Continued fraction for ![]()
There are various forms of continued fractions known for pi.
Its standard form would be

that is
in special form
or expressed as
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where the sequence will continue forever without any pattern.
It can be calculated by the following algorithm:
(1) Find b0 in x=b0+1/x1 where b0 is the largest integer that is less than or equal to x.
(2) Calculate the residual x-b0 and set it to 1/x1.
(3) Find b1 in x1=b1+1/x2 where b1 is the largest integer that is less than or equal to x1.
(4) Calculate the residual x1-b1 and set it to 1/x2 if it is not zero.
(5) If it is zero, stop there. Otherwise we repeat this for many times.
as in 0207 with all one numerators.
There are many more.

that is
in special form

that is
in special form
We have a continued fraction with all one denominators including the first constant.

that is
in special form
where we have a pattern with the numerators
of
for k=1,2,3,…
That is the products of floored value of (k+1)/2 and that of (k+2)/2.