Sum of arctangents equals to pi/4, part 2

In order the get the largest value of $d$ , the values of $a$ and $b$ must be as smallest as possible. Since $\arctan \frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$ , then the smallest values of $(a,b)=(2,4)$ . (The readers may check that $(a,b)=(3,3)$ could not gives the desired value $d$ .)

Now $\displaystyle \arctan\left(\frac{1}{2}\right)+\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{1}{c}\right)+\arctan\left(\frac{1}{d}\right)=\frac{\pi}{4}$ gives $cd-1=13(c+d)$ , which means that $d=\frac{13c+1}{c-13}=13+\frac{170}{c-13}$ . This implies that the largest value of $d$ is $13+170=183$ , which occurs when $c=14$ .

Remarks : I wonder if there is any efficient way to solve the general case $\sum_{k=1}^{n}\arctan\frac{1}{a_k}=\frac{\pi}{4}$ for positive integers $a_k$ and $a_i\le a_j$ for all $i \le j$ .