Counting primes by counting numbers of the form $6k \pm 1$ which are not prime Again, pondering on twin primes, I came upon the following result. It baffles me a bit, so could someone give more intuitive reasoning why it works. 
First, define a function $P_6$ as $$P_6(n)=\begin{cases}
0, \ \  6n-1 \not\in \mathbb P \wedge 6n+1 \not\in \mathbb P \\
1, \ \  (6n-1 \not\in \mathbb P \wedge 6n+1 \in \mathbb P) \vee (6n-1 \not\in \mathbb P \wedge 6n+1 \in \mathbb P)\\ 
2, \ \  6n-1 \in \mathbb P \wedge 6n+1 \in \mathbb P  
\end{cases}$$
So $P_6(n)$ has value $0$ if neither of the numbers around $6n$ is a prime, $1$ if either but not both are primes and $2$ is both are.  
Let sets $P_6^0,P_6^1,P_6^2$ be the corresponding sets of indexes where $P_6 = 0,1 \vee 2$, so, for example, $\forall n\in P_6^1, P_6(n) = 1$    
Define three new functions using the indicator functions of above sets:
\begin{cases}
\pi_{6\bullet}^0 (n) = \sum_{i=1}^n  1_{P_6^0}(i) \\
\pi_{6\bullet}^1 (n) = \sum_{i=1}^n  1_{P_6^1}(i) \\
\pi_{6\bullet}^2 (n) = \sum_{i=1}^n  1_{P_6^2}(i)
\end{cases}
So these functions tell how many such indexes $1 \leq s \leq n$ there are for whom the number of primes surrounding $6s$ is $i$, $i \in \{0,1,2\}$.
These functions have following relations: \begin{equation} \pi_{6\bullet}^0 (n)+\pi_{6\bullet}^1 (n)+\pi_{6\bullet}^2 (n) = n  \ \ \ \ \ (1) \end{equation} and \begin{equation} \pi(6n+1)-2 = \pi_{6\bullet}^1 (n)+2 \pi_{6\bullet}^2 (n) \ \ \ \ \ (2). \end{equation}
Here $\pi(n)$ is the prime counting function and it has argument $6n+1$ because the biggest number we test is indeed $6n+1$ and we have to remove $2$ because the first two primes are not reachable via number six.
Now, from (1) we get \begin{equation} \pi_{6\bullet}^2 (n) = n-\pi_{6\bullet}^0 (n)-\pi_{6\bullet}^1 (n)  \ \ \ \ \ (3). \end{equation}
Substituting to (2) we get $$\pi(6n+1) = 2n-2\pi_{6\bullet}^0 (n)-\pi_{6\bullet}^1 (n)+2.$$
This works. For example $\pi_{6\bullet}^0 (5000) = 2223,$ and $\pi_{6\bullet}^1 (5000) = 2311$, and $10000-2*2223-2311+2 =3245 = \pi(30001).$
Can someone offer a bit more intuition on how this works? The $(2\pi_{6\bullet}^0 (n)+\pi_{6\bullet}^1 (n))$ is the number of numbers of the form $6k-1 \vee 6k+1$ between 5 and $6n+1$ which are not prime and when this is subtracted from $2n$ we get number of primes between 5 and $6n+1$. How?
 A: An integer has a chance of being prime if it is not divisible by 2 or 3. Call such a number a potential prime. The number of potential primes between 5 and $6n+1$ (again, that means integers not divisible by 2 or 3) is $2n-2$, because numbers not divisible by 2 or 3 occur in pairs $(6k-1, 6k+1)$ around the multiples of 6.
To get the actual number of primes, we have to subtract from $2n$ the number of these potential primes that are not prime. For each $k \le n$, see if the two neighbors of $6k$ are primes, and if not, subtract from $2n$ accordingly. If neither $6k-1$ nor $6k+1$ is prime, subtract 2 to account for the two non-primes. If exactly one neighbor of $6k$ is prime, we subtract 1. We can subtract all the 2's at once by subtracting $2\pi_{6\bullet}^0 (n)$, and we can subtract all the 1's at once by subtracting $\pi_{6\bullet}^1 (n)$. This gives the number of primes between 5 and $6n+1$. Taking into account the primes 2 and 3, the result is the formula you wanted an explanation for, $$\pi(6n+1) = 2n-2\pi_{6\bullet}^0 (n)-\pi_{6\bullet}^1 (n)+2\textrm{.}$$
A: I proposed "matrix sieve" algorithm for finding primes: 
Positive integers which do not appear in both arrays $A1(i,j)=6i^2+(6i−1)(j−1)$ and $A2(i,j)=6i^2+(6i+1)(j−1)$
                        |  6   11    16     21   ...|
            A1(i,j) =   | 24   35     46    57   ...|
                        | 54   71     88   105   ...|
                        | 96  119    142   165   ...|
                        |...  ...  ...   ...     ...|


                         |  6    13   20    27   ...|
             A2(i,j) =   | 24    37   50    63   ...|
                         | 54    73   92   111   ...|
                         | 96   121  146   171   ...|
                         |...      ...       ...        ...   ...|

are indexes k of primes in the sequence $S1(k)=6k−1$ .
Positive integers which do not appear in both arrays $A3(i,j)=6i^2−2i+(6i−1)(j−1)$ and $A4(i,j)=6i^2+2i+(6i+1)(j−1)$
                               | 4       9     14       19.. |
                               |20      31     42       53...|
                               |48      65     82       99...|
                      A3(i,j)= |88     111     134     157...|
                               |...   ...      ...     ...   |

                        | 8      15      22     29 ..|
                        |28     41       54     67...|
               A4(i,j)= |60     79       98     117..|
                        |104   129      154    179...|
                        |...    ...     ...     ...  | 

are indexes k of primes in the sequence $S2(k)=6k+1$. Since all primes (except 2 and 3) are in one of two forms 6k−1 or 6k+1 so we
can find primes simply by picking up positive integers which do not appear in these arrays.(C++ code see http://www.planet-source-code.com/vb/scripts/BrowseCategoryOrSearchResults.asp?lngWId=3&blnAuthorSearch=TRUE&lngAuthorId=21687209&strAuthorName=Boris%20Sklyar&txtMaxNumberOfEntriesPerPage=25
