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 \(3\le p_n\le 2n-1\).
