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.

Vladimir Reshetnikov
  • 45,303
  • 7
  • 151
  • 282
  • The complexity of the expression for ${\frak F}(z)$ is somewhere in between that of [Euler's $\Gamma$ function](http://en.wikipedia.org/wiki/Gamma_function) and that of [Barnes' G function](http://en.wikipedia.org/wiki/Barnes_G-function). – Lucian Jun 26 '16 at 14:38

3 Answers3


(Main Result: I have a formula for what you want, its practical and it works!)

Since you ask multiple questions, I suppose what follows isn't a complete answer. I learned this trick the other day, so I'm more than happy to share it here,

If you want an integral that can generalize the recurrence,

$$(1) \quad \mathfrak F(0)=1,\quad \mathfrak F(n+1)=\mathfrak F(n)\times F(n+1).$$

You can do so by assuming the formula is of a certain form and that $k(t,n)$ satisfies a delay differential equation.

$$(2) \quad \mathfrak F(n)=\int_0^{\infty} e^{-t} \cdot k(t,n) \ dt$$

We can then demand that the eigenvalue for the differential operator is equal to $F(n+1)$.

$$(3) \quad \partial_t [k(t,n+1)]=F(n+1) \cdot k(t,n)$$

It then follows that applying integration by parts to $(2)$ yields,

$$\mathfrak F(n+1)=\int_0^{\infty} e^{-t} \cdot k(t,n+1) \ dt={F(n+1)} \cdot \int_0^{\infty} e^{-t} \cdot k(t,n) \ dt={F(n+1)} \cdot \mathfrak F(n)$$

$$\Rightarrow \mathfrak F(n+1)=\mathfrak F(n)\cdot F(n+1).$$

To obtain an explicit form for the integration kernel, we must solve $(3)$, which is classified as a delay differential equation. Whether or not an explicit solution exists is uncertain without further investigation. However, the kernel can certainly be found numerically.

However, we can certainly define the kernel that satisfies $(3)$ to be the Fibonacci kernel.

Tentative (Approximate) Solution

I didn't post this originally because its a bit cheap for a solution. Effectively, this solution indexes the new factorial differently. Set,


And also we have,

$$F(z)=(\phi^z-\cos(\pi \cdot z) \cdot \phi^{-z})/\sqrt{5}$$

Then we have,

$$\mathfrak F(n+1)=\mathfrak F(F^{-1}(F(n+1)-1)) \cdot F(n+1)$$

instead of,

$$\mathfrak F(n+1)=\mathfrak F(n) \cdot F(n+1)$$

Asymptotics and a Formula

I've switched notation since its the reason I came up with this. According to Wolfram Alpha, we have this relation for the asymptotics of the fibonorial,

$$(4) \quad n!_F \sim C \cdot \cfrac{\phi^{n \cdot (n+1)/2}}{5^{n/2}}$$

Given this, lets hope we can find a function $g(n,t)$ such that,

$$(5) \quad \lim_{n \rightarrow \infty} \cfrac{n!_F \cdot g(n,t)}{(n+t)!_F}=1$$

I find that, and I leave you to verify, that,

$$(6) \quad g(n,t)=5^{-t/2} \cdot \phi^{(t^2/2+t/2)} \cdot \phi^{n \cdot t}$$

Now that we know what $g(n,t)$ is, multiply by both sides by $t!_F$ and simplify, assuming $t$ is a positive integer. We get,

$$(7) \quad t!_F=\lim_{n \rightarrow \infty} \cfrac{n!_F \cdot g(n,t)}{F(1+t) \cdot F(2+t) \ ...\ F(n+t)}$$

Now we formally extend $t$ from $N \gt 0$ to $t \in C$. Using this, I can construct (perhaps the first?!) table of "F-Gamma" values using $n!_F=\mathfrak F(n+1)$,

$$ \begin{array}{c|lcr} n & \mathfrak F(n) & \ln(\mathfrak F(n))\\ \hline 1 & 1 & 0\\ 1.1 & 0.966975 & -0.033582\\ 1.2 & 0.951630 & -0.049579\\ 1.3 & 0.951521 & -0.049692\\ 1.4 & 0.963404 & -0.037283\\ 1.5 & 0.982610 & -0.017543\\ 1.6 & 1.003127 & 0.003122\\ 1.7 & 1.018567 & 0.018396\\ 1.8 & 1.023911 & 0.023630\\ 1.9 & 1.017193 & 0.017048\\ 2.0 & 1 & 0\\ \end{array} $$

Here's a graph (I didn't say it'd be pretty ;))

enter image description here

Check the recurrence, defined in $(1)$. It works!!!

  • 8,088
  • 1
  • 19
  • 50

Extending what Zach466920 said:

$$\partial_t[k(t,n)]=F(n+1)\cdot k(t,n+1)\tag1$$

We can develop a power series solution for this. Let $$k(t,n)=\sum_{m=0}^\infty \kappa(n, m)\ t^m$$

We equation $(1)$ as: $$\sum_{m=1}^\infty m\ \kappa(n, m)\ t^{m-1} =\sum_{m=0}^\infty(m+1)\ \kappa (n, m) \ t^m =F(n+1)\cdot\sum_{m=0}^\infty \kappa(n+1, m)\ t^m$$ Which gives us the recurrence relation: $$(m+1)\ \kappa(n, m)=F(n+1)\ \kappa (n+1,m)$$ I have no idea how to solve this recurrence equation or even if it is solvable. However if it is you should be able to get a series out of it. It then becomes a problem of solving that series until you can finally find a closed form for $k$.

With help from @Semiclassical we have the solution for $\kappa(n, m)$ as $$\kappa(n,m)=\left(\prod^n_{m=0}\frac{m+1}{F(n+1)}\right)\kappa(0, m)=\frac{(m+1)^{n+1}}{\mathfrak{F}(n+1)}\kappa(0, m)$$ The question is now finding $k(t, n)$ from this relation.

We have that $k(0, n) = 0$ or else there will be trouble with the convergence of $\mathfrak F$. This means that $$\mathfrak F(n) = \sum_{m=1}^\infty \kappa(n, m) \int_0^\infty e^{-t} t^m dt=\sum_{m=1}^\infty \kappa(n, m) \Gamma(m)$$ Consequently $$\sum_{m=1}^\infty \kappa(0, m) \Gamma(m)=1\tag 2$$ Which means that the seeding terms of $\kappa(n, m)$ seem to shrink faster than $\Gamma(m)$ grows as $m\to \infty$ and $n=0$.

Once a solution can be found for whatever $\kappa(0, m)$ is then a series for $k(t, n)$ can be established. I did some experimentation to see if $\kappa(0, m)$ could be guessed by letting $$\kappa(0, m)=\frac1{2^m\Gamma(m)}$$ as to satisfy condition $(2)$. This however lead to a value of $\mathfrak F(0)=\sqrt{3}$. This is clearly wrong.

If we let $\kappa(0, m)=\frac{a_m}{\Gamma(m)}$ we have $$\mathfrak F(n) = \sum_{m=1}^\infty \frac{(m+1)^{n+1}}{\mathfrak (n+1)}a_{m}\\ \implies \mathfrak F(n)^2F(n+1)=\sum_{m=1}^\infty (m+1)^{n+1} a_m$$ If we let $n=0$ we have $$1=\mathfrak F(0)=\sum_{m=1}^\infty a_m(m+1)=\sum_{m=1}^\infty m a_m + \underbrace{\sum_{m=1}^\infty a_m}_{=1\text{ by }(2)}$$ Therefore: $$\sum_{m=1}^\infty ma_m = 0$$ No idea what this means and I am unsure whether further manipulation will actually get anywhere. I feel quite lost carrying these last few out. I am open to ideas on how to determine $a_m$.

Ali Caglayan
  • 5,546
  • 9
  • 40
  • 72
  • 1
    How can we be sure that there is a unique $a_m$ satisfying the last statement? I've found a possible candidate: $$a_m = \frac {2m \cos(\pi m - \pi) -2 \cos(\pi m - \pi)}{2m-1-\cos(\pi m - \pi)}$$ which unfortunately blows up when $m=1$. Although one could maybe say $$\sum_{m\geq 1} ma_m = a_1 + \sum_{m\geq 2} ma_m = 0$$ which would mean that $a_1 = 1$. Have I gone wrong somewhere? – giobrach Jul 01 '16 at 17:15

I think I've found a suitable [see EDIT below] way to generalize the fibonorial in the reals! (I've taken inspiration from an answer to this question.)

The first terms of the fibonorial sequence $\mathfrak{F}(n)$ are: $$\mathfrak{F}_7(n)=\color{red}{1},1,2,6,30,240,3120,65620$$ where $\mathfrak{F}_7(n)$ is just $\mathfrak{F}(n)$ interrupted after the seventh term, for no particular reason. If we were to find the difference between consequent terms in this partial sequence we'd obtain $$\Delta\mathfrak{F}_7(n)=\color{red}{0},1,4,24,210,2880,62500$$ and if we were to go on, we'd arrive at the point where we'd obtain a single number, in this case $$\Delta^7\mathfrak{F}_7(n)=\color{red}{47844}$$ and if we were to go even further, we'd find an infinite sequence of $0$s. (Indeed this process is akin to differentiation of a polynomial.)

As you can notice, I've highlighted the first terms of each sequence of differences in red. The sequence of these red terms I define to be $c(n)=\Delta^n\mathfrak{F}(0)$, and by inspection and induction one obtains that $$c(n)=(-1)^n\sum_{k=0}^n(-1)^k\binom{n}{k}\mathfrak{F}(k)$$ Now the fibonorial sequence can be approximated as follows: $$\begin{align} T_q(n)&=\frac{c(0)}{0!}+\frac{c(1)}{1!}n+\frac{c(2)}{2!}n(n-1)+\cdots+\frac{c(q)}{q!}n(n-1)\cdots(n-q+1)\\ &=\sum_{j=0}^q \frac{c(j)}{j!}\frac{n!}{(n-j)!} \end{align}$$ This can of course be generalized to real $x$ by using the $\Gamma$ function: $$T_q(x)=\sum_{j=0}^q \frac{c(j)}{j!}\frac{\Gamma(x+1)}{\Gamma(x-j+1)}$$ which means that the limit of $T_q(x)$ as $q\to\infty$ should give us the Maclaurin series for a generalization of $\mathfrak{F}(n)$, which I shall call $\Phi(x)$: $$\Phi(x)=\sum_{j=0}^\infty \frac{c(j)}{j!}\frac{\Gamma(x+1)}{\Gamma(x-j+1)}=\sum_{j=0}^\infty \frac{(-1)^j}{j!}\sum_{k=0}^j\binom{j}{k}\frac{\Gamma(x+1)(-1)^k}{\Gamma(x-j+1)}\mathfrak{F}(k)$$

EDIT: unfortunately, I have just realized that this generalization is an oscillating function which happens to be equal to $\mathfrak{F}(n)$ when $x$ is natural, but is awfully off otherwise. I won't delete the answer because I want to know what has gone wrong!

  • 6,755
  • 15
  • 34