# Formula with prime-density 1 in the integers [closed]

The Hermite-grand-conjecture implies that f(k)=(2^(2^5^11^(7k+1))+1)/3 is prime for all natural numbers $k$.

Is there any explicit formula that has so far been proven to produce primes for all natural numbers?

If not, is there under some reasonable restriction of closed-form formula a possible non-constructive proof that there exist (or does not exist under even more restrictive conditions) a finite formula that produces primes for all natural integers?

If not, is there any formula that has been proven to output primes with a frequency approaching 1 for input naturals k approaching infinity?

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## closed as not a real question by Felipe Voloch, Will Jagy, Franz Lemmermeyer, Daniel Litt, Andrés CaicedoOct 9 '11 at 15:00

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Could you elaborate on the first line this seems surprising to me. For the rest yes prime generating functions exist, but they tend to be complex. Also, one can show that certain things cannot work, for example a polynomial. See en.wikipedia.org/wiki/Formula_for_primes – user9072 Oct 8 '11 at 17:25
The first line of your posting is nonsense in more than one way. – Franz Lemmermeyer Oct 8 '11 at 17:43
Please, be more specific. Define "explicit", and define "formula". For example, my formula f(x)=3 returns a prime for all natural numbers x. – Per Alexandersson Oct 8 '11 at 21:37

It depends a bit on what you accept as "explicit". E.g., there is a positive real number $A$ such that the integer part of $A^{3^n}$ is prime for all positive integers $n$. See Wikipedia on Mills' constant.
edit: This answer was for the original version, which said the "New Mersenne Conjecture" and gave the formula $f(k)=\frac{2^{2^k+1}+1}{3}$. At least that version gives integer results; the new one isn't an integer for any k.