Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer
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
Sorry, unless you calculate by brute force in any given instance, your question does not lend itself to exact answers. – Greg Martin Nov 25 '11 at 21:39

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 $\pm 1 \pmod 4.$

share|cite|improve this answer
Are you suggesting that those bounds are strong enough to settle the question? Would you give some details? – Gerry Myerson Nov 22 '11 at 6:26
@ 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. – Vassilis Parassidis Nov 23 '11 at 4:50
My question was directed to esofos. – Gerry Myerson Nov 23 '11 at 11:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.