Let $\gamma$ be the Euler-Mascheroni constant. Can there be natural numbers $a,b,c$ such that $\log a - \log b - \log \log \log c =\gamma$?

Question

Can there be integers satisfying $$\ \log a - \log b - \log \log \log c = \gamma \ \ \ ? $$

Answer 1

First let me say that I probably know far less about this topic than you. However, I would still like to post an observation I have about this problem. Let's use $$\log a - \log b - \log \log \log c =\log\left(\frac{a}{b \log \log c}\right)$$ Now let's suppose there are natural numbers $a,b,c$ where $$\log\left(\frac{a}{b \log \log c}\right) = \gamma = \lim_{n \to \infty} \left(\sum_{k=1}^n\frac{1}{k}-\log(n) \right) \\ \implies \frac{a}{b \log \log c} = \exp \left[ \lim_{n \to \infty} \left(\sum_{k=1}^n\frac{1}{k}-\log(n) \right)\right] \\ = \lim_{n \to \infty}\left[ \frac{1}{n}\prod_{k=1}^n e^{1/k}\right]$$ Now by manipulating the equality a bit we get $$\log \log c = \frac{a}{b} \lim_{n \to \infty}\left(\frac{1}{ \frac{1}{n}\prod_{k=1}^n e^{1/k}} \right) \\ = \frac{a}{b} \lim_{n \to \infty} \left(n\prod_{k=1}^n e^{-1/k}\right)$$

Next, I believe that $\log \log c$ is irrational for all $c \in \Bbb{N} - \{0,1 \}$. For if $\log \log c$ were rational, then there would be nonzero integers $x,y$ such that $$\log \log c = \frac{x}{y} \implies e^{e^{x/y}} \space \text{is a natural number.}$$ I think this can be established as impossible by playing around with the Lindemann-Weierstrass Theorem. http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem

If I am right that $\log \log c$ is always irrational, then by the equality obtained above, we deduce that $\lim_{n \to \infty} \left(n\prod_{k=1}^n e^{-1/k}\right)$ has to be irrational. However, that is unknown because we don't know if $\gamma$ is irrational or not. If it turns out that $\lim_{n \to \infty} \left(n\prod_{k=1}^n e^{-1/k}\right)$ is rational, then $\frac{a}{b} \lim_{n \to \infty} \left(n\prod_{k=1}^n e^{-1/k}\right)$ is rational, so we'd have a contradiction, meaning no natural numbers $a,b,c$ satisfy the equality. If $\lim_{n \to \infty} \left(n\prod_{k=1}^n e^{-1/k}\right)$ is irrational, the jury is still out. I realize I didn't prove anything in this post, but hopefully something in here can help you narrow down your options.