Let $F(n)$ denote the $n^{\text{th}}$ Fibonacci number$^{[1]}$$\!^{[2]}$$\!^{[3]}$. The Fibonacci numbers have a natural generalization to an analytic function of a complex argument:
$$F(z)=\left(\phi^z - \cos(\pi z)\,\phi^{-z}\right)/\sqrt5,\quad\text{where}\,\phi=\left(1+\sqrt5\right)/2.\tag1$$
This definition is used, for example, in *Mathematica*.$^{[4]}$ It produces real values for $z\in\mathbb R$, and preserves the usual functional equation for Fibonacci numbers for all $z\in\mathbb C$: $$F(z)=F(z-1) + F(z-2).\tag2$$

The fibonorial$^{[5]}$$\!^{[6]}$$\!^{[7]}$ is usually denoted as $n!_F$, but here we prefer a different notation $\mathfrak F(n)$. It is defined for non-negative integer $n$ inductively as $$\mathfrak F(0)=1,\quad \mathfrak F(n+1)=\mathfrak F(n)\times F(n+1).\tag3$$ In other words, the fibonorial $\mathfrak F(n)$ gives the product of the Fibonacci numbers from $F(1)$ to $F(n)$, inclusive. For example, $$\mathfrak F(5)=\prod_{m=1}^5F(m)=1\times1\times2\times3\times5=30.\tag4$$

Questions:Can the fibonorial be generalized in a natural way to an analytic function $\mathfrak F(z)$ of a complex (or, at least, positive real) variable, such that it preserves the functional equation $(3)$ for all arguments?Is there an integral, series or continued fraction representation of $\mathfrak F(z)$, or a representation in a closed form using known special functions?

Is there an efficient algorithm to calculate values of $\mathfrak F(z)$ at non-integer arguments to an arbitrary precision?

So, we can see that the fibonorial is to the Fibonacci numbers as the factorial is to natural numbers, and the analytic function $\mathfrak F(z)$ that I'm looking for is to the fibonorial as the analytic function $\Gamma(z+1)$ is to the factorial.

*Update:* While thinking on this question it occurred to me that perhaps we can use the same trick that is used to define the $\Gamma$-function using a limit involving factorials of integers:
$$\large\mathfrak F(z)=\phi^{\frac{z\,(z+1)}2}\cdot\lim_{n\to\infty}\left[F(n)^z\cdot\prod_{k=1}^n\frac{F(k)}{F(z+k)}\right]\tag5$$
or, equivalently,
$$\large\mathfrak F(z)=\frac{\phi^{\frac{z\,(z+1)}2}}{F(z+1)}\cdot\prod_{k=1}^\infty\frac{F(k+1)^{z+1}}{F(k)^z\,F(z+k+1)}\tag{$5'$}$$
This would give
$$\mathfrak F(1/2)\approx0.982609825013264311223774805605749109465380972489969443...\tag6$$
that appears to have a closed form in terms of the q-Pochhammer symbol:
$$\mathfrak F(1/2)=\frac{\phi^{3/8}}{\sqrt[4]{5}}\,\left(-\phi^{-2};-\phi^{-2}\right)_\infty\tag7$$
and is related to the Fibonacci factorial constant.