Conjecture: If a k − k π ( a k ) = 0 then a k + 1 / a k → e.A natural question to ask is for what n does π ( n ) ∣ n where π ( n ) is the prime counting function. It can be shown with a discrete variant of the intermediate value theorem that such an n exists for any k = n / π ( n ). (Intuition: f ( n ) = n − k π ( k ) can only increase by 1 and decrease by k − 1. Since f ( 2 ) &lt; 0 for k &gt; 3 and f ( n ) &gt; 0 for large n, there must be a zero).Computations suggest that all solutions to n − k π ( n ) = 0 are grouped at some n k in an interval of around n k width. But more surprisingly to me, for any pair of solutions a k and a k + 1 , with the former being in the k group and the latter in the k + 1 group, the ratio a k + 1 / a k seems to be about e.My conjecture: If a k − k π ( a k ) = 0 then a k + 1 / a k → e. In other words, the solutions of f ( n ) = n − k π ( k ) grow exponentially as e k .I am at a loss why this is the case. My guess is the gaps between primes [ n ! , n ! + n ] comes into play, giving us Stiring's formula and hence our e factor, but I really have no clue.Can someone either (1) provide a proof or disproof of this conjecture or (2) give an extremely plausible method of attack? I would also appreciate suggestions in the comments for ideas/theorems/techniques to investigate to help me make my own crack at the problem.

Question

Conjecture: If       a    k    −  k  π  (      a    k    )  =  0 then       a          k      +      1            /        a          k        →  e.A natural question to ask is for what n does   π  (  n  )  ∣  n where   π  (  n  ) is the prime counting function. It can be shown with a discrete variant of the intermediate value theorem that such an n exists for any   k  =  n      /    π  (  n  ). (Intuition:   f  (  n  )  =  n  −  k  π  (  k  ) can only increase by 1 and decrease by   k  −  1. Since   f  (  2  )  &amp;lt;  0 for   k  &amp;gt;  3 and   f  (  n  )  &amp;gt;  0 for large n, there must be a zero).Computations suggest that all solutions to   n  −  k  π  (  n  )  =  0 are grouped at some       n    k   in an interval of around             n      k       width. But more surprisingly to me, for any pair of solutions       a    k   and       a          k      +      1      , with the former being in the k group and the latter in the   k  +  1 group, the ratio       a          k      +      1            /        a    k   seems to be about e.My conjecture: If       a    k    −  k  π  (      a    k    )  =  0 then       a          k      +      1            /        a          k        →  e. In other words, the solutions of   f  (  n  )  =  n  −  k  π  (  k  ) grow exponentially as       e    k  .I am at a loss why this is the case. My guess is the gaps between primes   [  n  !  ,  n  !  +  n  ] comes into play, giving us Stiring&#039;s formula and hence our e factor, but I really have no clue.Can someone either (1) provide a proof or disproof of this conjecture or (2) give an extremely plausible method of attack? I would also appreciate suggestions in the comments for ideas/theorems/techniques to investigate to help me make my own crack at the problem.

Blaine Ross · Accepted Answer

Step 1This appears to be due to the distribution of the primes.Here&#039;s a heuristic.By the Prime Number Theorem,   π  (  x  )  ≈      x          log      ⁡      x      , so that       x          x              log        ⁡        x              ≈  log  ⁡  x and so to increase the value of       x          x              log        ⁡        x             by one, we need to increase x by a factor of e.

If a_k-k pi (a_k)=0 then a_{k+1}//a_k rightarrow e.

Open question

Answer & Explanation

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