Enumerating Assorted Arctans

From $\tan\left(2\arctan\dfrac17\right)=\dfrac{2\tan\left(\arctan\dfrac17\right)}{1-\tan^2\left(\arctan\dfrac17\right)}=\dfrac{2\left(\dfrac17\right)}{1-\left(\dfrac17\right)^2}=\dfrac27\times\dfrac{49}{48}=\dfrac7{24}$

and $\tan\left(4\arctan\dfrac17\right)=\dfrac{2\tan\left(2\arctan\dfrac17\right)}{1-\tan^2\left(2\arctan\dfrac17\right)}=\dfrac{2\left(\dfrac7{24}\right)}{1-\left(\dfrac7{24}\right)^2}=\dfrac7{12}\times\dfrac{576}{527}=\dfrac{336}{527}$

Thus,

$\begin{aligned}\tan\left(5\arctan\dfrac17\right)=&\tan\left(4\arctan\dfrac17+\arctan\dfrac17\right)=\dfrac{\tan\left(4\arctan\dfrac17\right)+\tan\left(\arctan\dfrac17\right)}{1-\tan\left(4\arctan\dfrac17\right)\tan\left(\arctan\dfrac17\right)}\\=&\dfrac{\dfrac{336}{527}+\dfrac17}{1-\left(\dfrac{336}{527}\right)\left(\dfrac17\right)}=\dfrac{2352+527}{3689}\times\dfrac{527}{479}=\dfrac{2879}{3353}\end{aligned}$

From $\tan\left(2\arctan\dfrac3k\right)=\dfrac{2\tan\left(\arctan\dfrac3k\right)}{1-\tan^2\left(\arctan\dfrac3k\right)}=\dfrac{2\left(\dfrac3k\right)}{1-\left(\dfrac3k\right)^2}=\dfrac6k\times\dfrac{k^2}{k^2-9}=\dfrac{6k}{k^2-9}$

From Equation

$\begin{aligned}\pi=&20\arctan\left(\dfrac17\right)+8\arctan\left(\dfrac3k\right)\\2\arctan\left(\dfrac3k\right)=&\dfrac{\pi}4-5\arctan\left(\dfrac17\right)\\\tan\left(2\arctan\dfrac3k\right)=&\tan\left(\dfrac{\pi}4-5\arctan\dfrac17\right)\\\dfrac{6k}{k^2-9}=&\dfrac{\tan\dfrac{\pi}4-\tan\left(5\arctan\dfrac17\right)}{1+\tan\dfrac{\pi}4\tan\left(5\arctan\dfrac17\right)}\\\dfrac{6k}{k^2-9}=&\dfrac{1-\dfrac{2879}{3353}}{1+\dfrac{2879}{3353}}\\\dfrac{6k}{k^2-9}=&\dfrac{474}{3353}\times\dfrac{3353}{6232}\\\dfrac{2k}{k^2-9}=&\dfrac{79}{3116}\\6232k=&79k^2-711\\79k^2-6232k-711=&0 \\(79k+9)(k-79)=&0\\k=&-\dfrac9{79},\ 79\end{aligned}$

but $k$ should be positive because $\arctan{(-x)}=-\arctan{x}$ . Thus, $k=\boxed{79}$ .

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Notes:

$\tan(\arctan{x})=x$

$\tan(2x)=\dfrac{2\tan{x}}{1-\tan^2x}$

$\tan(x+y)=\dfrac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$

$\tan(x-y)=\dfrac{\tan{x}-\tan{y}}{1+\tan{x}\tan{y}}$