is it true that this function:
f(n) = 3^(n)+2
will give a prime number for any natural value of n?
Nope, f(5) = 3^5 + 2 = 245 = 5 * 7^2.
Exercise: prove that f(n) assumes an infinite number of composite values.
To the best of my knowledge, there is no known algebraic expression that generates primes.
Well, you can get kind of close ;) http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html,
how about the function
f(n) = 3^(2n)+2
where n is a natural number
No. Have you even tried looking for a counterexample? One exists in the really small natural numbers.
in the examples cited from wolfram it sounds as if one may have no clue which inputs actually give primes (i.e. positive outputs) and which do not.
It may sound that way since it is true.
You can keep trying but you won't find a prime number function this way.
I think the only known single-parameter function that generates primes is the one involving Mill's constant : f(n) = [M^3^n]
what is mills constant? the 3^n th root of 3?
this does not sound promising Gokul. unless this "constant" is like my brother the engineers "fudge factor", i.e. the ratio between my answer and the right answer.
actually isn't it obvious no formula of this type, taking higher powers of the same thing, can ever give more than one prime?
or are you using brackets to mean something like the next smaller integer? even then I am highly skeptical. of course the rime number graph is convex, so has some sort of shape like an exponential, by the rpime number theorem, i guess, but what can you get out of that?
maybe asymptotically you might say something about a large number, unlikely even infinitely many, primes.
but i am a total novice here.
oh great, so "mills constant" is not even known. so the formula [M^(3^n)]. is not actually an explicit formula at all.
in fact apparently mills constant is computed by computing the primes instead.
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