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Radic
Radic is offline
#15
Sep29-03, 03:51 PM
P: 30
Well,
let use interval [0,24],and number 24.
We ca represent each even number as y=2*x where x is some positive integer,y is even positive integer and y>x.So as
y=2x=x+x=x+m+x-m,where m is such that x+m and x-m are primes.So the number of m ,called w where max(w)=n/2 (n is number of primes in interval [0,24]) defines number of two integer combinations that give the same number (in this case y).I call numbers x+m and x-m symmetric primes of x.So if in interval exists number such that w=n/2,other numbers could have maximaly (n-1)/2,(n-2)/2....1,or if we sum it then maximal number of combination that gives same numbers are n(n+1)/4.
Now about where comes n^2.There are n(n-1)/2 combinations of two different prime numbers,but there is 2+2,3+3,5+5 etc.. so u have more n combinations,and when you add it to n(n-1)/2 you'll get
(n^2+n)/2 which is actually number of combinations of two primes that gives even number.Now to get combinations in which pairs of primes gives the different even number,we have to do:
(4n^2+4n-n^2-n)/4=3(n^2+n)/4.....(A) combinations.
By the way in interval [0,a_n] there is a_n/2 even numbers and as
n^2>a_n then
n^2/2>a_n/2
and as
3(n^2+n)/4 - n^2/2=(n^2+3n)/4>0 or
3(n^2+n)/4>n^2/2
that means that number of combination of two primes which pairs give different even number(I meant on sum of two primes),equation (A) is
greater then number of even numbers in interval [0,a_n],or in other words for (ALL y=2k where k ELEMENT N)(EXISTS two primes) SUCH THAT
(y=a1+a2)