Yet another π formula

[Albert Chan]

(01-07-2021 12:57 PM)Gerson W. Barbosa Wrote:  Great! The continued fraction part is left unproved, however. But that would not be easy, I presume.

"Wasicki's formula" had first term converge to pi, correction converge to 0. QED

As to the gain of 25/12 digits per term, it tested OK even with 10,000 digits precision.

Code:

def test_pi(n, cf=0, c=0, s=0):
    cf += n         # extra cf "terms"
    b = 2*n*(n+1) + 4*cf*(cf+1) + mpfr(1.5)
    for k in range(cf,n,-1):
        t = mpfr(4*k*k-1)
        c = k*k*t/(b-c)
        b = b - 8*k
    for k in range(n,0,-1):
        t = mpfr(4*k*k-1)
        s = 1/t - s
        c = k*k*t/(b-c)
        b = b - 8*k
    return (s+0.5)*4 + (-1)**n/(b-c)

>>> from gmpy2 import *
>>> get_context().precision = 34000 # > 10000 dec. digits
>>> pi = const_pi()
>>>
>>> for n in range(1000,4001,1000): print n, format(pi - test_pi(n), 'g')
...
1000 2.19966e-2091
2000 2.92237e-4181
3000 3.88382e-6271
4000 5.16203e-8361