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 $4x+1$. $[(5^{2^{n-1}})^{1/n}]e^{1/n}<...>[(5^{2^{n-1}})^{1/n}]e^{-1/n}$
Prime of the form $4x+1$ within two bounds
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number-theory
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2And what might make you think such a thing? – 2011-11-19
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0If you answer this conjecture it will lead to a third way of counting primes besides the two other well known methods. – 2011-11-19
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0"the two other well known methods"? – 2011-11-23