[VA] SRC#002- Almost integers and other beasties

[Albert Chan]

(02-10-2019 12:48 PM)Gerson W. Barbosa Wrote:  Number of iterations = Ceil(W(10^n*ln(2))/ln(2)), where n = number of digits and W(x) is the Lambert W function.

50 -> 1.0000000000012374125757361102287196106466728742977

Above 50 digits numbers are confirmed corrrect.
Can you explain how the iteration count formula is derived ?

(01-23-2019 10:58 PM)Valentin Albillo Wrote:  \[ \frac{1}{12}+\frac{\Pi^2}{6Ln^2(2)}+ \frac{2}{Ln(2)} \sum_{k=1}^\infty \frac{(-1)^k}{k(2^k-1)} \]

I see that the sum gained about 1 bit precision with each iteration, so I use n / log10(2) ~ 3.322n
It is slightly over-estimated, but not by much.

Edit: the difference is the effect of 1/k, iterations ~ 3.322n - log2(3.322n)