More Prime Spiral Goldbach

 Consider,

 $2n=p+q$

where $p$ and $q$ are primes and $n$ an integer greater than 2.

$2n-q=p$

let $q=\cfrac{a}{b}$

$2n-\cfrac{a}{b}=p$ ---(*)

$\cfrac{2nb-a}{b}=p$

If p is prime,

$b=2nb-a$ or

$b=1$

When $b=1$,

$2n-a=p$ is prime

we have instead,

$2n=p+a$

so,

$a=q$

interesting result, $\rightarrow$备用

or from $b=2n_ob-a$

$a=2n_ob-b=b(2n_o-1)$

$\cfrac{a}{b}=2n_o-1$

for a specific number $n_o$ and $b$ not zero.

sub. into (*) we have,

$2n-2n_o+1=p$

$2(n-n_o)+1$ is prime as $p$ is prime and $n\gt n_o$ as primes are positive.

Let $2(n-n_{oo})+1=q_i$ where $n\gt n_{oo}$ for positive prime.

where $n$ is paired with $n_{oo}$ to give $q_{i}$ as $n$ and $n_o$ is paired to give $p$.  Setting $n_i$ instead of $n$ is not necessary.

In which case, we formulate,

$q_i+p=2(n-n_o)+1 + 2(n-n_{oo})+1$

and so,

$q_i+p=2(2n-n_o-n_{oo})+2=2n_j$

where $n_j=2n-n_o-n_{oo}+1$ which is just another integer and both $q_i$ and $p$ are prime. And when we set $2n=n_o+n_{oo}$ we start from $n_j=1$.

Similarly,

$p_i+q=2(n-n_g)+1 + 2(n-n_{gg})+1$

and so,

$p_i+q=2(2n-n_g-n_{gg})+2=2n_j$

where $n_j=2n-n_g-n_{gg}+1$ which is just another integer and both $p_i$ and $q$ are prime.

Is $n_j$ strictly non negative?  Yes. And When we set $2n=n_g+n_{gg}$ we start from $n_j=1$.

And we add the mess together,

$p_i+q_i+p+q=2(2n-n_g-n_{gg})+2(2n-n_o-n_{oo})$

since $p+q=2n$

$p_i+q_i=2(3n-n_g-n_{gg}-n_o-n_{oo})$

in order that 

$p_i+q_i=2(n+1)$

we set, 

$2(n+1)=2(3n-n_g-n_{gg}-n_o-n_{oo})$

by choosing appropriate values for $n_g$, $n_{gg}$, $n_o$ and $n_{oo}$.

$2n=n_g+n_{gg}+n_o+n_{oo}+1$

So, $p_i=2(n-n_{gg})+1$ and $q_1=2(n-n_{oo})+1$

where given $n$, $n_{gg}$ and $n_{oo}$ are chosen such that $p_i$ and $q_i$ are primes.

$n_g$ and $n_o$ are derived from the starting/previous primes $p$, $q$ and $n$.

$n_g$, $n_{gg}$, $n_o$ and $n_{oo}$ are not necessarily prime just integers greater than 2.

One way to simplify is to set $p=q$, but this series is troubled.

Given, $2n=n_g+n_{gg}+n_o+n_{oo}+1$,The set $[$ $n_g$, $n_{gg}$, $n_o$, $n_{oo}$$]$ has one odd or 3 odd numbers.  

If we go on further, the next equation is,

$2(n+1)=n_{gg}+n_{ggg}+n_{oo}+n_{ooo}+1$

$2n+1=n_{gg}+n_{ggg}+n_{oo}+n_{ooo}$

we have to introduce $n_{ggg}$ and $n_{ooo}$ and the set $[$ $n_{ggg}$, $n_{gggg}$, $n_{ooo}$, $n_{oooo}$$]$ has one odd or 3 odd numbers, as before.

and the next in the series after $n_{ggg}$ and $n_{ooo}$

$2(n+2)=n_{ggg}+n_{gggg}+n_{ooo}+n_{oooo}+1$

$2n+3=n_{ggg}+n_{gggg}+n_{oo}+n_{oooo}$

We gather the equations with a change in notation,

$2n-1=n_g+n_{g1}+n_o+n_{o1}$ 

$2n+1=n_{g1}+n_{g2}+n_{o1}+n_{o2}$

$2n+3=n_{g2}+n_{g3}+n_{o2}+n_{o3}$ 

$2n+5=n_{g3}+n_{g4}+n_{o3}+n_{o4}$ 

$2n+7=n_{g4}+n_{g5}+n_{o4}+n_{o5}$ 

and in general,

$2n+(2i-1)=n_{gi}+n_{gi+1}+n_{oi}+n_{oi+1}$ --- (*)

where $i$ is an integer starting at 0,  $n_{g0}=n_{g}$ and $n_{o0}=n_{o}$, given a start,

$2n=n_g+n_{g1}+n_o+n_{o1}+1$  from $p+q=2n$ and $p_i+q_i=2(n+1)$.

If we take the difference of the expressions to retain the pair $n_g$ and $n_o$ and the last pair $n_{gi}$ and $n_{oi}$

We have two series depending on $i$ being odd or even.

For $i=1$,

$n_{g2}+n_{o2}=n_g+n_o+2$

For $i=2$ substitute away $n_{g2}$ and $n_{o2}$

$2n+1=n_{o}+n_{g}+n_{o3}+n_{g3}$ 

which involves $n$.

For $i=3$, 

$n_{g4}+n_{o4}=n_{g}+n_{o}+4$ 

For $i=4$, substitute away $n_{g4}$ and $n_{o4}$

$2n+3=n_{g}+n_{o}+n_{g5}+n_{o5}$ 

Looking forward, when $i$ is odd,

$n_{g2}+n_{o2}=n_g+n_o+2$  for $i=1$

$n_{g4}+n_{o4}=n_{g}+n_{o}+4$  for $i=3$

So, in general,

$n_{gi+1}+n_{oi+1}=n_{g}+n_{o}+(i+1)$  when $i$ is odd.

When $i$ is even,

$n_{g3}+n_{o3}=(2n+1)-n_{g}-n_{o}$ for $i=2$ 

$n_{g5}+n_{o5}=(2n+3)-n_{g}-n_{o}$ for $i=4$

In general,

$n_{gi+1}+n_{oi+1}=(2n+(i-1))-n_{g}-n_{o}$  when $i$ is even

In summary,

$n_{gi+1}+n_{oi+1}=n_{g}+n_{o}+i+1$            $i$ is odd, 1, 3, 5...

predicts the next pair $n_{gi+1}$ and $n_{oi+1}$ from an odd position $i$ odd.

$n_{gi+1}+n_{oi+1}=(2n+i-1)-n_{g}-n_{o}$  $i$ is even, 2, 4, 6

predicts the next pair $n_{gi+1}$ and $n_{oi+1}$ from a even position $i$ even.

Consider starting from a odd position,

$n_{gi}+n_{oi}=n_{g}+n_{o}+i$   and the next equation,

$n_{gi+1}+n_{oi+1}=(2n+(i+1)-2)-n_{g}-n_{o}$ 

where next $i$ is $i+1$.

$n_{gi+1}+n_{oi+1}+n_{gi}+n_{oi}=(2n+(i+1)-2)+i=2n+2i-1$  

which is the expression (*).

Consider starting from a even position,

$n_{gi}+n_{oi}=(2n+i-2)-n_{g}-n_{o}$ and the next equation,

$n_{gi+1}+n_{oi+1}=n_{g}+n_{o}+i+1$  

where next $i$ is $i+1$.

$n_{gi+1}+n_{oi+1}+n_{gi}+n_{oi}=(2n+i-2)+i+1=2n+2i-1$  

which is again the expression (*).

This checks the expressions for predicting $n_{gi}$ and $n_{oi}$.

This is not proof of Goldbach's Conjecture but provide guidelines to predict $n_{gi+1}$ and $n_{oi+1}$ from $n_{gi}$ and $n_{oi}$ provided with a start $n_{g}$ and $n_{o}$ and some integer $n$ greater than 2.  

$n_{gi+1}$ and $n_{oi+1}$ gives an expression

$p_{i+1}+q_{i+1}=2(n+i+1)$

where $2(n+i+1)-2n_{oi+1}+1=p_{i+1}$ and $2(n+i+1)-2n_{gi+1}+1=q_{i+1}$

starting at $p+q=2n$ with $i=0$, $p$ and $q$ being prime and $n$ is an integer greater than 2.

$n_{g}$ and $n_{o}$ are derived from the expressions,

$2n-2n_o+1=p$ and $2n-2n_g+1=q$

where $n_{g}$ and $n_{o}$  is chosen such that $p$ and $q$ are primes.

And so primes can be analyzed using an expression,

$p_n=2n-2n_p+1$ where $n\in\mathbb{Z},\,\,n\gt 2$.  $n$ group primes into $n_p\le n$, given $n$, $n_p\gt 2$.

We can restrict membership in $p_n$ to $3\le p_n\le 2n-1$.