If you look at a table of Prime Numbers, you will notice that they aren’t evenly spaced which raises the question of how many primes precede any arbitrarily selected number. On the other hand, if you look at a table of prime numbers they seem to follow a pattern. You will also notice that the prime numbers get rarer, as you look at higher and higher numbers. Most formulas that have been developed to find the location of any prime number seem to be a hit and miss affair because the prime numbers aren’t evenly spaced. George Riemann, who formulated the Riemann Hypothesis, found that his formula produced the occasional zero as an answer which he hypothesized lay on the line ( y = ½ ). So far, calculating on an individual basis, it has been found that over a billion zeros using Riemann’s formula lie on this line but since Riemann hypothesized all zeros the search goes on. So far, no one has developed a proof that definitely establishes all zeros lie on the line ( y = ½ ). Here’s a way of resolving the conundrum concerning Riemann’s zeros (0’s) using the concept of strings. If we add the digits of any number (N) like ( 7919 ), for example, we get a total ( 7 + 9 + 1 + 9 = 26 ) which represents its’ string ( 26 ) or line ( y = 26 ). If we continue the addition of the digits ( 26 ) we get ( 2 + 6 = 8 ) or a line ( y = 8 ) for number ( 7919 ). Riemann had the idea that when his equation produced a zero (0) in the calculation, that zero for non- trivial numbers was on the line ( y = ½ ). Working in reverse, we can take any arbitrary line ( y = k ), with ( k = ( 0 to 9 )), because all numbers ( N ) sum to a number from ( 0 to 9 ) as one digit. Any line ( y = k ) can contain an infinite number of numbers, since zero ( 0 ) does not change the value of ( y = k ). For instance, if ( y = 2 ), it can hold the numbers ( 2, 11, 101, 110, etc. ) since the total digits do not exceed (2). It can readily be seen that all zeros lie on the line ( y = k ) no matter what the value of k whether it be one digit or multiple digits and thus Riemann was correct when he said all zeros lie on the line ( y = ½ ), since ( y = ½ ) is the reciprocal of ( y = 2 ) and contains the numbers ( ½, 1/11, 1/101, 1/110, etc.. The same conclusion can be applied to any arbitrary line ( y = k ) or ( y = 1/k ) for values of k from ( zero (0) to infinity ) of any number (N) which has an infinite number of digits (D) If Riemann had changed his formula to produce zeros ( 0’s ) for other non-trivial values for other lines ( y = k ) or ( y = 1/k ) he would have ended up hypothesizing that all zeros ( 0’s ) also lie on these lines. There is also the suggestion that the line ( y = ½ ) has a relationship to a formula for locating the position of any prime / number of primes preceding that particular prime. If ( y = 2 ), then the line ( y = 2 ) contains the numbers (2, 11, 101, 1001, 10001, 100001, etc.) since the sum of the digits does not exceed 2. If ( y = ½ ), then the line ( y = ½ ) contains the numbers ( ½, 1/11, 1/101, 1/1001, 1/10001, 1/100001, etc. ) since the sum of the digits does not exceed ½ and ½ is the reciprocal of 2 or ( y = 2 ) . Raise each number ( ½, 1/11, 1/101, 1/1001, 1/10001, 1/100001, etc. ) to the power of 2. Sum the numbers which have a limit of (0.258363501). The square root of (0.258363501) is (0.508294698). Similarly, If ( y = ½ ), then the line ( y = ½ ) contains the numbers ( ½, 1/11, 1/110, 1/1100, 1/11000, 1/110000, etc. ) since the sum of the digits does not exceed ½ and ½ is the reciprocal of 2 or ( y = 2 ) . Raise each number ( ½, 1/11, 1/110, 1/1100, 1/11000, 1/110000, etc. ) to the power of 2. Sum the numbers which have a limit of (0.258347942). The square root of (0.258347942) is (0.508279394). The important number is (.508) to 3 figures. The calculation of the location of any prime probably involves the adding of zeros (0’s) to the middle of (.508) to equal the number of digits in the probable prime after 3 digits. For instance ( 7919 has 4 digits so therefore extend ( .508 ) to ( .5008 )) by adding 1 zero to the middle and then raising (.5008) to the power equal to one less than the digit in the probable prime ending in ( 1, 3, 7, 9 ).
The general approach is the following:
Extend ( .508) to equal the number of digits in the probable prime starting after 3 digits and then raise ( .508 extended ) to a power equal to the number of digits in the probable prime minus 1. For instance ( 7919 ) contains 4 digits so extend ( .508 ) to 4 digits by adding zero (0) to the middle ( .5008 ). (7919) has 4 digits, so raise ( .5008 ) to the power of 3 ( 4 digits - 1 = 3 ). ( .5008 ^ 3 = .12560096 ) Multiply ( 7919 X .12560096 ) which equals ( 994.6340063 ). 7919 is the 1000th prime which means this method is approximately 5 short of 1000. The same method can be used using ( ½ or .5 ) but the answer is less accurate (7919 X (.5^3) = 989.875) which is ( 10.125 ) short of 1000 ( 7919 is the 1000th prime).
For numbers, 3 digits or under, such as ( 97 ) multiply ( 97 X (.508^2)) which equals approximately 25. As a matter of interest ( 97 ) is the 26th prime.
For 1 digit numbers multiply by ( .508 ).
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