Asymptotics for $f(z) = (z-1) \prod_n \frac{2}{z^{1/a_n}+1}$ with $1/a_1 + 1/a_2 + 1/a_3 + ... = 1.$

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

Why do we have $ \ln(z) = \frac{z-1}{e -1} \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {z^{2^{-n}} + 1} $

(it is equivalent since the "$e$ parts" cancel to 2's)

and also somewhat related :

$(3x- 3)\prod_{n = 1}^∞\frac{\exp(2^{-n})+1}{x^{2^{-n}}+1}=(x^3-1)\prod_{n = 1}^∞\frac{\exp(2^{-n})+1}{x^{3 \cdot 2^{-n}}+1}$

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