GCD Sum

April 22, 2016

Today’s exercise is inspired by A018804: Find the sum of the greatest common divisors gcd(k, n) for 1 ≤ kn.

Your task is to write a function that finds the sum of the greatest common divisors. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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9 Responses to “GCD Sum”

  1. Rutger said

    In Python, bruteforce solution:

    from fractions import gcd
    
    for n in range(20):
    	print n, ":", sum([gcd(n,k) for k in range(1,n+1)])
    
    # Result
    # 0 : 0
    # 1 : 1
    # 2 : 3
    # 3 : 5
    # 4 : 8
    # 5 : 9
    # 6 : 15
    # 7 : 13
    # 8 : 20
    # 9 : 21
    # 10 : 27
    # 11 : 21
    # 12 : 40
    # 13 : 25
    # 14 : 39
    # 15 : 45
    # 16 : 48
    # 17 : 33
    # 18 : 63
    # 19 : 37
    # [Finished in 0.2s]
    
  2. Paul said

    In Python, td_factors is from Programming Praxis primes essay and divisor_gen generates all divisors.

    def totient(n):
        product = n
        for prime in set(td_factors(n)):
            product -= product // prime
        return product
    
    def sumgcd(n):
        return sum(d * totient(n // d) for d in divisor_gen(n))
    
  3. matthew said

    http://oeis.org/A018804 also tells us that sumgcd (or Pillai’s arithmetic function) is multiplicative (f is multiplicative if f(nm) = f(n)f(m) for coprime n and m) and that its value for p^e is p^(e-1)*((p-1)e+p), for prime p – this suffices to calculate the function for any n, given a prime factorization. Here we use the Python primefac library, converting the output to prime-exponent pairs, and use that to define a generic multiplicative function. Fairly snappy, even for “colossally abundant” numbers with lots of divisors (has anyone seen the Ramanujan film yet?):

    from primefac import primefac
    from itertools import groupby
    from operator import mul
    
    def factors(n): return ((s[0], len(list(s[1]))) for s in groupby(primefac(n)))
    def multiplicative(f): return lambda n: reduce(mul, (f(p,e) for (p,e) in factors(n)))
    gcdsum = multiplicative(lambda p,e: p**(e-1)*((p-1)*e+p))
    
    print gcdsum(991*997)
    print gcdsum(224403121196654400) # 22nd colossally abundant number
    
    $ ./gcdsum.py 
    3948133
    4812563181512005920000
    
  4. Zack said

    Here is an alternative in Julia, without invoking a bunch of packages to do all the hard work:

    function gcd_(x::Int64, y::Int64)
    m = min(x,y)

    for d in m:-1:1
    if (x%d == 0) && (y%d == 0)
    return d
    end
    end
    end

    function gcd_sum(n::Int64)
    s = n + 1

    for k = 2:(n-1)
    s += gcd(k,n)
    end

    return s
    end

  5. Zack said

    @matthew: This CSS add-on doesn’t support formatting for Julia…

  6. Daniel said

    Here’s a Java solution that is similar in spirit to Sieve of Eratosthenes.

    static long gcdSum(int n) {
        int[] array = new int[n+1]; // array[0] ignored
        
        for (int i = 1; i <= n/2; i++) {
            if (n % i == 0) {
                for (int j = i; j <= n; j += i) {
                    array[j] = i;
                }
            }
        }
        
        long sum = n;
        for (int i = 1; i < n; i++) {
            sum += array[i];
        }
        
        return sum;
    }
    
  7. matthew said

    @Zack: you can just leave out the language specification, which will preserve indentation at least, eg “

    " at the start, "

    ” at the end:

    function gcd_(x::Int64, y::Int64)
      m = min(x,y)
    
      for d in m:-1:1
        if (x%d == 0) && (y%d == 0)
          return d
        end
      end
    end
    
    function gcd_sum(n::Int64)
      s = n + 1
    
      for k = 2:(n-1)
        s += gcd(k,n)
      end
    
      return s
    end
    
  8. matthew said

    Gah, I was attempting to show the formatting commands [code] and [/code] – I had a notion they only applied when on a new line, but evidently not. Using HTML entities for the square brackets here.

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