Question

Overall Problem: Project Euler 12 – What is the value of the first triangle number to have over five hundred divisors?

Focus of problem: The divisor function

Language: Python

Description: The function I used is brute and the time it take for the program to find a number with more divisors than x increases almost exponentially with each 10 or 20 numbers highers. I need to get to 500 or more divisors. I’ve identified that the divisor function is what is hogging down the program. The research I did lead me to divisor functions and specifically the divisor function which is supposed to be a function that will count all the divisors of any integer. Every page I’ve looked at seems to be directed toward mathematics majors and I only have high-school maths. Although I did come across some page that mentioned allot about primes and the Sieve of Atkins but I could not make the connection between primes and finding all the divisors of any integer nor find anything on the net about it.

Main Question: Could someone explain how to code the divisor function or even provide a sample? Maths concepts make more sense to me when I look at them with code. So much appreciated.

brute force divisor function:

def countdiv(a):
    count = 0
    for i in range(1,(a/2)+1):
        if a % i == 0:
            count += 1
    return count + 1    # +1 to account for number itself as a divisor

Answer 1

If you need a bruteforce function to calculate Number of Divisors (also known as tau(n))

Here’s what it looks like

def tau(n):
        sqroot,t = int(n**0.5),0
        for factor in range(1,sqroot+1):
                if n % factor == 0:
                        t += 2 # both factor and N/factor
        if sqroot*sqroot == n: t = t - 1 # if sqroot is a factor then we counted it twice, so subtract 1
        return t

The second method involves a decomposing n into its prime factors (and its exponents).

tau(n) = (e1+1)(e2+1)....(em+1) where n = p1^e1 * p2^e2 .... pm^em and p1,p2..pm are primes

More info here

The third method and much more simpler to understand is simply using a Sieve to calculate tau.

def sieve(N):
        t = [0]*(N+1)
        for factor in range(1,N+1):
                for multiple in range(factor,N+1,factor):
                        t[multiple]+=1
        return t[1:]

Here’s it in action at ideone