Is the Prime Numbers Function f(n) = 3^(n)+2 Always Prime?

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Discussion Overview

The discussion centers around the function f(n) = 3^(n) + 2 and its potential to generate prime numbers for natural values of n. Participants explore various functions and their properties related to prime generation, including f(n) = 3^(2n) + 2 and mention of Mills' constant.

Discussion Character

  • Debate/contested
  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether f(n) = 3^(n) + 2 produces prime numbers for all natural n.
  • Another participant provides a counterexample, noting that f(5) = 245 is composite, suggesting that f(n) produces infinitely many composite values.
  • Some participants express skepticism about the existence of a formula that consistently generates primes, referencing known polynomial expressions that approximate this goal.
  • Discussion includes the function f(n) = 3^(2n) + 2, with a participant asserting that it also does not yield primes, although counterexamples are suggested to exist.
  • There is mention of Mills' constant and a claim that it is the only known single-parameter function that generates primes, though the details of this constant are questioned.
  • Concerns are raised about the nature of formulas generating primes, with skepticism about their reliability and the implications of their mathematical properties.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the discussed functions can generate prime numbers. There are competing views on the validity of these functions and the existence of a reliable prime-generating formula.

Contextual Notes

Participants highlight the uncertainty surrounding the effectiveness of polynomial expressions in generating primes and the ambiguous nature of Mills' constant, which is not explicitly defined.

Anzas
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is it true that this function:
f(n) = 3^(n)+2

will give a prime number for any natural value of n?
 
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Nope, f(5) = 3^5 + 2 = 245 = 5 * 7^2.

Exercise: prove that f(n) assumes an infinite number of composite values.
 
Last edited:
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,

However, there exists a polynomial in 10 variables with integer coefficients such that the set of primes equals the set of positive values of this polynomial obtained as the variables run through all nonnegative integers, although it is really a set of Diophantine equations in disguise (Ribenboim 1991). Jones, Sato, Wada, and Wiens have also found a polynomial of degree 25 in 26 variables whose positive values are exactly the prime numbers (Flannery and Flannery 2000, p. 51).
 
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.
 
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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.
 
Anzas said:
how about the function
f(n) = 3^(2n)+2

where n is a natural number

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]
 
  • #10
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.
 
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  • #12
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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