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\centerline
{\bf Proof of Kyle Petersen's Amazing Conjecture Relating the q- and Fibonacci Analogs of n! }
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\centerline
{\it Doron ZEILBERGER}

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Let the {\it darga} of a polynomial
be the sum of its low-degree and (high) degree,
for example, 
$$
darga(2+q+q^2-q^3)=0+3=3 \quad , \quad darga(q^2+q^3)=2+3=5 \quad ,
\quad darga(q^3)=3+3=6 \quad .
$$
It is immediate that every {\it symmetric} polynomial of darga $d$ can be written,
uniquely, as a linear combination of 
$$
(1+q)^d \quad , \quad q(1+q)^{d-2} \quad , \quad q^2(1+q)^{d-4} \quad , \dots
$$
in symbols
$$
P(q)=\sum_{i=0}^{\lfloor d/2 \rfloor} \gamma_i \, q^i(1+q)^{d-2i} \quad ,
$$
and we will call the {\it Kyle Transform}, $\hat{P}(z)$, the polynomial, in $z$,
$$
\hat{P}(z) \, := \, \sum_{i=0}^{\lfloor d/2 \rfloor} \gamma_i \, \, z^i \quad .
$$
Kyle Petersen([P]) made the following elegant conjecture, that I will shortly
prove.

{\bf Fact}: Let $A_n(z)$ be the Kyle transform of 
$$
[n]!:=\prod_{i=1}^{n} \frac{1-q^i}{1-q} \quad ,
$$
then
$$
A_n(-1)=\prod_{i=1}^n F_i \quad ,
$$
where the $F_i$ are the famous {\it Fibonacci numbers} defined by
$F_0=0,F_1=1$, $F_i=F_{i-1}+F_{i-2}$ for $i \geq 2$.

{\bf Proof.} For a general symmetric $P(q)$ we have
$$
P(q)=\sum_{i=0}^{\lfloor d/2 \rfloor} \gamma_i \, q^i(1+q)^{d-2i} =
(1+q)^d \sum_{i=0}^{\lfloor d/2 \rfloor} \gamma_i \, (\frac{q}{(1+q)^2})^i \quad .
$$
Hence
$$
\frac{P(q)}{(1+q)^d}= \hat{P}(\frac{q}{(1+q)^2}) \quad.
$$
Letting $z=\frac{q}{(1+q)^2}$, we get the quadratic equation
$$
z q^2 + (2z-1)q+z =0 \quad ,
$$
whose (unique formal-power series) solution is
$$
q= \frac{1-2z - \sqrt{1-4z}}{2z}= C(z) -1 = zC(z)^2 \quad,
$$
where 
$$
C(z)=\frac{1-\sqrt{1-4z}}{2z} \quad ,
$$ 
is the unique formal power series solution
of $C(z)=1+zC(z)^2$ of Catalan fame.
Hence
$$
\hat{P}(z)= \frac{P(zC(z)^2)}{(1+zC(z)^2)^d}
= \frac{P(zC(z)^2)}{C(z)^d} \quad .
$$

Plugging-in $z=-1$ we get

$$
\hat{P}(-1) = \frac{P(-C(-1)^2)}{C(-1)^d} \quad .
$$
Now
$$
C(-1)=\frac{1-\sqrt{5}}{-2}=\frac{\sqrt{5}-1}{2}=\phi^{-1} \quad ,
$$ 
where $\phi=(\sqrt{5}+1)/2$ is the famous {\it golden ratio}. Hence
$$
\hat{P}(-1) = \frac{P(-\phi^{-2})}{\phi^{-d}} 
= \phi^d \, P(-\phi^{-2}) \quad .
$$
Applying this to $P(q)=[n]!$, we get
$$
A_n(-1)=
\phi^{n(n-1)/2}  \prod_{i=1}^{n} \frac{1-(-\phi^{-2})^i }{1+ \phi^{-2}} 
\, =\, \prod_{i=1}^{n} \frac{\phi^i-(-\phi^{-1})^i }{\phi +\phi^{-1}} =
\prod_{i=1}^{n} F_i \quad . \quad \halmos 
$$

{\bf Reference}

[P] Kyle Petersen, {\it Private communication} (via Richard P. Stanley) during the banquet 
in honor of Ira Gessel, Brandeis University, May 8, 2015.

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Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email: {\tt zeilberg at math dot rutgers dot edu}   \quad .
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May 11, 2015; Exclusively published in the Personal Journal of Shalosh B. Ekhad and Doron Zeilberger.

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