Saturday, November 12, 2011

The Largest Prime Number Conjecture



Yesterday (11/11/11) was a special day in terms of the calendar numerology. In China, many young people called this day as the “bachelor day of the century”, as every number in the calendar is “1”, single as bachelor. And, this won’t happen again until 100 years from now. In fact, the entire world celebrated this day in one way or the other.

Obviously, this calendar numerology has no scientific significance. Anyone who thinks it otherwise is, indeed, a solatic (meaning a lunatic in the daylight, a term invented by Matt Strassler, a theorist on physics). Of course, the calendar numerology does have some culture and psychological significances.

However, the simple numerology does have some great importance in modern science. Thus, on this special day, I would like to use a special number game (using only the number 0 and 1) to show this point.

The prime number game --- if X is the largest “known” prime number with m digits (such as, one billion digits),
then    Y1 = (10^n) + 1, and  n = m + 1.

          Y2 = (1111…1110) + 1, there are n digits in the bracketed number, and
                         n = m + 1 if m is an even number,
                         n = m + 2 if m is an odd number.
                       
           Then, n is checked with two steps.
  step 1. if  {3, 5, 7, 11 or 13} is a divisor of n, then n = n + 2

  step 2. if the square root of n [n^(1/2)],  [n^(1/3)], [n^(1/5)], [n^(1/7)], [n^(1/11)], or  [n^(1/13)]
                 i. is equal to an integer, then n = n + 2 and go back to step 1.
                 ii. is not equal to an integer, then n is chosen. 
           Thus, Y2 must have odd number digits and all its digits are “1”.

As this is only a game for this special (11/11/11) day, I will not try to prove a “Largest Prime Number” theorem but make it as a conjecture.

Largest Prime Number Conjecture:
 a. if X is the largest “known” prime number with m digits (such as, one billion digits), then one of the two numbers (Y1 or Y2) is a larger prime than X.


 b. if the Y2 above is not a prime, then let n = n + 2 and repeat the n-check steps to get a new Y2. There is always an Y2 which is a prime and Y2 > X.


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