I think I discovered a very (compared to other arctangent formulas) fast arctangent formula for pi. My question is: Is it the best? My formula starts: $$\frac{\pi}{4} = 1065\arctan(\frac{1}{1356}) + \arctan(\frac{1}{13211959}) + \arctan(\frac{1}{1661269101985727}) \pm \dots$$ There are $15$ arctangents in total. All of the arctangents except for the last one have a numerator of $1$. The last one has a numerator that is $3316$ digits long but it has a denominator that is $68680$ digits long so the numerator doesn't matter too much. All of the arctangents besides the first is either multiplied by $1$ or $-1$, whereas the first is multiplied by $1065$.
Because the denominators are all so big it makes it converge very fast. By estimating it's score (an approximate multiple of time) using $\sum \frac{1}{\ln(b_n)}$ where $b_n$ is the $n$-th denominator, we get a score of $\approx$0.1169, which is incredibly low! We would need to increase it a little though, to account for the fact that the last arctangent's numerator is not $1$. Although I don't think we would need to increase it by much because that arctangent's denominator is SO large ($\approx 10^{68679}$), but do I?
On this Wikipedia page it lists the most efficient currently know Machin-like formulas for pi and by applying the score formula we used earlier to them we get that the best one is $\approx 0.58232$. Because a lower score (meaning smaller time) is better mine seems to be the best arctangent formula for pi there is. Please tell me if I am correct or if I've made a mistake.
List of the terms in my formula:
- $1065 * \arctan(1 / 1356)$
- $\arctan(1 / 13211959)$
- $\arctan(1 / 1661269101985727)$
- $-1 * \arctan(1 / 45795729185495880262825011844107)$
- $-1 * \arctan(1 / 2700517276808431966374537126272309157148546111625235650501520960)$
- $\arctan(1 / \approx 10^{127})$
- $\arctan(1 / \approx 10^{254})$
- $-1 * \arctan(1 / \approx 10^{510})$
- $\arctan(1 / \approx 10^{1020})$
- $\arctan(1 / \approx 10^{2042})$
- $-1 * \arctan(1 / \approx 10^{4084})$
- $\arctan(1 / \approx 10^{8170})$
- $\arctan(1 / \approx 10^{16340})$
- $\arctan(1 / \approx 10^{32682})$
- $-1 * \arctan(\approx 10^{3315} / \approx 10^{68679})$
If you have any questions just ask! :)
PS: If this is confirmed to be correct then I can explain my method of getting this formula and how to get infinitely many formulas like it.