Experimental identities with Fibonacci series I obtained the following Fibonacci identities experimentally using SageMath code. Can these be proved theoretically?
Identity 1:
$$
\sum_{n = 0}^{\infty} \frac{F_{n}x^{n}}{n!}
+ e^x\sum_{n = 0}^{\infty} \frac{F_{n}(-x)^{n}}{n!} = 0
$$
Identity 2:
$$
52\sum_{n = 0}^{\infty} \frac{F_{4n}x^{4n}}{4n!}
= 156 \sum_{n = 0}^{\infty} \frac{F_{4n+1}x^{4n+1}}{(4n+1)!}
= 585 \sum_{n = 0}^{\infty} \frac{F_{4n+2}x^{4n+2}}{(4n+2)!}
= 630 \sum_{n = 0}^{\infty} \frac{F_{4n+3}x^{4n+3}}{(4n+3)!}
$$
 A: Identity 1:
Note by the power series product formula this identity
$$
\sum_{n = 0}^{\infty} \frac{F_{n}x^{n}}{n!}
+ e^x\sum_{n = 0}^{\infty} \frac{F_{n}(-x)^{n}}{n!} = 0
$$
can be written 
$$\sum_{n = 0}^{\infty} \frac{F_{n}x^{n}}{n!}
+ \sum_{n = 0}^{\infty}\left(\sum_{k=0}^n\binom{n}{k}\frac{F_{k}(-1)^{k}}{k!}\frac{1}{(n-k)!}\right)x^n = 0$$
So you'd need to show for all $n$
$$
\frac{F_{n}}{n!}=\sum_{k=0}^n\binom{n}{k}\frac{F_{k}(-1)^{k+1}}{n!}
$$
which simplifies 
$$
F_{n}=\sum_{k=0}^n\binom{n}{k}F_{k}(-1)^{k+1}
$$
Unfortunately I was unable to prove this without Binet's Formula: $F_k=\dfrac{\varphi^k-\varphi^{-k}}{\varphi-\varphi^{-1}}$ . So it's really a worse version of iqcd's proof but here it is anyway.
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}F_{k}(-1)^{k+1} &=\sum_{k=0}^n\binom{n}{k}(-1)^{k+1}\frac{\varphi^k-\varphi^{-k}}{\varphi-\varphi^{-1}}\\
&=\frac{1}{\varphi-\varphi^{-1}}\sum_{k=0}^n\binom{n}{k}(-1)^{k+1}\varphi^k-\sum_{k=0}^n\binom{n}{k}(-1)^{k+1}\varphi^{-k}\\
&=\frac{1}{\varphi-\varphi^{-1}}\sum_{k=0}^n\binom{n}{k}(-\varphi^{-1})^k-\sum_{k=0}^n\binom{n}{k}(-\varphi)^{k}\\
&= \frac{1}{\varphi-\varphi^{-1}}[(1-\varphi^{-1})^n-(1-\varphi)^n]\\
&= \frac{1}{\varphi-\varphi^{-1}}[(\varphi)^n-(\varphi^{-1})^n]\\
&= F_n
\end{align}
$$
That second to last line is because for the golden ratio $\varphi$, we have $1-\varphi=\varphi^{-1}$ similarly $1-\varphi^{-1}=\varphi$.
A: Using Binet's formula we have
$$\sum_{n}\frac{F_nx^n}{n!}
=\frac1{\sqrt5}\left(\sum_n\frac{(\phi x)^n}{n!}
-\sum_n\frac{(-\phi^{-1} x)^n}{n!}\right)
=\frac1{\sqrt5}(e^{\phi x}-e^{-\phi^{-1}x})$$
while, 
$$e^x\sum_{n}\frac{F_n(-x)^n}{n!}
=\frac{e^x}{\sqrt5}\left(\sum_n\frac{(-\phi x)^n}{n!}
-\sum_n\frac{(\phi^{-1} x)^n}{n!}\right)
=\frac1{\sqrt5}(e^{(1-\phi) x}-e^{(1+\phi^{-1})x})$$
and the first identity is consequence of the equality $1-\phi=-\phi^{-1}$.
EDIT: I've just checked numerically the second identity, and apparently it is false. The original question included a snippet of SAGE code, and I think the problem was that $4r$ is interpreted as $4$ by SAGE.
A: According to this paper:
$\sum_{n = 0}^{\infty} \frac{F_{n}x^{n}}{n!}= \frac{e^{\alpha x}-e^{\beta x}}{\alpha - \beta}$ where $\alpha$ and $\beta$ are roots of the Fibnocci polynomial $x^2-x-1$.
This makes the first identity equivalent to $G(x) = G(-x)e^{x}$ where $G(x) = \frac{e^{\alpha x}-e^{\beta x}}{\alpha - \beta}$. And that, in turn, is equivalent to $1-\alpha=\beta$ and $1-\beta =\alpha$, which are both equivalent to $\alpha+\beta =1$, which follows from the fact that the sum of roots of a quadratic polynomial add up to the coefficient of $x$ times negative one.
Note that the search term to use is "exponential Fibonocci generating function".
