# Prime of the form $4x+1$ within two bounds

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 $4x+1$. $[(5^{2^{n-1}})^{1/n}]e^{1/n}<...>[(5^{2^{n-1}})^{1/n}]e^{-1/n}$

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And what might make you think such a thing? –  mixedmath Nov 19 '11 at 19:01
Your two bounds are the same, have you noticed that? –  Patrick Da Silva Nov 19 '11 at 19:03
The two bounds are different –  Vassilis Parassidis Nov 19 '11 at 19:07
It might be good to edit it to read $e^{-1/n}$ instead of $/e^{1/n}$. Else it looks exactly the same at first glance. –  ADF Nov 19 '11 at 19:11
If you answer this conjecture it will lead to a third way of counting primes besides the two other well known methods. –  Vassilis Parassidis Nov 19 '11 at 19:16

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.
how many primes$4x+1$ exist between the above bounds when$n=16$ only precise numbers accepteted –  Vassilis Parassidis Nov 23 '11 at 23:39
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 $\pm 1 \pmod 4.$