identity:

∑_{k=0}^{n}(−1)^{k} \frac{\binom{n}{k}\binom{2n+1+k}{n+1+k}}{2n + 1 + k} = 0.

To continue, we invoke the role of the WZ method. Define the functions

F(n, k) := (−1)^{k} \frac{\binom{n}{k}\binom{2n+1+k}{n+1+k}}{2n+1+k} and also H(n, k) := −F(n, k) · \frac{k(n+1+k)}{n(2n+1)} to verify

F(n, k) = H(n, k + 1) − H(n, k). The rest is routine [3].

Our next step will actually find G[m_{1}, m_{2}]. For that we perform the division \frac{P_{n}(t_{1},t_{2})}{t_{1}+t_{2}}to obtain (algebraically) that

[t_{1}^{n−1−i}t_{2}^{i}] (\frac{P_{n}(t_{1}, t_{2})}{t_{1} + t_{2}})= ∑_{j=0}^{i} (−1)^{i−j} \frac{\binom{n}{j}\binom{2n+1+j}{n+1+j}}{2n + 1 + j}

= (−1)^{i}[H(n, i + 1) − H(n, 0)]

= (−1)^{i}H(n, i + 1)

= \frac{1}{2n + 1} \binom{n − 1}{i}\binom{2n + 1 + i}{n + 1 + i}

which leads to (an equivalent form of) the first conjecture [2] on G[m_{1}, m_{2}]. To wit:

Theorem 1.1. For non-negative integers m_{1} and m_{2}, we have

G[m_{1}, m_{2}] = \frac{1}{(2m_{1} + 2m_{2} + 3)(m_{1} + m_{2} + 1)} \frac{(2m_{1} + 3m_{2} + 3)!}{(m_{1} + 2m_{2} + 2)!m_{1}!m_{2}!}.

2. On the second conjecture

Now that the reader, hopefully, is getting accustomed to our proof-procedure as depicted in Section 1, let’s move on to next conjecture [2, p. 399] which does generalize the one we just finished proving. For brevity, denote G̃ = G̃[m_{a}, m_{a+1}] = G[0, 0, …, m_{a}, m_{a+1}]. Again, we revive the Lagrange Inversion (1.2). Suppose n = m_{a} + m_{a+1}. Then the total content of such

4 T. AMDEBERHAN AND D. ZEILBERGER

monomials is encapsulated by

∑_{m_{a}+m_{a+1}=n} G̃t_{a}^{m_{2}}t_{a+1}^{m_{3}}= \frac{[Y^{n}]}{t_{a} + t_{a+1}} ∑_{k=1}^{(a+1)n+1} \frac{1}{k} [z^{k−1}] (1 + Yt_{a}z^{a} + Yt_{a+1}z^{a+1})^{k}

= \frac{[Y^{n}]}{t_{a} + t_{a+1}} ∑_{m_{a},m_{a+1}≥0} \frac{\binom{1+am_{a}+(a+1)m_{a+1}}{m_{a},m_{a+1},1+(a−1)m_{a}+am_{a+1}}Y^{m_{a}+m_{a+1}}t_{a}^{m_{a}}t_{a+1}^{m_{a+1}}}{1 + am_{a} + (a + 1)m_{a+1}}

= ∑_{\begin{matrix}m_{a},m_{a+1}≥0 \\ m_{a}+m_{a+1}=n \end{matrix}} \frac{\binom{1+am_{a}+(a+1)m_{a+1}}{m_{a},m_{a+1},1+(a−1)m_{a}+am_{a+1}}}{1 + am_{a} + (a + 1)m_{a+1}} \frac{t_{a}^{m_{a}}t_{a+1}^{m_{a+1}}}{t_{a} + t_{a+1}}

= ∑_{m_{a+1}=0}^{n} \frac{\binom{1+an+m_{a+1}}{n−m_{a+1},m_{a+1},1+(a−1)n+m_{3}}}{1 + an + m_{a+1}} \frac{t_{a}^{n−m_{a+1}}t_{a+1}^{m_{a+1}}}{t_{a} + t_{a+1}}

= ∑_{k=0}^{n} \frac{\binom{n}{k}\binom{an+1+k}{(a−1)n+1+k}}{an + 1 + k} \frac{t_{a}^{n−k}t_{a+1}^{k}}{t_{a} + t_{a+1}}.

As a first step, we justify that the linear term t_{a}+t_{a+1} divides the polynomial

P_{n}(t_{a}, t_{a+1}) := ∑_{k=0}^{n} \frac{\binom{n}{k}\binom{an+1+k}{(a−1)n+1+k}}{an + 1 + k}t_{a}^{n−k}t_{a+1}^{k}.

This is tantamount to P_{n}(−t_{a+1}, t_{a+1}) = 0 which is equivalent to the identity that

∑_{k=0}^{n}(−1)^{k} \frac{\binom{n}{k}\binom{an+1+k}{(a−1)n+1+k}}{an + 1 + k} = 0.

Again, apply the Wilf-Zeilberger approach with F(n, k) := \frac{(−1)^{k}\binom{n}{k}\binom{an+1+k}{(a−1)n+1+k}}{an+1+k} and H(n, k) := −F(n, k)· \frac{k((a−1)n+1+k)}{n(an+1)} to verify F(n, k) = H(n, k+1)−H(n, k). The rest is trivial. Our next step will actually determine G̃[m_{a}, m_{a+1}]. To this effect, let’s divide \frac{P_{n}(t_{a},t_{a+1})}{t_{a}+t_{a+1}}to obtain (routinely) that

[t_{a}^{n−1−i}t_{a+1}^{i}] (\frac{P_{n}(t_{a}, t_{a+1})}{t_{a} + t_{a+1}})= ∑_{j=0}^{i} (−1)^{i−j} \frac{\binom{n}{j}\binom{an+1+j}{(a−1)n+1+j}}{an + 1 + j}

= (−1)^{i}[H(n, i + 1) − H(n, 0)] = (−1)^{i}H(n, i + 1)

= \frac{1}{an + 1} \binom{n − 1}{i}\binom{an + 1 + i}{(a − 1)n + 1 + i}

which proves the desired conjecture on G̃[m_{a}, m_{a+1}]. To wit:

Theorem 2.1. Denote m = m_{a}+m_{a+1}. For integers m_{a}, m_{a+1} ≥ 0 there holds

G̃[m_{a}, m_{a+1}] = \frac{(am_{a} + (a + 1)(m_{a+1} + 1))!}{(a(m + 1) + 1)(m + 1)((a − 1)m_{a} + a(m_{a+1} + 1))!m_{a}!m_{a+1}!} .

3. On the third conjecture

The proof of the last conjecture [2, p. 399] is a bit more complicated.

Theorem 3.1. For the 2a-variate case, we have

G G G[−f, f, …, −f, f] = ∑_{n} a^{n}f^{n}.

Proof. To begin, we make a slight alteration by writing (−1)^{i}t_{i} instead of the customary plain t_{i} [2]. Thanks to the Lagrange Inversion (1.2), we have

[Y^{n}] (∑_{k=1}^{∞} \frac{1}{k} [z^{k−1}] (1 − Yt_{1}z^{2} + Yt_{2}z^{3} − ⋯ − Yf t_{2a−1}z^{2a} + Yt_{2a}z^{2a+1})^{k})

=[Y^{n}] ∑_{m_{1},...,m_{2a}≥0} \frac{(−1)^{m_{1}+··· +m_{2a−1}} \binom{1+2m_{1}+3m_{2}+··· +(2a+1)m_{2a}}{m_{1},m_{2},...,m_{2a},1+m_{1}+2m_{2}+··· +(2a)m_{2a}}(Yt_{1})^{m_{1}}⋯ (Yt_{2a})^{m_{2a}}}{1 + 2m_{1} + 3m_{2} + ⋯ + (2a + 1)m_{2a}}

= ∑_{\begin{matrix}m_{1},...,m_{2a}≥0 \\ m_{1}+··· +m_{2a}=n \end{matrix}} \frac{(−1)^{m_{1}+··· +m_{2a−1}} \binom{1+2m_{1}+3m_{2}+··· +(2a+1)m_{2a}}{m_{1},m_{2},...,m_{2a},1+m_{1}+2m_{2}+··· +(2a)m_{2a}}t_{1}^{m_{1}}⋯ t_{2a}^{m_{2a}}}{1 + 2m_{1} + 3m_{2} + ⋯ + (2a + 1)m_{2a}}.

First, consider the case a = 1 and refer back to Theorem 1.1 (and its proof), to gather that if t_{1} = −f and t_{2} = f then, as expected, we arrive at

f^{n−1} ∑_{m=0}^{n−1} \frac{(−1)^{n−1−m}}{2n + 1} \binom{n − 1}{m}\binom{2n + 1 + m}{n + 1 + m} = f^{n−1}

as justified by the WZ-certificate [3] given by

R(n, m) := \frac{m(8mn + 10n^{2} + 6m + 15n + 6)}{2(2n + 3)(n + 1)(n − m)}.

Second, we go back to study the above-posed calculations when a > 1. To set the stage, substitute t_{1} = t_{2} = ⋯= t_{2a−1} = f while leaving out t_{2a} as an indeterminate. The outcome takes the form

∑_{\begin{matrix}m_{1},...,m_{2a}≥0 \\ m_{1}+··· +m_{2a}=n \end{matrix}} \frac{(−1)^{m_{1}+m_{3}+··· +m_{2a−1}} \binom{1+2m_{1}+3m_{2}+··· +(2a+1)m_{2a}}{m_{1},m_{2},...,m_{2a},1+m_{1}+2m_{2}+··· +(2a)m_{2a}}f^{n−m_{2a}}t_{2a}^{m_{2a}}}{1 + 2m_{1} + 3m_{2} + ⋯ + (2a + 1)m_{2a}} .

At this point, divide out the current polynomial (in t_{2a}) by the linear factor

−t_{1} + t_{2} − ⋯ − t_{2a−3} + t_{2a−1} − t_{2a−1} + t_{2a} = t_{2a} − f

and then replace t_{2a} by f. That leads to the sum

f^{n−1} ∑_{i=0}^{n−1} ∑_{m_{2a}=0}^{i} ∑_{\begin{matrix}m_{1},...,m_{2a}≥0 \\ m_{1}+··· +m_{2a}=n \end{matrix}} \frac{(−1)^{1+m_{1}+m_{3}+··· +m_{2a−1}} \binom{1+2m_{1}+3m_{2}+··· +(2a+1)m_{2a}}{m_{1},m_{2},...,m_{2a},1+m_{1}+2m_{2}+··· +(2a)m_{2a}}}{1 + 2m_{1} + 3m_{2} + ⋯ + (2a + 1)m_{2a}} .

Therefore, our main task that remains is to prove the identity declared by

∑_{i=0}^{n−1} ∑_{\begin{matrix}m_{1},...,m_{2a−1}≥0 \\ m_{1}+··· +m_{2a}=n \\ 0≤m_{2a}≤i \end{matrix}} \frac{(−1)^{1+m_{1}+m_{3}+··· +m_{2a−1}} \binom{1+2m_{1}+3m_{2}+··· +(2a+1)m_{2a}}{m_{1},m_{2},...,m_{2a},1+m_{1}+2m_{2}+··· +(2a)m_{2a}}}{1 + 2m_{1} + 3m_{2} + ⋯ + (2a + 1)m_{2a}} . = a^{n−1}.

To put this more succinctly, introduce some notation. Let P denote the set of all integer partitions λ, written as λ = (λ_{1}, λ_{2}, …) or λ = 1^{m_{1}}2^{m_{2}}…(2a)^{m_{2a}}. The size of λ is denoted by |λ| = λ_{1} + λ_{2} + ⋯= m_{1} + 2m_{2} + ⋯ + (2a)m_{2a} while we use ℓ(λ) = m_{1} + m_{2} + ⋯ + m_{2a} for the length of the partition. So, the claim stands at

∑_{\begin{matrix}λ∈P \\ ℓ(λ)=n \\ λ_{1}≤2a \end{matrix}}(−1)^{1+|λ|} · \frac{(n − m_{2a}) \binom{n}{m_{1},...,m_{2a}}\binom{|λ|+n+1}{|λ|+1}}{|λ| + n + 1} = a^{n−1}. (3.1)

We find it more convenient to split up this assertion into two separate claims

(3.2)  (−1)^{1} ∑_{\begin{matrix}λ∈P \\ ℓ(λ)=n \\ λ_{1}≤2a \end{matrix}}(−1)^{|λ|} \binom{n}{m_{1}, …, m_{2a}}\binom{|λ| + n}{|λ| + 1} = 0,

(3.3)  ∑_{\begin{matrix}μ∈P \\ ℓ(μ)=n−1 \\ μ_{1}≤2a \end{matrix}}(−1)^{|μ|} \binom{n − 1}{m_{1}, …, m_{2a}}\binom{|μ| + 2a + n}{|μ| + 2a + 1} = a^{n−1}.

One arrives at (3.2) due to \frac{n \binom{|λ+n+1}{|λ|+1}}{|λ+n+1} = \binom{|λ+n}{|λ|+1} and (3.3) arises because of m_{2a}\binom{n}{m_{1},...,m_{2a}} \frac{(|λ+n)!}{(|λ+1)!n!} = \binom{n−1}{m_{1},...,m_{2a}−1}\binom{|λ|+n}{|λ|+1} and then we reindex m′_{2a} = m_{2a} − 1 to convert |λ| = |μ| + 2a where ℓ(μ) = n − 1. In fact, let’s generalize (3.2) and (3.3) by introducing an extra parameter x. Claim 1: For positive integers n, a and an indeterminate x, we have

∑_{\begin{matrix}λ∈P \\ ℓ(λ)=n \\ λ_{1}≤2a \end{matrix}}(−1)^{|λ|} \binom{n}{m_{1}, …, m_{2a}}\binom{|λ| + n + x}{n − 1} = 0.

Claim 2: For positive integers n, a and an indeterminate x, we have

∑_{\begin{matrix}λ∈P \\ ℓ(λ)=n−1 \\ λ_{1}≤2a \end{matrix}}(−1)^{|λ|} \binom{n − 1}{m_{1}, …, m_{2a}}\binom{|λ| + n + x}{n − 1} = a^{n−1}.

Claim 2 implies Claim 1: We apply the multinomial recurrence (assume n = k_{1} + ⋯ + k_{r})

(3.4)  \binom{n}{k_{1}, …, k_{r}} = \binom{n − 1}{k_{1} − 1, …, k_{r}} + ⋯ + \binom{n − 1}{k_{1}, …, k_{r} − 1}

followed by appropriate reindexing so that

∑_{\begin{matrix}λ∈P \\ ℓ(λ)=n \\ λ_{1}≤2a \end{matrix}}(−1)^{|λ|} \binom{n}{m_{1}, …, m_{2a}}\binom{|λ| + n + x}{n − 1}

= ∑_{i=1}^{2a} ∑_{\begin{matrix}λ∈P \\ ℓ(λ)=n \\ λ_{1}≤2a \end{matrix}}(−1)^{|λ|} \binom{n − 1}{m_{1}, …, m_{i} − 1, …m_{2a}}\binom{|λ| + n + x}{n − 1}

= ∑_{i=1}^{2a} ∑_{\begin{matrix}μ∈P \\ ℓ(μ)=n−1 \\ μ_{1}≤2a \end{matrix}}(−1)^{|μ|+i} \binom{n − 1}{m_{1}, …, m′_{i}, …m_{2a}}\binom{|μ| + n + (x + i)}{n − 1}

= ∑_{i=1}^{2a} (−1)^{i} ∑_{\begin{matrix}μ∈P \\ ℓ(μ)=n−1 \\ μ_{1}≤2a \end{matrix}}(−1)^{|μ|} \binom{n − 1}{m_{1}, …, m′_{i}, …m_{2a}}\binom{|μ| + n + (x + i)}{n − 1}

=a^{n−1} ∑_{i=1}^{2a} (−1)^{i}
= 0.

Proof of Claim 2: Let’s now utilize the multinomial theorem (1.1) and constant-term extraction. Start by noting the constant-term extraction

\binom{|λ| + n + x}{n − 1} = \binom{m_{1} + 2m_{2} + ⋯ + (2a)m_{2a} + n + x}{n − 1}

= CT CT CT_{z}[\frac{(1 + z)^{m_{1}+2m_{2}+··· +(2a)m_{2a}+n+x}}{z^{n−1}} ].

Insert this into the left-hand side of Claim 2, take CT CT CT_{z} outside the sum, factor out the inside and reapply the multinomial theorem in reverse (1.1) to get

∑_{\begin{matrix}λ∈P \\ ℓ(λ)=n−1 \\ λ_{1}≤2a \end{matrix}}(−1)^{|λ|} \binom{n − 1}{m_{1}, …, m_{2a}}\binom{|λ| + n + x}{n − 1}

=CT CT CT_{z}[\frac{(1 + z)^{n+x}}{z^{n−1}} ∑ \binom{n − 1}{m_{1}, …, m_{2a}}(−1 − z)^{m_{1}}(−1 − z)^{2m_{2}}⋯ (−1 − z)^{(2a)m_{2a}}]

=CT CT CT_{z}[\frac{(1 + z)^{n+x}}{z^{n−1}} \{−(1 + z)^{1} + (1 + z)^{2} − (1 + z)^{3} + ⋯ + (1 + z)^{2a}\} ^{n−1}].

Next, follow through with the geometric series expansion to obtain

∑_{\begin{matrix}λ∈P \\ ℓ(λ)=n−1 \\ λ_{1}≤2a \end{matrix}}(−1)^{|λ|} \binom{n − 1}{m_{1}, …, m_{2a}}\binom{|λ| + n + x}{n − 1}

=CT CT CT_{z} [(−1)^{n−1} \frac{(1 + z)^{2n+x−1}}{z^{n−1}} \{ \frac{1 − (1 + z)^{2a}}{2 + z} \}^{n−1} ]

=CT CT CT_{z} [\frac{(1 + z)^{2n+x−1}}{(2z)^{n−1}} \{ \frac{z ∑ _{k=1}^{2a} \binom{2a}{k}z^{k−1}}{1 + \frac{z}{2}}\}^{n−1} ]= a^{n−1}.

The proof is indeed complete.

Remark 3.2. On [2, p. 399], it is stated that “With k − 2 leading zeros, we conjecture that G[0, …, m_{k}] is a two-parameter Fuss-Catalan number." In light of the conjectures we already proved, the current claim is rather obvious (for further discussion on the topic the reader is directed to [1]).

References

[1] S. R. Mane, Multiparameter Fuss–Catalan numbers with application to algebraic equations, https://arxiv.org/pdf/1607.04144

[2] N. J. Wildberger, D.Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly, 132, (2025), no. 5, 383-402. https: //www.tandfonline.com/doi/full/10.1080/00029890.2025.2460966

[3] H. S. Wilf, D. Zeilberger, Rational functions certify combinatorial identities, J. Amer. Math. Soc. 3 (1990), no. 1, 147-158. https://www.ams.org/journals/jams/ 1990-03-01/S0894-0347-1990-1007910-7/S0894-0347-1990-1007910-7. pdf

[4] D. Zeilberger, Lagrange Inversion Without Tears (Analysis) (based on Henrici) , https://sites.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimPDF/ lag.pdf Department of Mathematics, Tulane University, New Orleans, LA 70118, USA E-mail address: tamdeber@tulane.edu

Rutgers University, Department of Mathematics, 110 Frelinghuysen Rd, Piscataway, NJ 08854, USA

E-mail address: DoronZeil@gmail.com