"""
Berechnung von Zeta(11)
(c) Jürgen Meier
www.3d-meier.de
11.07.2024
"""

import math


a = 0
b = 0
c = 0
aa = 92992/4009.0
bb = -1617613/64144.0
cc = 1453/64144.0

n = 11

k = 3

for i in range(1, k + 1):
    a = a + 1.0/((i**n)*(math.exp(1*math.pi*i)-1))
    b = b + 1.0/((i**n)*(math.exp(2*math.pi*i)-1))
    c = c + 1.0/((i**n)*(math.exp(4*math.pi*i)-1))

print(aa*a + bb*b + cc*c)