Let $Re(z) > 1$
Consider functions such as
$$f(z) = (z-1) ([2 / (z^{1/a_1} + 1)] [2 / (z^{1/a_2} + 1)][2 / (z^{1/a_3} + 1)][2 / (z^{1/a_4} + 1)][2 / (z^{1/a_5} + 1)]...)$$
or more formally
$$f(z) = (z-1) \prod_{n>0} \frac{2}{z^{1/a_n}+1}$$
Where
$$1/a_1 + 1/a_2 + 1/a_3 + ... = 1.$$
and $a_n$ is a rising integer sequence.
Now clearly this function does not grow like any power of $x$.
For instance if $a_n = 2^n$ we get $f(z) = \ln(z)$.(*)
So what are the good asymptotics of $f(z)$ ?
Let $\pi(n)$ be the counting function of integers $a_n$ between $0$ and $n$.
Then it seems a good estimate for $f(z)$ when $z$ is close to the positive real line is
$$f(z) = \exp(C \pi(\ln(z)))$$
For some value $C$.
Is this correct ? And how good is it exactly ?
For instance if $a_n = n(n+1)$ do we get
$$f(z) = O(\exp(\sqrt\ln(z))) $$?
And if $a_n$ is sylvester's sequence do we get
$$f(z) = O(\exp(C \pi(\ln(z))))$$
where $\pi$ counts the sylvester numbers ?
For instance let $z = \exp(400)$
and let $a_n = n(n+1)$.
Then $f(z) = \exp(q)$
where $20 < q < 21$.
thereby supporting the estimate
$$f(z) = O(\exp(\sqrt\ln(z))) $$
See the proof for the $\ln(x)$ case here
(it is equivalent since the "$e$ parts" cancel to 2's)
and also somewhat related :
I must also thank my mentor for this idea.
I will add a proof for the first link that I received from my mentor.
https://sites.google.com/site/tommy1729/special-products