where $\phi = \frac{1+\sqrt{5}}{2}$ and $k!_F$ is the fibonorial of $k$, or the product of the first $k$ Fibonacci numbers? My hunch is that, this can be represented as a function in terms of the constant in the case where x=1, so I tried the following: $$ \begin{split} S = \sum_{k=0}^{\infty}{\frac{\phi^k}{k!_F}} & =\sum_{k=0}^{\infty}{\frac{F_k\phi+F_{k-1}}{k!_F}} \\ &=\sum_{k=0}^{\infty}\Bigg (\phi\frac{F_{k}}{k!_F}+\frac{F_{k-1}}{k!_F}\Bigg)\\ & =\phi \displaystyle \sum_{k=0}^{\infty}\frac{F_{k}}{k!_F}+\sum_{k=0}^{\infty}\frac{F_{k-1}}{k!_F} \end{split} $$ How should I continue from here?. I'm asking for a closed form expression but any alternate representations as an integral or another series are greatly appreciated as well.

(Fibonacci numbers as defined by the recurence $ F_k = F_{k-1}+ F_{k-2} $ and $ F_0=F_1=1 $)

(Edit: The first couple of terms of $ \displaystyle \sum_{k=0}^{\infty}\frac{1}{(k-2)!_F}\frac{1}{F_k} $ and the first term of $ \displaystyle \sum_{k=0}^{\infty}\frac{1}{(k-1)!_F} $ aren't defined so I changed it back to $ \displaystyle \sum_{k=0}^{\infty}\frac{F_{k-1}}{k!_F} $ and $ \displaystyle \sum_{k=0}^{\infty}\frac{F_{k}}{k!_F}$ respectively )

(Edit: It seems that $\displaystyle \sum_{k=0}^{\infty} \frac{F_k}{k!_F} = \sum_{k=0}^{\infty} \frac{1}{k!_F}$)

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    $\begingroup$ I think it is always a good idea to give your definition of Fibonacci numbers: There are two natural indexations: $F_0=0,F_1=1$ or $F_0=F_1=1$. I guess you want the second one. $\endgroup$ Aug 21, 2021 at 16:11
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    $\begingroup$ The first term in the last summation is ${1\over(-2)!_F}{1\over F_0}$, and I'm having trouble seeing what $(-2)!_F$ could mean. If you start at $k=2$, you get a rapidly converging series, so you could compute a few decimals and then see whether you can conjecture a closed form for the result. $\endgroup$ Aug 21, 2021 at 23:12
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    $\begingroup$ It's not equivalent, because it has terms that don't make sense. I think you have to split of the terms with $k=0$ and $k=1$. $\endgroup$ Aug 22, 2021 at 8:53
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    $\begingroup$ Right, but me simplifying $ \displaystyle \sum_{k=0}^{\infty} \frac{F_{k-1}}{k!_F}$ to $ \displaystyle \sum_{k=0}^{\infty} \frac{1}{(k-2)!_F}\frac{1}{F_k} $ was probably a mistake. @GerryMyerson $\endgroup$ Aug 22, 2021 at 9:00
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    $\begingroup$ $\sum_0{F_k\over k!_F}=1+\sum_1{F_k\over k!_F}=1+\sum_1{1\over(k-1)!_F}=1+\sum_0{1\over k!_F}$. $\endgroup$ Aug 22, 2021 at 12:51


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