def A(n, j):
if j == 1:
return Float(1)
s = Float(0)
pi = pi_float()
for h in xrange(1, j):
if gcd(h,j) == 1:
s += cos((g(h,j)-2*h*n)*pi/j)
return s
def D(n, j):
pi = pi_float()
a = sqrt(Float(2)/3) * pi / j
b = Float(n) - Float(1)/24
c = sqrt(b)
expa = exp(a*c)
iexpa = Float(1)/expa
ch = (expa + iexpa)*0.5
sh = (expa - iexpa)*0.5
return sqrt(j) / (2*sqrt(2)*b*pi) * (a*ch-sh/c)
def npartitions(n):
"""
Calculate the partition function P(n), i.e. the number of ways that
n can be written as a sum of positive integers.
P(n) is computed using a straightforward implementation of the
Hardy-Ramanujan-Rademacher formula, described e.g. at
http://mathworld.wolfram.com/PartitionFunctionP.html
The speed is decent up to n = 10**5 or so. The function has
been tested to give the correct result for n = 10**6.
"""
n = int(n)
if n < 0:
return 0
if n <= 5:
return [1, 1, 2, 3, 5, 7][n]
from sympy.core.numbers import gcd
from sympy.numerics import Float
from sympy.numerics.functions import pi_float, sqrt, exp, log, cos
def frac(x):
return x - int(x)
def D(n, j):
pi = pi_float()
a = sqrt(Float(2)/3) * pi / j
b = Float(n) - Float(1)/24
c = sqrt(b)
expa = exp(a*c)
iexpa = Float(1)/expa
ch = (expa + iexpa)*0.5
sh = (expa - iexpa)*0.5
return sqrt(j) / (2*sqrt(2)*b*pi) * (a*ch-sh/c)
def A(n, j):
if j == 1:
return Float(1)
s = Float(0)
pi = pi_float()
for h in xrange(1, j):
if gcd(h,j) == 1:
s += cos((g(h,j)-2*h*n)*pi/j)
return s
def g(h, j):
if j < 3:
return Float(0)
s = Float(0)
for k in xrange(1, j):
s += k*(frac(h*Float(k)/j)-0.5)
return s
# estimate number of digits in p(n)
pdigits = int((pi_float()*sqrt(2.0*n/3)-log(4*n))/log(10)+1)
Float.store()
Float.setdps(pdigits*1.1 + 10)
s = Float(0)
M = max(6, int(0.24*sqrt(n)+4))
for q in xrange(1, M):
s += A(n,q) * D(n,q)
p = int(s + 0.5)
Float.revert()
return p