Where to download a lot of digits of pi?

Question

Where could I find a large amount of digits of pi? I have already calculated 3.14 billion using PiFast (works well under wine).

I don't care about slow download speeds.

Answer 1

I know you say you don't care, but I seriously suspect your cpu can calculate them faster than your network card is capable of downloading them.

Given the last digit and the current state of the calculator used to generate it, the next digit can be found in constant time. It doesn't get progressively harder like finding the next prime does.

Answer 2

Adding on to Joel's comment, SuperPi is one of the most popular tools for this. It's also used for stress testing.

Answer 3

On Ubuntu, you can sudo apt-get install pi

and then:

$ pi 100 3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067

It calculates arbitrary precision given the number of digits to calculate.

Answer 4

If you want to use Python to calculate it, here's an extremely fast method (using Python and the gmpy2 library):

http://www.craig-wood.com/nick/articles/pi-chudnovsky/

Here's the code with a small fix:

"""
Python3 program to calculate Pi using python long integers, binary
splitting and the Chudnovsky algorithm

See: http://www.craig-wood.com/nick/articles/pi-chudnovsky/ for more
info

Nick Craig-Wood <[e-mail]>
"""

import math
from gmpy2 import mpz
from time import time
import gmpy2

def pi_chudnovsky_bs(digits):
    """
    Compute int(pi * 10**digits)

    This is done using Chudnovsky's series with binary splitting
    """
    C = 640320
    C3_OVER_24 = C**3 // 24
    def bs(a, b):
        """
        Computes the terms for binary splitting the Chudnovsky infinite series

        a(a) = +/- (13591409 + 545140134*a)
        p(a) = (6*a-5)*(2*a-1)*(6*a-1)
        b(a) = 1
        q(a) = a*a*a*C3_OVER_24

        returns P(a,b), Q(a,b) and T(a,b)
        """
        if b - a == 1:
            # Directly compute P(a,a+1), Q(a,a+1) and T(a,a+1)
            if a == 0:
                Pab = Qab = mpz(1)
            else:
                Pab = mpz((6*a-5)*(2*a-1)*(6*a-1))
                Qab = mpz(a*a*a*C3_OVER_24)
            Tab = Pab * (13591409 + 545140134*a) # a(a) * p(a)
            if a & 1:
                Tab = -Tab
        else:
            # Recursively compute P(a,b), Q(a,b) and T(a,b)
            # m is the midpoint of a and b
            m = (a + b) // 2
            # Recursively calculate P(a,m), Q(a,m) and T(a,m)
            Pam, Qam, Tam = bs(a, m)
            # Recursively calculate P(m,b), Q(m,b) and T(m,b)
            Pmb, Qmb, Tmb = bs(m, b)
            # Now combine
            Pab = Pam * Pmb
            Qab = Qam * Qmb
            Tab = Qmb * Tam + Pam * Tmb
        return Pab, Qab, Tab
    # how many terms to compute
    DIGITS_PER_TERM = math.log10(C3_OVER_24/6/2/6)
    N = int(digits/DIGITS_PER_TERM + 1)
    # Calclate P(0,N) and Q(0,N)
    P, Q, T = bs(0, N)
    one_squared = mpz(10)**(2*digits)
    #sqrtC = (10005*one_squared).sqrt()
    sqrtC = gmpy2.isqrt(10005*one_squared)
    return (Q*426880*sqrtC) // T

# The last 5 digits or pi for various numbers of digits
check_digits = {
        100 : 70679,
       1000 :  1989,
      10000 : 75678,
     100000 : 24646,
    1000000 : 58151,
   10000000 : 55897,
}

if __name__ == "__main__":
    digits = 100
    pi = pi_chudnovsky_bs(digits)
    print(pi)
    #raise SystemExit
    for log10_digits in range(1,9):
        digits = 10**log10_digits
        start =time()
        pi = pi_chudnovsky_bs(digits)
        print("chudnovsky_gmpy_mpz_bs: digits",digits,"time",time()-start)
        if digits in check_digits:
            last_five_digits = pi % 100000
            if check_digits[digits] == last_five_digits:
                print("Last 5 digits %05d OK" % last_five_digits)
                open("%s_pi.txt" % log10_digits, "w").write(str(pi))
            else:
                print("Last 5 digits %05d wrong should be %05d" % (last_five_digits, check_digits[digits]))