0
$\begingroup$

Let's have the number $5^{2^{n-1}}$ where $n$ any non-zero natural number. I conjecturally say that between the following two bounds we will always obtain $n$ primes of the form $4x+1$. $[(5^{2^{n-1}})^{1/n}]e^{1/n}<...>[(5^{2^{n-1}})^{1/n}]e^{-1/n}$

  • 2
    And what might make you think such a thing?2011-11-19
  • 0
    If you answer this conjecture it will lead to a third way of counting primes besides the two other well known methods.2011-11-19
  • 0
    "the two other well known methods"?2011-11-23

2 Answers 2

3

The interval in question is essentially of the form$$\bigg[T\bigg(1-\frac c{\log\log T}\bigg),T\bigg(1+\frac c{\log\log T}\bigg)\bigg]$$ for $T=5^{2^{n-1}/n}$. Chebyshev-type bounds will not be strong enough to establish that there are primes in this interval. However, the prime number theorem for arithmetic progressions is strong enough to show that there are asymptotically $cT/(\log T \log\log T)$ primes in that interval that are 1 (mod 4). Therefore your conjecture is true when $n$ is sufficiently large, and could in principle be confirmed for all $n$ by a finite calculation.

  • 0
    how many primes$ 4x+1$ exist between the above bounds when$ n=16$ only precise numbers accepteted2011-11-23
  • 0
    Sorry, unless you calculate by brute force in any given instance, your question does not lend itself to exact answers.2011-11-25
2

Erdos proved the Chebyshev bounds for the prime counting function at a very young age. You can find these proofs in Hardy-Wright. After a short period of time he also proved the analogous result for the counting function of the primes $\pm 1 \pmod 4.$

  • 0
    Are you suggesting that those bounds are strong enough to settle the question? Would you give some details?2011-11-22
  • 0
    @ Gerry: If the conjecture is true then we can develop a method to find the number of primes that are contained up to the upper bound (see above). That makes the other methods for the distribution of primes only of historical interest. Until this is settled I have nothing more to add.2011-11-23
  • 0
    My question was directed to esofos.2011-11-23