Let me give an example of what I mean by the solution of The Strong Goldbach Conjecture. Consider the case where e=44;
The prime just less than the root of '44' is 5, so only the factor sets Z3 and Z5 exist up to 44;
[Z3] = 1/6 (44) = 7 1/3. But since there can be no fraction of a number, the answer must be either 7 or 8. In this case, listing the numbers according to
Z3:=> (6d -3) ,1< n<44 gives, {3;9;15;21;27;33;39} with '7' as the number of elements of Z3.
[Z5]= 2/30 (44) = 3, Z5:=>(30d-25)^(30d-5) , 1<n< 45 gives {5;25;35}
The number of prime pairs, according to the solution for the STRONG Goldbach Conjecture, is given by;
p(pairs) =(1/2) (44/4) (1 - {[Z3] +[Z5]}[Zr5]) ,
where [Zr5] = (8/30 )(45)=12
But(1/6 + 2/30) =7/30.
This is the portion of Zr5 that is taken over by the factor sets Z3 and Z5 in sharing; The portion left over is therefore (23/30)
(23/30)(11) = 8.433
=> 8<(answer)<9
=> p(pairs) = (1/2)(8) or (1/2)(9)
=4 or 4 1/2
The pairs are, actually, (from Zr5:=> (3od-r5) ; r5 ={ 29;23;19;17;13;11;7;1}
which gives,
for 1<n<44, {1;7;11;13;17;19;23;29;31;37;41;43})
{(43+1);(41+3);(37+7);(31+13)}
Found to be 4, which raises questions about where the so-called 'Simultaneous equations' got their origins.
The prime just less than the root of '44' is 5, so only the factor sets Z3 and Z5 exist up to 44;
[Z3] = 1/6 (44) = 7 1/3. But since there can be no fraction of a number, the answer must be either 7 or 8. In this case, listing the numbers according to
Z3:=> (6d -3) ,1< n<44 gives, {3;9;15;21;27;33;39} with '7' as the number of elements of Z3.
[Z5]= 2/30 (44) = 3, Z5:=>(30d-25)^(30d-5) , 1<n< 45 gives {5;25;35}
The number of prime pairs, according to the solution for the STRONG Goldbach Conjecture, is given by;
p(pairs) =(1/2) (44/4) (1 - {[Z3] +[Z5]}[Zr5]) ,
where [Zr5] = (8/30 )(45)=12
But(1/6 + 2/30) =7/30.
This is the portion of Zr5 that is taken over by the factor sets Z3 and Z5 in sharing; The portion left over is therefore (23/30)
(23/30)(11) = 8.433
=> 8<(answer)<9
=> p(pairs) = (1/2)(8) or (1/2)(9)
=4 or 4 1/2
The pairs are, actually, (from Zr5:=> (3od-r5) ; r5 ={ 29;23;19;17;13;11;7;1}
which gives,
for 1<n<44, {1;7;11;13;17;19;23;29;31;37;41;43})
{(43+1);(41+3);(37+7);(31+13)}
Found to be 4, which raises questions about where the so-called 'Simultaneous equations' got their origins.