Finding a recursive formula/sum for $\pi$ I'm very much aware of the
$$\pi = 4 \left(1 - \frac13 + \frac15 - \frac17 + \frac19 -\cdots \right)$$ and
$$\pi = \sqrt{6\left(1 + \frac14 + \frac19 + \frac{1}{16}+ \cdots \right)}$$ 
and even less common $$\frac\pi2 = \frac21 \cdot \frac23 \cdot \frac43 \cdot \frac45 \cdot \frac65 \cdot \frac67 \cdot \frac87 \cdot \cdots$$  

What I'm wondering is if there is a RECURSIVE (not summative) way to find $\pi$. 

An example for phi ($\phi$) is: $$f(1) = 1; \quad f(n) = 1+\frac1{f(n-1)}$$

By recursive function, I mean a function that calculates its next value based on the current or previous (or even further back) value.
There is no rational recurrence of the form

$$\pi_{n+1}=\frac{P(\pi_n)}{Q(\pi_n)}$$ where $P,Q$ are polynomials with algebraic coefficients, because that would mean that $\pi$ is itself algebraic (it would be solution of $x Q(x)-P(x)=0$).

If transcendental functions are allowed, you can use the interesting

$$\pi_{n+1}=\pi_n+\sin\pi_n$$

but it is simpler to directly use $\pi_n=4\arctan 1$.

(This essentially means that there is no suitable recurrence.)
Your question is not very clear, in case of $\phi$ we calculate $\phi(n)$ but we do not calculate $\pi(n)$. $\pi$ is a constant, and you want recursive formula, so from the context i assume you want to calculate the value of $\pi$ using a recursion. To calculate value of $\pi$ using recursion you can use any of the formula listed above.
e.g.
$\pi = \displaystyle \prod_{n=1}^{\infty}  \cfrac{2n}{2n -1} \cfrac{2n}{2n - 1} $
You can do like this
$\pi =\lim_{z \to \infty} \displaystyle \prod_{n=1}^{z}  \cfrac{2n}{2n -1}   
\cfrac{2n}{2n - 1} $
$\pi(z) = \displaystyle \prod_{n=1}^{z}  \cfrac{2n}{2n -1}   
\cfrac{2n}{2n - 1} \implies \pi(z) = \pi(z-1)  \cfrac{2z}{2z -1}   
\cfrac{2z}{2z - 1}$
More the value of  z more accurate value of $\pi$ you will get.