Just noticed something interesting. Can you prove it or find a counterexample? Any interesting generalizations?

The Fibonacci numbers are given by
[tex: F_0 = 0]
[tex: F_1 = 1]
[tex: F_{n+2} = F_n + F_{n+1}].

For even [tex: n \ge 2],
[tex: \arctan(1/F_n) = \arctan(1/F_{n+1}) + \arctan(1/F_{n+2})].

[tex: tan(a+b)=\frac{tan(a)+ban(b)}{1-tan(a)tan(b)}]
Thus, your formula would be correct if and only if [tex: \frac{1}{F_n} = \frac{\frac{1}{F_{n+1}}+\frac{1}{F_{n+2}}}{1-\frac{1}{F_{n+1}F_{n+2}}}] which can be simplified to [tex: P_n=F_{n+1}^2 - F_{n+1}F_n-F_n^2 = 1]

It seems that [tex: P_n=1] for n even and [tex: P_n=-1] for n odd.
As it turns out, [tex: P_{n+1} = F_{n+2}^2 - F_{n+2} F_{n+1} - F_{n+1}^2 = F_n^2 + F_{n+1} F_n - F_{n+1}^2 = -P_n], so this is indeed true.

P7391
Nice. I came across this while looking for ways to compute pi = 4 arctan(1) by repeatedly decomposing arctangents. We have

arctan(1/n) = arctan(1/a) + arctan(1/b)

if and only if [tex: (a-n)(b-n) = n^2 + 1],

so you can decompose arctan(1/n) into arctangents of smaller unit fractions by factoring [tex: n^2 + 1].

For odd n we get
[tex: \arctan(1/F_n) = \arctan(1/F_{n+1}) + \arctan(1/F_{n+2}) - \arctan(1/G_{n+1})]
where [tex: G_n = \frac{F_n^3 + 3F_n}{2}].

P7467
which can be rewritten as
[tex: \arctan(1/F_n) = \arctan(2/F_{n+1}) - \arctan(1/F_{n+1}) + \arctan(1/F_{n+2})]