"""
Berechnung von pi über arctan(x)
(c) Jürgen Meier
www.3d-meier.de
26.06.2024
"""

import math
import decimal


# Anzahl Kommastellen
decimal.getcontext().prec = 105


a1 = decimal.Decimal(1)
a2 = decimal.Decimal(2)
a4 = decimal.Decimal(4)
a12 = decimal.Decimal(12)
a32 = decimal.Decimal(32)
a68 = decimal.Decimal(68)
a100 = decimal.Decimal(100)
a183 = decimal.Decimal(183)
a239 = decimal.Decimal(239)
a1023 = decimal.Decimal(1023)
a5832 = decimal.Decimal(5832)
a110443 = decimal.Decimal(110443)
a4841182 = decimal.Decimal(4841182)
a6826318 = decimal.Decimal(6826318)

#print(a1023**2+1)

m1 = decimal.Decimal(-1)

x1 = a1/a239
x2 = a1/a1023
x3 = a1/a5832
x4 = a1/a110443
x5 = a1/a4841182
x6 = a1/a6826318


# Reihenentwicklung für arctan(x)
k = 300

t1 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x1**(a2*n + a1)))/(a2*n + a1)
    t1 = t1 + a

t2 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x2**(a2*n + a1)))/(a2*n + a1)
    t2 = t2 + a

t3 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x3**(a2*n + a1)))/(a2*n + a1)
    t3 = t3 + a

t4 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x4**(a2*n + a1)))/(a2*n + a1)
    t4 = t4 + a
    
t5 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x5**(a2*n + a1)))/(a2*n + a1)
    t5 = t5 + a
        
t6 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x6**(a2*n + a1)))/(a2*n + a1)
    t6 = t6 + a
        
    
    
p = a183*t1 + a32*t2 - a68*t3 +a12* t4 - a12*t5 - 100*t6
print(4*p)

print(((complex(239, 1)/complex(239, -1))**183)*
((complex(1023, 1)/complex(1023, -1))**32) *
((complex(5832, 1)/complex(5832, -1))**-68) *
((complex(110443, 1)/complex(110443, -1))**12) *
((complex(4841182, 1)/complex(4841182, -1))**-12) *
((complex(6826318, 1)/complex(6826318, -1))**-100))