GC: Every even number (N) is the sum of two odd primes (P_i) & (P_j).
Conditions: P_i < P_j, P_i + P_j => N, for all P_i < N, P_n < N, P_(n+1) > N
General Statistics of Prime Numbers.
Number of Primes Less than N = PI ~ N/LN(N).
Average Interval Between Primes (in the vicinty of N) = LN(5*N). Note this is a
conservative estimate based on the premise that the average interval between all
primes less than N is LN(N) and the supposition that the interval in the vicinity
of N is should be slightly greater than LN(N); but less than the average interval between all primes less than 10*N. ie, LN(10*N).
HYPOTHESIS: For every P_i in the range 3 to P_i > P_j, there is a P_j such that
P_i + P_(j-1) < N <= P_i + P_j
GOAL. To predict the probability that P_i + P_j = N. For any given prime pair,
the sum is close to N but not necessarily equal to N. Consider all of the PP's
whose sums are in the range N to N + 8. This means that the sum of a pair in this
range must be N, N+2, N+4, N+6 or N+8. A priori, it is reasonable to assume that
the sums are evenly distributed amomg these 5 numbers.
Let P_i + P_j = N_k.
We need an acceptable value for k. K cannot be too high or P_i + P_(j-1) > N_0
If k is too low, the probabilities will be skewed.
The nominal interval between successive primes is I = LN(N). This means that
for approximately half of the prime pairs; the range for N_k is approximately
LN(N).
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QUESTION: For a specific N, how many P_i's;ie (I) are there in the specified range.
ANSWer: There are nominally PI/2 = N/(2*(LN(N)) = I such primes.
SIMPLE EXAMPLE: If N = 50, then
P_i P_(j-1) P_j
3 43 47
5 43 47
7 41 43
11 37 43
13 31 37
17 31 37
19 29 31
23 23 29
The expected values are; PI = 50/LN(50) = 12.7. Actual = 14.
Note that there are more than 7 P_i's. This is natural as there are typically more primes less than N/2 than there are primes greater than N/2.
Let's consider a very large N; say 10*10^17. Now we must use statistical properties
in our analysis.
PI = (10*10*17)/LN(10*10^17) = 2.413*10^16
I = 1.207*10^16
b