Mathematica to Python

Question

How can this Mathematica code be ported to Python? I do not know the Mathematica syntax and am having a hard time understanding how this is described in a more traditional language.

Source (pg 5): http://subjoin.net/misc/m496pres1.nb.pdf

Answer 1

This cannot be ported to Python directly as the definition a[j] uses the Symbolic Arithmetic feature of Mathematica.

a[j] is basically the coefficient of x^j in the series expansion of that rational function inside Apart.

Assume you have a[j], then f[n] is easy. A Block in Mathematica basically introduces a scope for variables. The first list initializes the variable, and the rest is the execution of the code. So

from __future__ import division
def f(n):
  v = n // 5
  q = v // 20
  r = v % 20
  return sum(binomial(q+5-j, 5) * a[r+20*j] for j in range(5))

(binomial is the Binomial coefficient.)

Answer 2

The symbolics can be done with sympy. Combined with KennyTM's answer, something like this might be what you want:

from __future__ import division
from sympy import Symbol, apart, binomial

x = Symbol('x')
poly = (1-x**20)**5 / ((1-x)**2 * (1-x**2) * (1-x**5) * (1-x**10))
poly2 = apart(poly,x)

def a(j):
    return poly2.coeff(x**j)

def f(n):
    v = n // 5
    q = v // 20
    r = v % 20
    return sum(binomial(q+5-j, 5)*a(r+20*j) for j in range(5))

Although I have to admit that f(n) does not work (I'm not very good at Python).