Finding infinite sequences with pairwise relatively prime outputs. I am looking for a formula which for every element in $\mathbb{Z}$ as an input, gives pairwise relatively prime outputs. That is for example thanks to Greg Martin's suggestion the positive outputs of 
$2^{2^{n}}+1$ seem to all be pairwise relatively prime.
I understand that the infinite set of primes is a solution however there is no known formula for primes. 
I have also tried the product
($\prod_{i=1}^ni)+1$ Which I am 90% sure is correct. In the best of scenarios I can find a working solution of the form $y=f(x)$ and maybe a proof solidifying my guess. So far I have been unsuccessful but would like to hear your suggestions or answers. Thank you.
 A: I was just working on something similar myself which might help; this isn't a single formula so much as a description of a set of sequences with that property, and although there are much smaller-valued solution sets, an easy one to describe is $\Phi_{p_k}(x)$ for $k\in\mathbb{N}$, where $p_k$ is the $k$th prime, and $\Phi_i$ represents the $i$th cyclotomic polynomial.
E.g., a set like
$$
\begin{array}
\\ x+1,\\ x^2+x+1,\\ x^4+x^3+x^2+x+1,\\x^6+x^5+x^4+x^3+x^2+x+1,\\\ldots
\end{array}$$
will be pairwise prime for any $x$.
I also suspect you can do it with an infinite number of quadratics, but they're not as easily generated. Linear recurrences would also be a good place to try.
Oh right, and if nobody's mentioned it, Sylvester's sequence and its variants has this exact property.
A: The Fermat numbers $2^{2^n}+1$ are pairwise relatively prime - this is not just an observation but is in fact provable.
It's easy to turn this into a function $f\colon \Bbb Z\to\Bbb N$ where the outputs are pairwise relatively prime:
$$
f(n) = \begin{cases}
2^{2^{2n}}+1, &\text{if }n\ge0; \\
2^{2^{2|n|-1}}+1, &\text{if }n\le-1.
\end{cases}
$$
Or, if you prefer something defined without cases, just let $f(n) = 2^{2^{|3n-1|}}+1$. Or if you even want an analytic function: $f(n) = 2^{2^{(3n-1)^2}}+1$.
