Maybe you like this more (I derived it yesterday) $$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_{2n} H_{2n+2}}{(2n+1)(2n+2)}=$$
$$\frac{29}{64}\zeta(3)+\frac{\pi^2}{12}+\frac{5\pi^3}{192}+\log(2)+\frac{\pi}{4}\log(2)+\frac{1}{24}\log^3(2)$$
$$-G-\frac{\pi}{2}-\frac{\pi^2}{24}\log(2)-\frac{1}{4}\log^2(2)-\frac{\pi}{16}\log^2(2)-\frac14\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right)$$
The question asks, what can be said about $\gcd(a,b)$ and $\gcd(3a, b)$ with $a,b\in\mathbb{Z}$.
They are obviously not equal in general, as $\gcd(ax, bx)=|x|\gcd(a,b)$. One thing I tried is letting $a=n\gcd(a,b)$ and $b=k\gcd(a,b)$ so that $3a=3n\gcd(a,b)$. Then $k$ and $n$ will be coprime. I ...
