## 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.

Pages: 1 2 3

### 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, len(list(s))) 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 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.