@Will: I am totally lost as to what you mean by 'smallest subinterval $[p-1,p]$ ... which after $p$ is that number ends at $-1$.' Also, you refer to primes where $Pi(x=p)$ is odd. However, the unresolved 'pattern' relates to $Pi(x=p)$ being even. If thats what $p_1$ and $p_2$ deal with, I'm really sorry. May I ask for a complete example, so I can understand better? Thanks!

$-1$ cannot occur, is more personal opinion and evidence based guessing than proven fact. Maybe I didn't make that clear. Making the assumption seems to make the problem 'almost' susceptible to inductive methods, heuristics, or some witch's brew of everything we need :)

@quid: Yes, conjecture is equivalent to $Z_f(P({2n}))=1$ Also, so far, $-1$ hasn't shown up for me (and I assume, for others who checked via programs, otherwise they would have posted). It is very possible that there is some prime number with $Z_f$ equal to $-1$. However, I find it unlikely as every odd number so far has terminated in $\{0,1,2\}$ only. Check the page on M.SE, I think it has a better explanation.

About the whole $-2$ thing, I applied the algorithm wrong for $9$. And didn't double check. I'm sorry guys!

I would like a formal proof. Knowing the initial parity cannot be used to predict a result. The final parity is still dependent on $\pi(x)$.

Also, $Z_f(5)=0$ and $Z_f(31)=2$

At math.stackexchange.com/questions/176394/a-prime-number-patte‌​rn @alex has already shown (using above parity argument) that starting positions between -2 to 2 are valid. To resolve the conjecture as proposed, only the -1 case needs scrutiny.

'If n is prime, it is assumed accounted for by the first step' $19-17-13+11+7-5-3+2=1$

Also, while we inductively prove for the next prime $q$, we also need to prove that $z \in \{0,1,2\}$ for all $2r>k>2q$ where $r$ is the next prime (before we attempt induction on $r$).

This question has also been posted here.(math.stackexchange.com/questions/176394/…). Thanks to @Gerry Myerson

The above induction wouldn't suffice to prove the regularity in pattern, i.e. only every other prime, beginning with 3, reaches 1.