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Longest repeated sequence in the prime counting function

  1. Jun 19, 2009 #1
    Is there a longest repeated sequence (congruency) in the prime counting function [tex] \pi (x) [/tex] (that which gives the number of primes less than or equal to x)?

    Recall that [tex] \pi (x) [/tex], although infinite, may not be random, and itself starts out with an unrepeated sequence [tex] \pi (2)=1 [/tex] and [tex] \pi (3)=2 [/tex] (with a "slope" of 1).
     
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  3. Jun 19, 2009 #2

    CRGreathouse

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    I really can't tell what you mean. Could you explain what you mean more carefully, maybe with an example?
     
  4. Jun 19, 2009 #3
    There are arbitrarily large gaps in the prime numbers. This means that [tex]\pi(n)[/tex] can be constant over an arbitrarily large interval.

    Consider the n-1 numbers [tex]n!+k[/itex] where [itex]k=2,3,\cdots n[/tex]
     
  5. Jun 19, 2009 #4
    I think that Simon-M answered my question, and with a basic example - that the prime counting function as graphed can repeat itself indefinitely (such as when constant over an arbitrarily large interval). Another example would include the interjection of one prime into such an arbitrarily large sequence, which then could be repeated.
     
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